Section 1.1 Definition of the Integral
Arguably the easiest way to introduce integration is by considering the area between the graph of a given function and the \(x\)-axis, between two specific vertical lines — such as is shown in the figure above. We'll follow this route by starting with a motivating example.
Subsection 1.1.1 A Motivating Example
Let us find the area under the curve \(y=e^x\) (and above the \(x\)-axis) for \(0\le x\le 1\text{.}\) That is, the area of \(\big\{\ (x,y)\ \big|\ 0\le y\le e^x\text{,}\) \(0\le x\le 1\ \big\}\text{.}\)
This area is equal to the “definite integral”
\begin{align*}
\text{Area} &= \int_0^1 e^x \dee{x}
\end{align*}
Do not worry about this notation or terminology just yet. We discuss it at length below. In different applications this quantity will have different interpretations — not just area. For example, if \(x\) is time and \(e^x\) is your velocity at time \(x\text{,}\) then we'll see later (in Example 1.1.18) that the specified area is the net distance travelled between time \(0\) and time \(1\text{.}\) After we finish with the example, we'll mimic it to give a general definition of the integral \(\int_a^b f(x) \dee{x}\text{.}\)
We wish to compute the area of \(\big\{\ (x,y)\ \big|\ 0\le y\le e^x\text{,}\) \(0\le x\le 1\ \big\}\text{.}\) We know, from our experience with \(e^x\) in differential calculus, that the curve \(y=e^x\) is not easily written in terms of other simpler functions, so it is very unlikely that we would be able to write the area as a combination of simpler geometric objects such as triangles, rectangles or circles.
So rather than trying to write down the area exactly, our strategy is to approximate the area and then make our approximation more and more precise 1 . We choose 2 to approximate the area as a union of a large number of tall thin (vertical) rectangles. As we take more and more rectangles we get better and better approximations. Taking the limit as the number of rectangles goes to infinity gives the exact area 3 .
As a warm up exercise, we'll now just use four rectangles. In Example 1.1.2, below, we'll consider an arbitrary number of rectangles and then take the limit as the number of rectangles goes to infinity. So
- subdivide the interval \(0\le x\le 1\) into \(4\) equal subintervals each of width \(\frac{1}{4}\text{,}\) and
- subdivide the area of interest into four corresponding vertical strips, as in the figure below.
The area we want is exactly the sum of the areas of all four strips.
Each of these strips is almost, but not quite, a rectangle. While the bottom and sides are fine (the sides are at right-angles to the base), the top of the strip is not horizontal. This is where we must start to approximate. We can replace each strip by a rectangle by just levelling off the top. But now we have to make a choice — at what height do we level off the top?
Consider, for example, the leftmost strip. On this strip, \(x\) runs from \(0\) to \(\frac{1}{4}\text{.}\) As \(x\) runs from \(0\) to \(\frac{1}{4}\text{,}\) the height \(y\) runs from \(e^0\) to \(e^{\frac{1}{4}}\text{.}\) It would be reasonable to choose the height of the approximating rectangle to be somewhere between \(e^0\) and \(e^{\frac{1}{4}}\text{.}\) Which
height should we choose? Well, actually it doesn't matter. When we eventually take the limit of infinitely many approximating rectangles all of those different choices give exactly the same final answer. We'll say more about this later.
In this example we'll do two sample computations.
- For the first computation we approximate each slice by a rectangle whose height is the height of the left hand side of the slice.
- On the first slice, \(x\) runs from \(0\) to \(\frac{1}{4}\text{,}\) and the height \(y\) runs from \(e^0\text{,}\) on the left hand side, to \(e^{\frac{1}{4}}\text{,}\) on the right hand side.
- So we approximate the first slice by the rectangle of height \(e^0\) and width \(\frac{1}{4}\text{,}\) and hence of area \(\frac{1}{4}\,e^0 =\frac{1}{4}\text{.}\)
- On the second slice, \(x\) runs from \(\frac{1}{4}\) to \(\frac{1}{2}\text{,}\) and the height \(y\) runs from \(e^{\frac{1}{4}}\) and \(e^{\frac{1}{2}}\text{.}\)
- So we approximate the second slice by the rectangle of height \(e^{\frac{1}{4}}\) and width \(\frac{1}{4}\text{,}\) and hence of area \(\frac{1}{4}\,e^{\frac{1}{4}}\text{.}\)
- And so on.
- All together, we approximate the area of interest by the sum of the areas of the four approximating rectangles, which is\begin{gather*} \big[1+ e^{\frac{1}{4}} + e^{\frac{1}{2}} +e^{\frac{3}{4}}\big]\frac{1}{4} =1.5124 \end{gather*}
- This particular approximation is called the “left Riemann sum approximation to \(\int_0^1 e^x\dee{x}\) with \(4\) subintervals”. We'll explain this terminology later.
- This particular approximation represents the shaded area in the figure on the left below. Note that, because \(e^x\) increases as \(x\) increases, this approximation is definitely smaller than the true area.
- For the second computation we approximate each slice by a rectangle whose height is the height of the right hand side of the slice.
- On the first slice, \(x\) runs from \(0\) to \(\frac{1}{4}\text{,}\) and the height \(y\) runs from \(e^0\text{,}\) on the left hand side, to \(e^{\frac{1}{4}}\text{,}\) on the right hand side.
- So we approximate the first slice by the rectangle of height \(e^{\frac{1}{4}}\) and width \(\frac{1}{4}\text{,}\) and hence of area \(\frac{1}{4}\,e^{\frac{1}{4}}\text{.}\)
- On the second slice, \(x\) runs from \(\frac{1}{4}\) to \(\frac{1}{2}\text{,}\) and the height \(y\) runs from \(e^{\frac{1}{4}}\) and \(e^{\frac{1}{2}}\text{.}\)
- So we approximate the second slice by the rectangle of height \(e^{\frac{1}{2}}\) and width \(\frac{1}{4}\text{,}\) and hence of area \(\frac{1}{4}\,e^{\frac{1}{2}}\text{.}\)
- And so on.
- All together, we approximate the area of interest by the sum of the areas of the four approximating rectangles, which is\begin{gather*} \big[e^{\frac{1}{4}} + e^{\frac{1}{2}} +e^{\frac{3}{4}}+e^1\big]\frac{1}{4} =1.9420 \end{gather*}
- This particular approximation is called the “right Riemann sum approximation to \(\int_0^1 e^x\dee{x}\) with \(4\) subintervals”.
- This particular approximation represents the shaded area in the figure on the right above. Note that, because \(e^x\) increases as \(x\) increases, this approximation is definitely larger than the true area.
Now for the full computation that gives the exact area.
Example 1.1.2. Computing an area exactly.
Recall that we wish to compute the area of
\begin{gather*}
\big\{\ (x,y)\ \big|\ 0\le y\le e^x,\ 0\le x\le 1\ \big\}
\end{gather*}
and that our strategy is to approximate this area by the area of a union of a large number of very thin rectangles, and then take the limit as the number of rectangles goes to infinity. In Example 1.1.1, we used just four rectangles. Now we'll consider a general number of rectangles, that we'll call \(n\text{.}\) Then we'll take the limit \(n\rightarrow\infty\text{.}\) So
- pick a natural number \(n\) and
- subdivide the interval \(0\le x\le 1\) into \(n\) equal subintervals each of width \(\frac{1}{n}\text{,}\) and
- subdivide the area of interest into corresponding thin strips, as in the figure below.
The area we want is exactly the sum of the areas of all of the thin strips.
Each of these strips is almost, but not quite, a rectangle. As in Example 1.1.1, the only problem is that the top is not horizontal. So we approximate each strip by a rectangle, just by levelling off the top. Again, we have to make a choice — at what height do we level off the top?
Consider, for example, the leftmost strip. On this strip, \(x\) runs from \(0\) to \(\frac{1}{n}\text{.}\) As \(x\) runs from \(0\) to \(\frac{1}{n}\text{,}\) the height \(y\) runs from \(e^0\) to \(e^{\frac{1}{n}}\text{.}\) It would be reasonable to choose the height of the approximating rectangle to be somewhere between \(e^0\) and \(e^{\frac{1}{n}}\text{.}\) Which height should we choose?
Well, as we said in Example 1.1.1, it doesn't matter. We shall shortly take the limit \(n\rightarrow\infty\) and, in that limit, all of those different choices give exactly the same final answer. We won't justify that statement in this example, but there will be an (optional) section shortly that provides the justification. For this example we just, arbitrarily, choose the height of each rectangle to be the height of the graph \(y=e^x\) at the smallest value of \(x\) in the corresponding strip 4 . The figure on the left below shows the approximating rectangles when \(n=4\) and the figure on the right shows the approximating rectangles when \(n=8\text{.}\)
Now we compute the approximating area when there are \(n\) strips.
- We approximate the leftmost strip by a rectangle of height \(e^0\text{.}\) All of the rectangles have width \(\frac{1}{n}\text{.}\) So the leftmost rectangle has area \(\frac{1}{n}e^0\text{.}\)
- On strip number \(2\text{,}\) \(x\) runs from \(\frac{1}{n}\) to \(\frac{2}{n}\text{.}\) So the smallest value of \(x\) on strip number \(2\) is \(\frac{1}{n}\text{,}\) and we approximate strip number \(2\) by a rectangle of height \(e^{\frac{1}{n}}\) and hence of area \(\frac{1}{n}e^{\frac{1}{n}} \text{.}\)
- And so on.
- On the last strip, \(x\) runs from \(\frac{n-1}{n}\) to \(\frac{n}{n}=1\text{.}\) So the smallest value of \(x\) on the last strip is \(\frac{n-1}{n}\text{,}\) and we approximate the last strip by a rectangle of height \(e^{\frac{(n-1)}{n}}\) and hence of area \(\frac{1}{n}e^{\frac{(n-1)}{n}} \text{.}\)
The total area of all of the approximating rectangles is
\begin{align*}
\text{Total approximating area} &= \frac{1}{n}e^0
+ \frac{1}{n}e^{\frac{1}{n}}
+ \frac{1}{n}e^{\frac{2}{n}}
+ \frac{1}{n}e^{\frac{3}{n}}
+ \cdots
+ \frac{1}{n}e^{\frac{(n-1)}{n}}\\
&= \frac{1}{n}\Big( 1+ e^{\frac{1}{n}} +e^{\frac{2}{n}}+e^{\frac{3}{n}}
+\cdots+ e^{\frac{(n-1)}{n}}\Big)
\end{align*}
Now the sum in the brackets might look a little intimidating because of all the exponentials, but it actually has a pretty simple structure that can be easily seen if we rename \(e^{\frac{1}{n}}=r\text{.}\) Then
- the first term is 1 = \(r^0\) and
- the second term is \(e^{\frac{1}{n}}=r^1\) and
- the third term is \(e^{\frac{2}{n}}=r^2\) and
- the fourth term is \(e^{\frac{3}{n}}=r^3\) and
- and so on and
- the last term is \(e^{\frac{(n-1)}{n}}=r^{n-1}\text{.}\)
So
\begin{align*}
\text{Total approximating area}
&= \frac{1}{n}\left( 1+ r +r^2 +\cdots+ r^{n-1}\right)
\end{align*}
The sum in brackets is known as a geometric sum and satisfies a nice simple formula:
Equation 1.1.3. Geometric sum.
\begin{gather*}
1+ r +r^2 +\cdots+ r^{n-1} =\frac{r^n-1}{r-1} \qquad\text{provided $r\ne 1$}
\end{gather*}
The derivation of the above formula is not too difficult. So let's derive it in a little aside.
Now we can go back to our area approximation armed with the above result about geometric sums.
\begin{align*}
\text{Total approximating area}
&= \frac{1}{n}\left( 1+ r +r^2 +\cdots+ r^{n-1}\right)\\
&= \frac{1}{n} \frac{r^n-1}{r-1} \qquad\qquad \text{remember that $r=e^{1/n}$}\\
&= \frac{1}{n} \frac{e^{n/n} - 1}{e^{1/n}-1}\\
&= \frac{1}{n} \frac{e - 1}{e^{1/n}-1}
\end{align*}
To get the exact area 5 all we need to do is make the approximation better and better by taking the limit \(n\rightarrow \infty\text{.}\) The limit will look more familiar if we rename \(\frac{1}{n}\) to \(X\text{.}\) As \(n\) tends to infinity, \(X\) tends to \(0\text{,}\) so
\begin{align*}
\text{Area}&=\lim_{n\rightarrow\infty} \frac{1}{n}\ \frac{e-1}{e^{1/n}-1}\\
&=(e-1)\lim_{n\rightarrow\infty} \frac{1/n}{e^{1/n}-1}\\
&=(e-1)\lim_{X\rightarrow 0} \frac{X}{e^X-1}
&\text{(with $X=\frac{1}{n}$)}
\end{align*}
Examining this limit we see that both numerator and denominator tend to zero as \(X\to 0\text{,}\) and so we cannot evaluate this limit by computing the limits of the numerator and denominator separately and then dividing the results. Despite this, the limit is not too hard to evaluate; here we give two ways:
- Perhaps the easiest way to compute the limit is by using l'Hôpital's rule 6 . Since both numerator and denominator go to zero, this is a \(\frac00\) indeterminate form. Thus\begin{align*} \lim_{X\rightarrow 0} \frac{X}{e^X-1} &=\lim_{X\rightarrow 0} \frac{\diff{}{X}X}{\diff{}{X}(e^X-1)} =\lim_{X\rightarrow 0} \frac{1}{e^X}=1 \end{align*}
- Another way 7 to evaluate the same limit is to observe that it can be massaged into the form of the limit definition of the derivative. First notice that\begin{align*} \lim_{X\rightarrow 0} \frac{X}{e^X-1} &= \left[\lim_{X\rightarrow 0} \frac{e^X-1}{X} \right]^{-1} \end{align*}provided this second limit exists and is nonzero 8 . This second limit should look a little familiar:\begin{align*} \lim_{X\rightarrow 0} \frac{e^X-1}{X} &= \lim_{X\rightarrow 0} \frac{e^X-e^0}{X-0} \end{align*}which is just the definition of the derivative of \(e^x\) at \(x=0\text{.}\) Hence we have\begin{align*} \lim_{X\rightarrow 0} \frac{X}{e^X-1} &=\left[\lim_{X\rightarrow 0}\, \frac{e^X-e^0}{X-0} \right]^{-1}\\ &=\left[\diff{}{X}e^X\Big|_{X=0} \right]^{-1}\\ &=\left[e^X\big|_{X=0}\right]^{-1}\\ &=1 \end{align*}
So, after this short aside into limits, we may now conclude that
\begin{align*}
\text{Area} &=(e-1)\lim_{X\rightarrow 0} \frac{X}{e^X-1}\\
&=e-1
\end{align*}
Subsection 1.1.2 Optional — A more rigorous area computation
In Example 1.1.1 above we considered the area of the region \(\big\{\ (x,y)\ \big|\ 0\le y\le e^x\text{,}\) \(0\le x\le 1\ \big\}\text{.}\) We approximated that area by the area of a union of \(n\) thin rectangles. We then claimed that upon taking the number of rectangles to infinity, the approximation of the area became the exact area. However we did not justify the claim. The purpose of this optional section is to make that calculation rigorous.
The broad set-up is the same. We divide the region up into \(n\) vertical strips, each of width \(\frac1n\) and we then approximate those strips by rectangles. However rather than an uncontrolled approximation, we construct two sets of rectangles — one set always smaller than the original area and one always larger. This then gives us lower and upper bounds on the area of the region. Finally we make use of the squeeze theorem 11 to establish the result.
- To find our upper and lower bounds we make use of the fact that \(e^x\) is an increasing function. We know this because the derivative \(\diff{}{x}e^x=e^x\) is always positive. Consequently, the smallest and largest values of \(e^x\) on the interval \(a\le x\le b\) are \(e^a\) and \(e^b\text{,}\) respectively.
-
In particular, for \(0\le x\le \frac{1}{n}\text{,}\) \(e^x\) takes values only between \(e^0\) and \(e^{\frac{1}{n}}\text{.}\) As a result, the first strip\begin{gather*} \big\{\ (x,y)\ \big|\ 0\le x\le \frac{1}{n},\ 0\le y\le e^x\ \big\} \end{gather*}
- contains the rectangle of \(0\le x\le \frac{1}{n}\text{,}\) \(0\le y\le e^0\) (the lighter rectangle in the figure on the left below) and
- is contained in the rectangle \(0\le x\le \frac{1}{n}\text{,}\) \(0\le y\le e^{\frac{1}{n}}\) (the largest rectangle in the figure on the left below).
Hence\begin{gather*} \frac{1}{n}e^{0} \le {\rm Area} \big\{\ (x,y)\ \big|\ 0\le x\le \frac{1}{n},\ 0\le y\le e^x\ \big\} \le \frac{1}{n}e^{\frac{1}{n}} \end{gather*}
- Similarly, for the second, third, …, last strips, as in the figure on the right above,\begin{align*} \frac{1}{n}e^{\frac{1}{n}} &\le {\rm Area}\big\{\ (x,y)\ \big|\ \frac{1}{n}\le x\le \frac{2}{n},\ 0\le y\le e^x\ \big\} \ \ \ \le \frac{1}{n}e^{\frac{2}{n}}\\ \frac{1}{n}e^{\frac{2}{n}} &\le {\rm Area}\big\{\ (x,y)\ \big|\ \frac{2}{n}\le x\le \frac{3}{n},\ 0\le y\le e^x\ \big\} \ \ \ \le \frac{1}{n}e^{\frac{3}{n}}\\ & \vdots\\ \frac{1}{n}e^{\frac{(n-1)}{n}} &\le {\rm Area}\big\{\ (x,y)\ \big|\ \frac{(n-1)}{n}\le x\le \frac{n}{n},\ 0\le y\le e^x\ \big\} \le\frac{1}{n}e^{\frac{n}{n}} \end{align*}
- Adding these \(n\) inequalities together gives\begin{align*} &\frac{1}{n}\left(1+e^{\frac{1}{n}}+\cdots+e^{\frac{(n-1)}{n}}\right)\\ &\le {\rm Area}\big\{\ (x,y)\ \big|\ 0\le x\le 1,\ 0\le y\le e^x\ \big\}\\ &\le \frac{1}{n}\left(e^{\frac{1}{n}}+e^{\frac{2}{n}}+\cdots+ e^{\frac{n}{n}}\right) \end{align*}
- We can then recycle equation 1.1.3 with \(r=e^{\frac1n}\text{,}\) so that \(r^n=\left(e^{\frac{1}{n}}\right)^n=e\text{.}\) Thus we have\begin{gather*} \frac{1}{n}\frac{e-1}{e^{\frac{1}{n}}-1} \le {\rm Area}\big\{\ (x,y)\ \big|\ 0\le x\le 1,\ 0\le y\le e^x\ \big\} \le \frac{1}{n}e^{\frac{1}{n}}\frac{e-1}{e^{\frac{1}{n}}-1} \end{gather*}where we have used the fact that the upper bound is a simple multiple of the lower bound:\begin{align*} \left(e^{\frac{1}{n}}+e^{\frac{2}{n}}+\cdots+ e^{\frac{n}{n}}\right) &= e^{\frac{1}{n}}\left(1+e^{\frac{1}{n}}+\cdots +e^{\frac{(n-1)}{n}}\right). \end{align*}
- We now apply the squeeze theorem to the above inequalities. In particular, the limits of the lower and upper bounds are \(\lim_{n\rightarrow\infty}\frac{1}{n}\frac{e-1}{e^{\frac{1}{n}}-1}\) and \(\lim_{n\rightarrow\infty}\frac{1}{n}e^{\frac{1}{n}}\frac{e-1}{e^{\frac{1}{n}}-1}\text{,}\) respectively. As we did near the end of Example 1.1.2, we make these limits look more familiar by renaming \(\frac{1}{n}\) to \(X\text{.}\) As \(n\) tends to infinity, \(X\) tends to \(0\text{,}\) so the limits of the lower and upper bounds are\begin{align*} \lim_{n\rightarrow\infty}\frac{1}{n}\frac{e-1}{e^{\frac{1}{n}}-1} &=(e-1)\lim_{X=\frac{1}{n}\rightarrow 0}\frac{X}{e^X-1} =e-1 \end{align*}(by l'Hôpital's rule) and\begin{align*} \lim_{n\rightarrow\infty}\frac{1}{n}e^{\frac{1}{n}}\frac{e-1}{e^{\frac{1}{n}}-1} &=(e-1)\lim_{X=\frac{1}{n}\rightarrow 0}\cdot \frac{Xe^X}{e^X-1}\\ &=(e-1)\lim_{X\to 0}e^X \cdot \lim_{X=\to 0}\frac{X}{e^X-1}\\ &=(e-1) \cdot 1 \cdot 1 \end{align*}Thus, since the exact area is trapped between the lower and upper bounds, the squeeze theorem then implies that\begin{align*} \text{Exact area} &= e-1. \end{align*}
Subsection 1.1.3 Summation notation
As you can see from the above example (and the more careful rigorous computation), our discussion of integration will involve a fair bit of work with sums of quantities. To this end, we make a quick aside into summation notation. While one can work through the material below without this notation, proper summation notation is well worth learning, so we advise the reader to persevere.
Writing out the summands explicitly can become quite impractical — for example, say we need the sum of the first 11 squares:
\begin{gather*}
1 + 2^2 + 3^2 + 4^2+ 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2
\end{gather*}
This becomes tedious. Where the pattern is clear, we will often skip the middle few terms and instead write
\begin{gather*}
1 + 2^2 + \cdots + 11^2.
\end{gather*}
A far more precise way to write this is using \(\Sigma\) (capital-sigma) notation. For example, we can write the above sum as
\begin{gather*}
\sum_{k=1}^{11} k^2
\end{gather*}
This is read as
The sum from \(k\) equals 1 to 11 of \(k^2\text{.}\)
More generally
Definition 1.1.4.
Let \(m\leq n\) be integers and let \(f(x)\) be a function defined on the integers. Then we write
\begin{gather*}
\sum_{k=m}^n f(k)
\end{gather*}
to mean the sum of \(f(k)\) for \(k\) from \(m\) to \(n\text{:}\)
\begin{gather*}
f(m) + f(m+1) + f(m+2) + \cdots + f(n-1) + f(n).
\end{gather*}
Similarly we write
\begin{gather*}
\sum_{i=m}^n a_i
\end{gather*}
to mean
\begin{gather*}
a_m+a_{m+1}+a_{m+2}+\cdots+a_{n-1}+a_n
\end{gather*}
for some set of coefficients \(\{ a_m, \ldots, a_n \}\text{.}\)
Consider the example
\begin{gather*}
\sum_{k=3}^7 \frac{1}{k^2}=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+
\frac{1}{6^2}+\frac{1}{7^2}
\end{gather*}
It is important to note that the right hand side of this expression evaluates to a number 12 ; it does not contain “\(k\)”. The summation index \(k\) is just a “dummy” variable and it does not have to be called \(k\text{.}\) For example
\begin{gather*}
\sum_{k=3}^7 \frac{1}{k^2}
=\sum_{i=3}^7 \frac{1}{i^2}
=\sum_{j=3}^7 \frac{1}{j^2}
=\sum_{\ell=3}^7 \frac{1}{\ell^2}
\end{gather*}
Also the summation index has no meaning outside the sum. For example
\begin{gather*}
k\sum_{k=3}^7 \frac{1}{k^2}
\end{gather*}
has no mathematical meaning; it is gibberish.
A sum can be represented using summation notation in many different ways. If you are unsure as to whether or not two summation notations represent the same sum, just write out the first few terms and the last couple of terms. For example,
\begin{align*}
\sum_{m=3}^{15} \frac{1}{m^2}
&=\overbrace{\frac{1}{3^2}}^{m=3}
+\overbrace{\frac{1}{4^2}}^{m=4}
+\overbrace{\frac{1}{5^2}}^{m=5}
+\cdots
+\overbrace{\frac{1}{14^2}}^{m=14}
+\overbrace{\frac{1}{15^2}}^{m=15}\\
\sum_{m=4}^{16} \frac{1}{(m-1)^2}
&=\overbrace{\frac{1}{3^2}}^{m=4}
+\overbrace{\frac{1}{4^2}}^{m=5}
+\overbrace{\frac{1}{5^2}}^{m=6}
+\cdots
+\overbrace{\frac{1}{14^2}}^{m=15}
+\overbrace{\frac{1}{15^2}}^{m=16}
\end{align*}
are equal.
Here is a theorem that gives a few rules for manipulating summation notation.
Theorem 1.1.5. Arithmetic of Summation Notation.
Let \(n\ge m\) be integers. Then for all real numbers \(c\) and \(a_i,b_i\text{,}\) \(m\le i\le n\text{.}\)
- \(\displaystyle \sum\limits_{i=m}^nca_i = c\bigg(\sum\limits_{i=m}^na_i\bigg)\)
- \(\displaystyle \sum\limits_{i=m}^n(a_i+b_i) = \bigg(\sum\limits_{i=m}^na_i\bigg) + \bigg(\sum\limits_{i=m}^nb_i\bigg)\)
- \(\displaystyle \sum\limits_{i=m}^n(a_i-b_i) = \bigg(\sum\limits_{i=m}^na_i\bigg) - \bigg(\sum\limits_{i=m}^nb_i\bigg)\)
Proof.
We can prove this theorem by just writing out both sides of each equation, and observing that they are equal, by the usual laws of arithmetic 13 . For example, for the first equation, the left and right hand sides are
\begin{gather*}
\sum_{i=m}^nca_i = ca_m+ca_{m+1}+\cdots+ca_n\\
\quad\text{and}\quad
c\bigg(\sum\limits_{i=m}^na_i\bigg) = c(a_m+a_{m+1}+\cdots+a_n)
\end{gather*}
They are equal by the usual distributive law. The “distributive law” is the fancy name for \(c(a+b)=ca+cb\text{.}\)
Not many sums can be computed exactly 14 . Here are some that can. The first few are used a lot.
Theorem 1.1.6.
- \(\sum\limits_{i=0}^n ar^i = a\frac{1-r^{n+1}}{1-r}\text{,}\) for all real numbers \(a\) and \(r\ne 1\) and all integers \(n\ge 0\text{.}\)
- \(\sum\limits_{i=1}^n 1 = n\text{,}\) for all integers \(n\ge 1\text{.}\)
- \(\sum\limits_{i=1}^n i = \frac{1}{2}n(n+1)\text{,}\) for all integers \(n\ge 1\text{.}\)
- \(\sum\limits_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1)\text{,}\) for all integers \(n\ge 1\text{.}\)
- \(\sum\limits_{i=1}^n i^3 = \Big[\frac{1}{2}n(n+1)\Big]^2\text{,}\) for all integers \(n\ge 1\text{.}\)
Subsubsection 1.1.3.1 Proof of Theorem 1.1.6 (Optional)
Proof.
- The first sum is\begin{gather*} \sum_{i=0}^n ar^i =ar^0 + ar^1 + ar^2 + \cdots + ar^n \end{gather*}which is just the left hand side of equation 1.1.3, with \(n\) replaced by \(n+1\) and then multiplied by \(a\text{.}\)
- The second sum is just \(n\) copies of \(1\) added together, so of course the sum is \(n\text{.}\)
- The third and fourth sums are discussed in the appendix of the CLP-1 text. In that discussion certain “tricks” are used to compute the sums with only simple arithmetic. Those tricks do not easily generalise to the fifth sum.
-
Instead of repeating that appendix, we'll derive the third sum using a trick that generalises to the fourth and fifth sums (and also to higher powers). The trick uses the generating function 15 \(S(x)\text{:}\)
Equation 1.1.7. Finite geometric sum.
\begin{align*} S(x) = 1+x+x^2+\cdots+x^n &= \frac{x^{n+1}-1}{x-1} \end{align*}Notice that this is just the geometric sum given by equation 1.1.3 with \(n\) replaced by \(n+1\text{.}\)Now, consider the limit\begin{align*} \lim_{x\to 1} S(x) &= \lim_{x\to 1} \left(1+x+x^2+\cdots+x^n\right) = n+1 \qquad \text{but also}\\ &= \lim_{x\to 1} \frac{x^{n+1}-1}{x-1} \qquad\qquad\qquad \text{now use l'Hôpital's rule}\\ &= \lim_{x\to 1} \frac{(n+1)x^n}{1} = n+1. \end{align*}This is not so hard (or useful). But now consider the derivative of \(S(x)\text{:}\)\begin{align*} S'(x) &= 1 +2x + 3x^2 + \cdots + n x^{n-1}\\ &= \diff{}{x} \left[\frac{x^{n+1}-1}{x-1}\right] \qquad\qquad\qquad\qquad \text{use the quotient rule}\\ &= \frac{(x-1)\cdot (n+1)x^n - (x^{n+1}-1)\cdot 1}{(x-1)^2} \qquad \text{now clean it up}\\ &= \frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}. \end{align*}Hence if we take the limit of the above expression as \(x\to 1\) we recover\begin{align*} \lim_{x\to 1} S'(x) &= 1 +2 +3+\cdots+n\\ &=\lim_{x\to 1} \frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2} \qquad\qquad \text{now use l'Hôpital's rule}\\ &=\lim_{x\to 1} \frac{n(n+1)x^{n}-n(n+1)x^{n-1}}{2(x-1)} \qquad \text{l'Hôpital's rule again}\\ &=\lim_{x\to 1} \frac{n^2(n+1)x^{n-1}-n(n+1)(n-1)x^{n-2}}{2}\\ &= \frac{n^2(n+1) - n(n-1)(n+1)}{2} = \frac{n(n+1)}{2} \end{align*}as required. This computation can be done without l'Hôpital's rule, but the manipulations required are a fair bit messier. - The derivation of the fourth and fifth sums is similar to, but even more tedious than, that of the third sum. One takes two or three derivatives of the generating functional.
Subsection 1.1.4 The Definition of the Definite Integral
In this section we give a definition of the definite integral \(\ds \int_a^b f(x)\dee{x}\) generalising the machinery we used in Example 1.1.1. But first some terminology and a couple of remarks to better motivate the definition.
Definition 1.1.8.
We are here just going to introduce some notation and terminology. The definition of the definite integral will come later, in Definition 1.1.9. The symbol \(\ds \int_a^b f(x)\dee{x}\) is read “the definite integral of the function \(f(x)\) from \(a\) to \(b\)”. The function \(f(x)\) is called the integrand of \(\int_a^b f(x)\dee{x}\) and \(a\) and \(b\) are called 16 the limits of integration. The interval \(a\le x \le b\) is called the interval of integration and is also called the domain of integration.
Before we explain more precisely what the definite integral actually is, a few remarks (actually — a few interpretations) are in order.
-
If \(f(x)\ge 0\) and \(a\le b\text{,}\) one interpretation of the symbol \(\ds \int_a^b f(x)\dee{x}\) is “the area of the region \(\big\{\ (x,y)\ \big|\ a\le x\le b,\ 0\le y\le f(x)\ \big\}\)”.
In this way we can rewrite the area in Example 1.1.1 as the definite integral \(\int_0^1 e^x \dee{x}\text{.}\) - This interpretation breaks down when either \(a \gt b\) or \(f(x)\) is not always positive, but it can be repaired by considering “signed areas”.
-
If \(a\le b\text{,}\) but \(f(x)\) is not always positive, one interpretation of \(\int_a^b f(x)\dee{x}\) is “the signed area between \(y=f(x)\) and the \(x\)-axis for \(a\le x\le b\)”. For “signed area” (which is also called the “net area”), areas above the \(x\)-axis count as positive while areas below the \(x\)-axis count as negative. In the example below, we have the graph of the function\begin{align*} f(x)=\begin{cases} -1 & \text{if }1\le x\le 2\\ 2 & \text{if }2 \lt x\le 4 \\ 0 & \text{otherwise} \end{cases} \end{align*}The \(2\times 2\) shaded square above the \(x\)-axis has signed area \(+2\times 2=+4\text{.}\) The \(1\times 1\) shaded square below the \(x\)-axis has signed area \(-1\times 1=-1\text{.}\) So, for this \(f(x)\text{,}\)\begin{gather*} \int_0^5 f(x)\dee{x} = +4-1=3 \end{gather*}
- We'll come back to the case \(b \lt a\) later.
We're now ready to define \(\int_a^b f(x)\dee{x}\text{.}\) The definition is a little involved, but essentially mimics what we did in Example 1.1.1 (which is why we did the example before the definition). The main differences are that we replace the function \(e^x\) by a generic function \(f(x)\) and we replace the interval from \(0\) to \(1\) by the generic interval 17 from \(a\) to \(b\text{.}\)
- We start by selecting any natural number \(n\) and subdividing the interval from \(a\) to \(b\) into \(n\) equal subintervals. Each subinterval has width \(\frac{b-a}{n}\text{.}\)
- Just as was the case in Example 1.1.1 we will eventually take the limit as \(n\to\infty\text{,}\) which squeezes the width of each subinterval down to zero.
- For each integer \(0\le i\le n\text{,}\) define \(x_i = a + i \cdot\frac{b-a}{n}\text{.}\) Note that this means that \(x_0=a\) and \(x_n = b\text{.}\) It is worth keeping in mind that these numbers \(x_i\) do depend on \(n\) even though our choice of notation hides this dependence.
-
Subinterval number \(i\) is \(x_{i-1} \leq x \leq x_i\text{.}\) In particular, on the first subinterval, \(x\) runs from \(x_0=a\) to \(x_1=a+\frac{b-a}{n}\text{.}\) On the second subinterval, \(x\) runs from \(x_1\) to \(x_2=a+2\frac{b-a}{n}\text{.}\)
-
On each subinterval we now pick \(x_{i,n}^*\) between \(x_{i-1}\) and \(x_i\text{.}\) We then approximate \(f(x)\) on the \(i^\mathrm{th}\) subinterval by the constant function \(y=f(x_{i,n}^*)\text{.}\) We include \(n\) in the subscript to remind ourselves that these numbers depend on \(n\text{.}\)Geometrically, we're approximating the region\begin{gather*} \big\{\ (x,y)\ \big|\ \text{$x$ is between $x_{i-1}$ and $x_i$, and $y$ is between $0$ and $f(x)$} \ \big\} \end{gather*}by the rectangle\begin{gather*} \big\{\ (x,y)\ \big|\ \text{$x$ is between $x_{i-1}$ and $x_i$, and $y$ is between $0$ and $f(x_{i,n}^*)$} \ \big\} \end{gather*}
In Example 1.1.1 we chose \(x_{i,n}^* = x_{i-1}\) and so we approximated the function \(e^x\) on each subinterval by the value it took at the leftmost point in that subinterval. - So, when there are \(n\) subintervals our approximation to the signed area between the curve \(y=f(x)\) and the \(x\)-axis, with \(x\) running from \(a\) to \(b\text{,}\) is\begin{gather*} \sum_{i=1}^n f(x_{i,n}^*)\cdot \frac{b-a}{n} \end{gather*}We interpret this as the signed area since the summands \(f(x_{i,n}^*)\cdot\frac{b-a}{n}\) need not be positive.
- Finally we define the definite integral by taking the limit of this sum as \(n\rightarrow\infty\text{.}\)
Oof! This is quite an involved process, but we can now write down the definition we need.
Definition 1.1.9.
Let \(a\) and \(b\) be two real numbers and let \(f(x)\) be a function that is defined for all \(x\) between \(a\) and \(b\text{.}\) Then we define
\begin{gather*}
\int_a^b f(x)\dee{x} =\lim_{n\rightarrow\infty}\sum_{i=1}^n f(x_{i,n}^*)\cdot\frac{b-a}{n}
\end{gather*}
when the limit exists and takes the same value for all choices of the \(x_{i,n}^*\)'s. In this case, we say that \(f\) is integrable on the interval from \(a\) to \(b\text{.}\)
Of course, it is not immediately obvious when this limit should exist. Thankfully it is easier for a function to be “integrable” than it is for it to be “differentiable”.
Theorem 1.1.10.
Let \(f(x)\) be a function on the interval \([a,b]\text{.}\) If
- \(f(x)\) is continuous on \([a,b]\text{,}\) or
- \(f(x)\) has a finite number of jump discontinuities on \([a,b]\) (and is otherwise continuous)
then \(f(x)\) is integrable on \([a,b]\text{.}\)
We will not justify this theorem. But a slightly weaker statement is proved in (the optional) Section 1.1.7. Of course this does not tell us how to actually evaluate any definite integrals — but we will get to that in time.
Some comments:
- Note that, in Definition 1.1.9, we allow \(a\) and \(b\) to be any two real numbers. We do not require that \(a \lt b\text{.}\) That is, even when \(a \gt b\text{,}\) the symbol \(\int_a^b f(x)\dee{x}\) is still defined by the formula of Definition 1.1.9. We'll get an interpretation for \(\int_a^b f(x)\dee{x}\text{,}\) when \(a \gt b\text{,}\) later.
- It is important to note that the definite integral \(\int_a^b f(x)\dee{x}\) represents a number, not a function of \(x\text{.}\) The integration variable \(x\) is another “dummy” variable, just like the summation index \(i\) in \(\sum_{i=m}^n a_i\) (see Section 1.1.3). The integration variable does not have to be called \(x\text{.}\) For example\begin{gather*} \int_a^b f(x)\dee{x} = \int_a^b f(t)\dee{t} = \int_a^b f(u)\dee{u} \end{gather*}Just as with summation variables, the integration variable \(x\) has no meaning outside of \(f(x)\dee{x}\text{.}\) For example\begin{gather*} x\int_0^1 e^x\dee{x}\qquad\text{and}\qquad \int_0^x e^x\dee{x} \end{gather*}are both gibberish.
The sum inside definition 1.1.9 is named after Bernhard Riemann 18 who made the first rigorous definition of the definite integral and so placed integral calculus on rigorous footings.
Definition 1.1.11.
The sum \(\ds \) inside definition 1.1.9
\begin{gather*}
\sum_{i=1}^n f(x_{i,n}^*)\,\frac{b-a}{n}
\end{gather*}
is called a Riemann sum. It is also often written as
\begin{gather*}
\sum_{i=1}^n f(x_i^*)\,\De x
\end{gather*}
where \(\De x = \frac{b-a}{n}\text{.}\)
- If we choose each \(x_{i,n}^* = x_{i-1}=a+(i-1)\frac{b-a}{n}\) to be the left hand end point of the \(i^{\rm th}\) interval, \([x_{i-1},x_i]\text{,}\) we get the approximation\begin{gather*} \sum_{i=1}^n f\left(a+(i-1)\frac{b-a}{n}\right)\,\frac{b-a}{n} \end{gather*}which is called the “left Riemann sum approximation to \(\int_a^b f(x)\dee{x}\) with \(n\) subintervals”. This is the approximation used in Example 1.1.1.
- In the same way, if we choose \(x_{i,n}^* = x_{i}=a+i\frac{b-a}{n}\) we obtain the approximation\begin{gather*} \sum_{i=1}^n f\left(a+i\frac{b-a}{n}\right)\,\frac{b-a}{n} \end{gather*}which is called the “right Riemann sum approximation to \(\int_a^b f(x)\dee{x}\) with \(n\) subintervals”. The word “right” signifies that, on each subinterval \([x_{i-1},x_i]\) we approximate \(f\) by its value at the right-hand end-point, \(x_i=a+i\frac{b-a}{n}\text{,}\) of the subinterval.
- A third commonly used approximation is\begin{gather*} \sum_{i=1}^n f\left(a+(i-\frac12)\frac{b-a}{n}\right)\,\frac{b-a}{n} \end{gather*}which is called the “midpoint Riemann sum approximation to \(\int_a^b f(x)\dee{x}\) with \(n\) subintervals”. The word “midpoint” signifies that, on each subinterval \([x_{i-1},x_i]\) we approximate \(f\) by its value at the midpoint, \(\frac{x_{i-1}+x_i}{2} =a+(i-\frac{1}{2})\frac{b-a}{n}\text{,}\) of the subinterval.
In order to compute a definite integral using Riemann sums we need to be able to compute the limit of the sum as the number of summands goes to infinity. This approach is not always feasible and we will soon arrive at other means of computing definite integrals based on antiderivatives. However, Riemann sums also provide us with a good means of approximating definite integrals — if we take \(n\) to be a large, but finite, integer, then the corresponding Riemann sum can be a good approximation of the definite integral. Under certain circumstances this can be strengthened to give rigorous bounds on the integral. Let us revisit Example 1.1.1.
Example 1.1.12. Upper and lower bounds on area.
Let's say we are again interested in the integral \(\int_0^1 e^x\dee{x}\text{.}\) We can follow the same procedure as we used previously to construct Riemann sum approximations. However since the integrand \(f(x)=e^x\) is an increasing function, we can make our approximations into upper and lower bounds without much extra work.
More precisely, we approximate \(f(x)\) on each subinterval \(x_{i-1}\le x\le x_i\)
- by its smallest value on the subinterval, namely \(f(x_{i-1})\text{,}\) when we compute the left Riemann sum approximation and
- by its largest value on the subinterval, namely \(f(x_i)\text{,}\) when we compute the right Riemann sum approximation.
This is illustrated in the two figures below. The shaded region in the left hand figure is the left Riemann sum approximation and the shaded region in the right hand figure is the right Riemann sum approximation.
We can see that exactly because \(f(x)\) is increasing, the left Riemann sum describes an area smaller than the definite integral while the right Riemann sum gives an area larger 19 than the integral.
When we approximate the integral \(\int_0^1 e^x\dee{x}\) using \(n\) subintervals, then, on interval number \(i\text{,}\)
- \(x\) runs from \(\frac{i-1}{n}\) to \(\frac{i}{n}\) and
- \(y=e^x\) runs from \(e^{\frac{(i-1)}{n}}\text{,}\) when \(x\) is at the left hand end point of the interval, to \(e^{\frac{i}{n}}\text{,}\) when \(x\) is at the right hand end point of the interval.
Consequently, the left Riemann sum approximation to \(\int_0^1 e^x\dee{x}\) is \(\sum_{i=1}^n e^{\frac{(i-1)}{n}}\,\frac{1}{n}\) and the right Riemann sum approximation is \(\sum_{i=1}^n e^{\frac{i}{n}}\cdot\frac{1}{n}\text{.}\) So
\begin{gather*}
\sum_{i=1}^n e^{\frac{(i-1)}{n}}\,\frac{1}{n}
\ \le\ \int_0^1 e^x\dee{x}\ \le\
\sum_{i=1}^n e^{\frac{i}{n}}\cdot\frac{1}{n}
\end{gather*}
Thus \(L_n=\sum_{i=1}^n e^{\frac{(i-1)}{n}}\,\frac{1}{n}\text{,}\) which for any \(n\) can be evaluated by computer, is a lower bound on the exact value of \(\int_0^1 e^x\dee{x}\) and \(R_n=\sum_{i=1}^n e^{\frac{i}{n}}\,\frac{1}{n}\text{,}\) which for any \(n\) can also be evaluated by computer, is an upper bound on the exact value of \(\int_0^1 e^x\dee{x}\text{.}\) For example, when \(n=1000\text{,}\) \(L_n= 1.7174\) and \(R_n=1.7191\) (both to four decimal places) so that, again to four decimal places,
\begin{gather*}
1.7174 \le \int_0^1 e^x\dee{x} \le 1.7191
\end{gather*}
Recall that the exact value is \(e-1 = 1.718281828\dots\text{.}\)
Subsection 1.1.5 Using Known Areas to Evaluate Integrals
One of the main aims of this course is to build up general machinery for computing definite integrals (as well as interpreting and applying them). We shall start on this soon, but not quite yet. We have already seen one concrete, if laborious, method for computing definite integrals — taking limits of Riemann sums as we did in Example 1.1.1. A second method, which will work for some special integrands, works by interpreting the definite integral as “signed area”. This approach will work nicely when the area under the curve decomposes into simple geometric shapes like triangles, rectangles and circles. Here are some examples of this second method.
Example 1.1.13. A very simple integral and a very simple area.
The integral \(\int_a^b 1\dee{x}\) (which is also written as just \(\int_a^b\dee{x}\)) is the area of the shaded rectangle (of width \(b-a\) and height \(1\)) in the figure on the right below. So
\(\int_a^b\dee{x} = (b-a)\times (1)=b-a\)
Example 1.1.14. Another simple integral.
Let \(b \gt 0\text{.}\) The integral \(\int_0^b x\dee{x}\) is the area of the shaded triangle (of base \(b\) and of height \(b\)) in the figure on the right below. So
\(\int_0^b x\dee{x} = \frac{1}{2}b\times b=\frac{b^2}{2}\)
The integral \(\int_{-b}^0 x\dee{x}\) is the signed area of the shaded triangle (again of base \(b\) and of height \(b\)) in the figure on the right below. So
\(\int_{-b}^0 x\dee{x} = -\frac{b^2}{2} \)
Notice that it is very easy to extend this example to the integral \(\int_0^b c x\dee{x}\) for any real numbers \(b,c \gt 0\) and find
\begin{align*}
\int_0^b c x\dee{x} &= \frac{c}{2} b^2.
\end{align*}
Example 1.1.15. Evaluating \(\int_{-1}^1 \left(1-|x|\right)\dee{x}\).
In this example, we shall evaluate \(\int_{-1}^1 \left(1-|x|\right)\dee{x}\text{.}\) Recall that
\begin{align*}
|x|=\begin{cases} -x &\text{if $x\le 0$} \\
x &\text{if $x\ge 0$}
\end{cases}
\end{align*}
so that
\begin{align*}
1-|x|=\begin{cases} 1+x &\text{if $x\le 0$}\\
1-x &\text{if $x\ge 0$}
\end{cases}
\end{align*}
To picture the geometric figure whose area the integral represents observe that
- at the left hand end of the domain of integration \(x=-1\) and the integrand \(1-|x|=1-|-1|=1-1=0\) and
- as \(x\) increases from \(-1\) towards \(0\text{,}\) the integrand \(1-|x|=1+x\) increases linearly, until
- when \(x\) hits \(0\) the integrand hits \(1-|x|=1-|0|=1\) and then
- as \(x\) increases from \(0\text{,}\) the integrand \(1-|x|=1-x\) decreases linearly, until
- when \(x\) hits \(+1\text{,}\) the right hand end of the domain of integration, the integrand hits \(1-|x|=1-|1|=0\text{.}\)
So the integral \(\int_{-1}^1 \left(1-|x|\right)\dee{x}\) is the area of the shaded triangle (of base \(2\) and of height \(1\)) in the figure on the right below and
\(\int_{-1}^1 \left(1-|x|\right)\dee{x} = \frac{1}{2}\times 2\times 1 =
1\)
Example 1.1.16. Evaluating \(\int_0^1 \sqrt{1-x^2}\dee{x}\).
The integral \(\int_0^1 \sqrt{1-x^2}\dee{x}\) has integrand \(f(x)=\sqrt{1-x^2}\text{.}\) So it represents the area under \(y=\sqrt{1-x^2}\) with \(x\) running from \(0\) to \(1\text{.}\) But we may rewrite
\begin{align*}
y&=\sqrt{1-x^2} &\text{as}&& x^2+y^2&= 1, y\geq 0
\end{align*}
But this is the (implicit) equation for a circle — the extra condition that \(y\geq0\) makes it the equation for the semi-circle centred at the origin with radius 1 lying on and above the \(x\)-axis. Thus the integral represents the area of the quarter circle of radius \(1\text{,}\) as shown in the figure on the right below. So
\(\int_0^1 \sqrt{1-x^2}\dee{x} = \frac{1}{4}\pi(1)^2 = \frac{\pi}{4}\)
This next one is a little trickier and relies on us knowing the symmetries of the sine function.
Example 1.1.17. Integrating sine.
The integral \(\int_{-\pi}^\pi \sin x\dee{x}\) is the signed area of the shaded region in the figure on the right below. It naturally splits into two regions, one on either side of the \(y\)-axis. We don't know the formula for the area of either of these regions (yet), however the two regions are very nearly the same. In fact, the part of the shaded region below the \(x\)-axis is exactly the reflection, in the \(x\)-axis, of the part of the shaded region above the \(x\)-axis. So the signed area of part of the shaded region below the \(x\)-axis is the negative of the signed area of part of the shaded region above the \(x\)-axis and
\(\int_{-\pi}^\pi \sin x\dee{x} = 0\)
Subsection 1.1.6 Another Interpretation for Definite Integrals
So far, we have only a single interpretation 20 for definite integrals — namely areas under graphs. In the following example, we develop a second interpretation.
Example 1.1.18. A moving particle.
Suppose that a particle is moving along the \(x\)-axis and suppose that at time \(t\) its velocity is \(v(t)\) (with \(v(t) \gt 0\) indicating rightward motion and \(v(t) \lt 0\) indicating leftward motion). What is the change in its \(x\)-coordinate between time \(a\) and time \(b \gt a\text{?}\)
We'll work this out using a procedure similar to our definition of the integral. First pick a natural number \(n\) and divide the time interval from \(a\) to \(b\) into \(n\) equal subintervals, each of width \(\frac{b-a}{n}\text{.}\) We are working our way towards a Riemann sum (as we have done several times above) and so we will eventually take the limit \(n\rightarrow\infty\text{.}\)
- The first time interval runs from \(a\) to \(a+\frac{b-a}{n}\text{.}\) If we think of \(n\) as some large number, the width of this interval, \(\frac{b-a}{n}\) is very small and over this time interval, the velocity does not change very much. Hence we can approximate the velocity over the first subinterval as being essentially constant at its value at the start of the time interval — \(v(a)\text{.}\) Over the subinterval the \(x\)-coordinate changes by velocity times time, namely \(v(a) \cdot \frac{b-a}{n}\text{.}\)
- Similarly, the second interval runs from time \(a+\frac{b-a}{n}\) to time \(a+2\frac{b-a}{n}\text{.}\) Again, we can assume that the velocity does not change very much and so we can approximate the velocity as being essentially constant at its value at the start of the subinterval — namely \(v\left(a+\frac{b-a}{n}\right)\text{.}\) So during the second subinterval the particle's \(x\)-coordinate changes by approximately \(v\left(a+\frac{b-a}{n}\right) \frac{b-a}{n}\text{.}\)
- In general, time subinterval number \(i\) runs from \(a+(i-1)\frac{b-a}{n}\) to \(a+i\frac{b-a}{n}\) and during this subinterval the particle's \(x\)-coordinate changes, essentially, by\begin{gather*} , v\left(a+(i-1)\frac{b-a}{n}\right) \frac{b-a}{n}. \end{gather*}
So the net change in \(x\)-coordinate from time \(a\) to time \(b\) is approximately
\begin{align*}
&v(a)\,\frac{b-a}{n} + v\Big(a+\frac{b-a}{n}\Big)\,\frac{b-a}{n} +\cdots
+v\Big(a+(i-1)\frac{b-a}{n}\Big)\,\frac{b-a}{n} + \cdots\\
&
+v\Big(a+(n-1)\frac{b-a}{n}\Big)\,\frac{b-a}{n}\\
&= \sum_{i=1}^n v\Big(a+(i-1)\frac{b-a}{n}\Big)\,\frac{b-a}{n}
\end{align*}
This exactly the left Riemann sum approximation to the integral of \(v\) from \(a\) to \(b\) with \(n\) subintervals. The limit as \(n\rightarrow\infty\) is exactly the definite integral \(\int_a^b v(t)\dee{t}\text{.}\) Following tradition, we have called the (dummy) integration variable \(t\) rather than \(x\) to remind us that it is time that is running from \(a\) to \(b\text{.}\)
The conclusion of the above discussion is that if a particle is moving along the \(x\)-axis and its \(x\)-coordinate and velocity at time \(t\) are \(x(t)\) and \(v(t)\text{,}\) respectively, then, for all \(b \gt a\text{,}\)
\begin{gather*}
x(b) - x(a) = \int_a^b v(t)\dee{t}.
\end{gather*}
Subsection 1.1.7 Optional — careful definition of the integral
In this optional section we give a more mathematically rigorous definition of the definite integral \(\ds \int_a^b f(x)\dee{x}\text{.}\) Some textbooks use a sneakier, but equivalent, definition. The integral will be defined as the limit of a family of approximations to the area between the graph of \(y=f(x)\) and the \(x\)-axis, with \(x\) running from \(a\) to \(b\text{.}\) We will then show conditions under which this limit is guaranteed to exist. We should state up front that these conditions are more restrictive than is strictly necessary — this is done so as to keep the proof accessible.
The family of approximations needed is slightly more general than that used to define Riemann sums in the previous sections, though it is quite similar. The main difference is that we do not require that all the subintervals have the same size.
- We start by selecting a positive integer \(n\text{.}\) As was the case previously, this will be the number of subintervals used in the approximation and eventually we will take the limit as \(n \to \infty\text{.}\)
- Now subdivide the interval from \(a\) to \(b\) into \(n\) subintervals by selecting \(n+1\) values of \(x\) that obey\begin{gather*} a=x_0 \lt x_1 \lt x_2 \lt \cdots \lt x_{n-1} \lt x_n=b. \end{gather*}The subinterval number \(i\) runs from \(x_{i-1}\) to \(x_i\text{.}\) This formulation does not require the subintervals to have the same size. However we will eventually require that the widths of the subintervals shrink towards zero as \(n\to\infty\text{.}\)
- Then for each subinterval we select a value of \(x\) in that interval. That is, for \(i=1,2,\dots,n\text{,}\) choose \(x_i^*\) satisfying \(x_{i-1} \leq x_i^* \leq x_i\text{.}\) We will use these values of \(x\) to help approximate \(f(x)\) on each subinterval.
-
The area between the graph of \(y=f(x)\) and the \(x\)-axis, with \(x\) running
from \(x_{i-1}\) to \(x_i\text{,}\) i.e. the contribution, \(\int_{x_{i-1}}^{x_i} f(x)\dee{x}\text{,}\) from interval number \(i\) to the integral, is approximated by the area of a rectangle. The rectangle has width \(x_i-x_{i-1}\) and height \(f(x_i^*)\text{.}\)
- Thus the approximation to the integral, using all \(n\) subintervals, is\begin{gather*} \int_a^b f(x)\dee{x} \approx f(x_1^*)[x_1-x_0]+f(x_2^*)[x_2-x_1]+\cdots+ f(x_n^*)[x_n-x_{n-1}] \end{gather*}
- Of course every different choice of \(n\) and \(x_1,x_2,\cdots,x_{n-1}\) and \(x_1^*, x_2^*,\cdots,x_n^*\) gives a different approximation. So to simplify the discussion that follows, let us denote a particular choice of all these numbers by \(\bbbp\text{:}\)\begin{gather*} \bbbp=\left(n,x_1,x_2,\cdots,x_{n-1},x_1^*, x_2^*, \cdots, x_n^*\right). \end{gather*}Similarly let us denote the resulting approximation by \(\cI(\bbbp)\text{:}\)\begin{gather*} \cI(\bbbp)=f(x_1^*)[x_1-x_0]+f(x_2^*)[x_2-x_1]+\cdots+ f(x_n^*)[x_n-x_{n-1}] \end{gather*}
- Let's be more precise. We can take the limit of these approximations in two equivalent ways. Above we did this by taking the number of subintervals \(n\) to infinity. When we did this, the width of all the subintervals went to zero. With the formulation we are now using, simply taking the number of subintervals to be very large does not imply that they will all shrink in size. We could have one very large subinterval and a large number of tiny ones. Thus we take the limit we need by taking the width of the subintervals to zero. So for any choice \(\bbbp\text{,}\) we define\begin{gather*} M(\bbbp)=\max\big\{ x_1-x_0\ ,\ x_2-x_1\ ,\ \cdots\ ,\ x_n-x_{n-1}\big\} \end{gather*}that is the maximum width of the subintervals used in the approximation determined by \(\bbbp\text{.}\) By forcing the maximum width to go to zero, the widths of all the subintervals go to zero.
- We then define the definite integral as the limit\begin{gather*} \int_a^b f(x)\dee{x}=\lim_{M(\bbbp)\rightarrow 0}\cI(\bbbp). \end{gather*}
Of course, one is now left with the question of determining when the above limit exists. A proof of the very general conditions which guarantee existence of this limit is beyond the scope of this course, so we instead give a weaker result (with stronger conditions) which is far easier to prove.
For the rest of this section, assume
- that \(f(x)\) is continuous for \(a\le x\le b\text{,}\)
- that \(f(x)\) is differentiable for \(a \lt x \lt b\text{,}\) and
- that \(f'(x)\) is bounded — ie \(|f'(x)|\leq F\) for some constant \(F\text{.}\)
We will now show that, under these hypotheses, as \(M(\bbbp)\) approaches zero, \(\cI(\bbbp)\) always approaches the area, \(A\text{,}\) between the graph of \(y=f(x)\) and the \(x\)-axis, with \(x\) running from \(a\) to \(b\text{.}\)
These assumptions are chosen to make the argument particularly transparent. With a little more work one can weaken the hypotheses considerably. We are cheating a little by implicitly assuming that the area \(A\) exists. In fact, one can adjust the argument below to remove this implicit assumption.
-
Consider \(A_j\text{,}\) the part of the area coming from \(x_{j-1}\le x\le x_j\text{.}\)
We have approximated this area by \(f(x_j^*)[x_j-x_{j-1}]\) (see figure left). - Let \(f({\overline x}_j)\) and \(f({\underline x}_j)\) be the largest and smallest values 23 of \(f(x)\) for \(x_{j-1}\le x\le x_j\text{.}\) Then the true area is bounded by\begin{gather*} f({\underline x}_j)[x_j-x_{j-1}] \leq A_j \leq f({\overline x}_j)[x_j-x_{j-1}]. \end{gather*}(see figure right).
- Now since \(f({\underline x}_j) \leq f(x_j^*) \leq f({\overline x}_j)\text{,}\) we also know that\begin{gather*} f({\underline x}_j)[x_j-x_{j-1}] \leq f(x_j^*)[x_j-x_{j-1}] \leq f({\overline x}_j)[x_j-x_{j-1}]. \end{gather*}
- So both the true area, \(A_j\text{,}\) and our approximation of that area \(f(x_j^*)[x_j - x_{j-1}]\) have to lie between \(f({\overline x}_j)[x_j-x_{j-1}]\) and \(f({\underline x}_j)[x_j-x_{j-1}]\text{.}\) Combining these bounds we have that the difference between the true area and our approximation of that area is bounded by\begin{gather*} \big|A_j-f(x_j^*)[x_j-x_{j-1}]\big| \le[f({\overline x}_j)-f({\underline x}_j)]\cdot[x_j-x_{j-1}]. \end{gather*}(To see this think about the smallest the true area can be and the largest our approximation can be and vice versa.)
- Now since our function, \(f(x)\) is differentiable we can apply one of the main theorems we learned in CLP-1 — the Mean Value Theorem 24 . The MVT implies that there exists a \(c\) between \({\underline x}_j\) and \({\overline x}_j\) such that\begin{gather*} f({\overline x}_j)-f({\underline x}_j) =f'(c)\cdot [{\overline x}_j-{\underline x}_j] \end{gather*}
- By the assumption that \(|f'(x)|\le F\) for all \(x\) and the fact that \({\underline x}_j\) and \({\overline x}_j\) must both be between \(x_{j-1}\) and \(x_j\)\begin{gather*} \big|f({\overline x}_j)-f({\underline x}_j)\big| \le F\cdot \big|{\overline x}_j-{\underline x}_j\big| \le F\cdot [x_j-x_{j-1}] \end{gather*}Hence the error in this part of our approximation obeys\begin{gather*} \big|A_j-f(x_j^*)[x_j-x_{j-1}]\big| \le F\cdot [x_j-x_{j-1}]^2. \end{gather*}
- That was just the error in approximating \(A_j\text{.}\) Now we bound the total error by combining the errors from approximating on all the subintervals. This gives\begin{align*} \left| A-\cI(\bbbp)\right| &= \left| \sum_{j=1}^n A_j - \sum_{j=1}^n f(x_j^*)[x_j-x_{j-1}] \right|\\ &= \left| \sum_{j=1}^n \left(A_j - f(x_j^*)[x_j-x_{j-1}] \right) \right| &\text{triangle inequality}\\ &\leq \sum_{j=1}^n\left|A_j - f(x_j^*)[x_j-x_{j-1}]\right|\\ &\leq \sum_{j=1}^n F\cdot [x_j-x_{j-1}]^2 & \text{from above}\\ \end{align*}
Now do something a little sneaky. Replace one of these factors of \([x_j-x_{j-1}]\) (which is just the width of the \(j^\mathrm{th}\) subinterval) by the maximum width of the subintervals:
\begin{align*} &\leq \sum_{j=1}^n F\cdot M(\bbbp)\cdot [x_j-x_{j-1}] &\text{$F$ and $M(\bbbp)$ are constant}\\ &\leq F\cdot M(\bbbp)\cdot \sum_{j=1}^n [x_j-x_{j-1}] & \text{sum is total width}\\ & = F\cdot M(\bbbp)\cdot (b-a). \end{align*} - Since \(a\text{,}\) \(b\) and \(F\) are fixed, this tends to zero as the maximum rectangle width \(M(\bbbp)\) tends to zero.
Thus, we have proven
Theorem 1.1.19.
Assume that \(f(x)\) is continuous for \(a\le x\le b\text{,}\) and is differentiable for all \(a \lt x \lt b\) with \(|f'(x)|\le F\text{,}\) for some constant \(F\text{.}\) Then, as the maximum rectangle width \(M(\bbbp)\) tends to zero, \(\cI(\bbbp)\) always converges to \(A\text{,}\) the area between the graph of \(y=f(x)\) and the \(x\)-axis, with \(x\) running from \(a\) to \(b\text{.}\)
Exercises 1.1.8 Exercises
Exercises — Stage 1 .
1.
Give a range of possible values for the shaded area in the picture below.
2.
Give a range of possible values for the shaded area in the picture below.
3.
Using rectangles, find a lower and upper bound for \(\displaystyle\int_1^3 \dfrac{1}{2^x}\dee{x}\) that differ by at most 0.2 square units.
4.
Let \(f(x)\) be a function that is decreasing from \(x=0\) to \(x=5\text{.}\) Which Riemann sum approximation of \(\displaystyle\int_0^5 f(x)\dee{x}\) is the largest--left, right, or midpoint?
5.
Give an example of a function \(f(x)\text{,}\) an interval \([a,b]\text{,}\) and a number \(n\) such that the midpoint Riemann sum of \(f(x)\) over \([a,b]\) using \(n\) intervals is larger than both the left and right Riemann sums of \(f(x)\) over \([a,b]\) using \(n\) intervals.
6.
Express the following sums in sigma notation:
- \(\displaystyle 3+4+5+6+7\)
- \(\displaystyle 6+8+10+12+14\)
- \(\displaystyle 7+9+11+13+15\)
- \(\displaystyle 1+3+5+7+9+11+13+15\)
7.
Express the following sums in sigma notation:
- \(\displaystyle \frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
- \(\displaystyle \frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}\)
- \(\displaystyle -\frac{2}{3}+\frac{2}{9}-\frac{2}{27}+\frac{2}{81}\)
- \(\displaystyle \frac{2}{3}-\frac{2}{9}+\frac{2}{27}-\frac{2}{81}\)
8.
Express the following sums in sigma notation:
- \(\displaystyle \frac{1}{3}+\frac{1}{3}+\frac{5}{27}+\frac{7}{81}+\frac{9}{243}\)
- \(\displaystyle \frac{1}{5}+\frac{1}{11}+\frac{1}{29}+\frac{1}{83}+\frac{1}{245}\)
- \(\displaystyle 1000+200+30+4+\frac{1}{2}+\frac{3}{50}+\frac{7}{1000}\)
9.
- \(\displaystyle \displaystyle\sum_{i=0}^{100} \left(\dfrac{3}{5}\right)^i\)
- \(\displaystyle \displaystyle\sum_{i=50}^{100} \left(\dfrac{3}{5}\right)^i\)
- \(\displaystyle \displaystyle\sum_{i=1}^{10} \left(i^2-3i+5\right)\)
- \(\displaystyle\sum_{n=1}^{b}\left[ \left(\frac{1}{e}\right)^n+en^3\right]\text{,}\) where \(b\) is some integer greater than 1.
10.
Evaluate the following sums. You might want to use the formulas from Theorem 1.1.6.
- \(\displaystyle \displaystyle\sum_{i=50}^{100} (i-50)+\displaystyle\sum_{i=0}^{50} i\)
- \(\displaystyle \displaystyle\sum_{i=10}^{100} \left(i-5\right)^3\)
- \(\displaystyle \displaystyle\sum_{n=1}^{11} (-1)^n\)
- \(\displaystyle \displaystyle\sum_{n=2}^{11} (-1)^{2n+1}\)
11.
In the picture below, draw in the rectangles whose (signed) area is being computed by the midpoint Riemann sum \(\displaystyle\sum_{i=1}^4 \dfrac{b-a}{4}\cdot f\left(a+\left(i-\frac{1}{2}\right)\dfrac{b-a}{4}\right)\text{.}\)
12. (✳).
\(\displaystyle \sum_{k=1}^4 f(1+k)\cdot 1\) is a left Riemann sum for a function \(f(x)\) on the interval \([a,b]\) with \(n\) subintervals. Find the values of \(a\text{,}\) \(b\) and \(n\text{.}\)
13.
Draw a picture illustrating the area given by the following Riemann sum.
\begin{equation*}
\sum_{i=1}^3 2\cdot\left(5+2i\right)^2
\end{equation*}
14.
Draw a picture illustrating the area given by the following Riemann sum.
\begin{equation*}
\sum_{i=1}^5 \frac{\pi}{20}\cdot \tan\left(\frac{\pi (i-1)}{20}\right)
\end{equation*}
15. (✳).
Fill in the blanks with right, left, or midpoint; an interval; and a value of n.
- \(\sum\limits_{k=0}^3 f (1.5 + k) \cdot 1\) is a \(\underline{\ \ \ \ \ \ \ \ \ \ \ \ }\) Riemann sum for \(f\) on the interval \([\,\underline{\ \ \ \ \ \ }\ ,\ \underline{\ \ \ \ \ \ }\,]\) with \(n =\underline{\ \ \ \ \ }\text{.}\)
16.
Evaluate the following integral by interpreting it as a signed area, and using geometry:
\begin{equation*}
\int_0^5 x \,\dee{x}
\end{equation*}
17.
Evaluate the following integral by interpreting it as a signed area, and using geometry:
\begin{equation*}
\int_{-2}^5 x \,\dee{x}
\end{equation*}
Exercises — Stage 2 .
Remember that a definite integral is a signed area between a curve and the \(x\)-axis. We'll spend a lot of time learning strategies for evaluating definite integrals, but we already know lots of ways to find area of geometric shapes. In Questions 28 through 33, use your knowledge of geometry to find the signed areas described by the integrals given.
18. (✳).
Use sigma notation to write the midpoint Riemann sum for \(f(x)=x^8\) on \([5,15]\) with \(n=50\text{.}\) Do not evaluate the Riemann sum.
19. (✳).
Estimate \(\displaystyle\int_{-1}^5 x^3\,\,\dee{x}\) using three approximating rectangles and left hand end points.
20. (✳).
Let \(f\) be a function on the whole real line. Express \(\displaystyle\int_{-1}^{7}f(x)\,\,\dee{x}\) as a limit of Riemann sums, using the right endpoints.
21. (✳).
The value of the following limit is equal to the area below a graph of \(y=f(x)\text{,}\) integrated over the interval \([0,b]\text{:}\)
\begin{gather*}
\lim_{n \to \infty} \sum_{i=1}^{n} \frac{4}{n}
\left[ \sin \left( 2 + \frac{4i}{n}\right)\right]^2
\end{gather*}
Find \(f(x)\) and \(b\text{.}\)
22. (✳).
For a certain function \(f(x)\text{,}\) the following equation holds:
\begin{equation*}
\lim_{n\rightarrow\infty}\sum\limits_{k=1}^n
\frac{k}{n^2}\sqrt{1-\frac{k^2}{n^2}}
=\int_0^1 f(x)\ \,\dee{x}
\end{equation*}
Find \(f(x)\text{.}\)
23. (✳).
Express \(\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^{n}
\frac{3}{n} e^{-i/n} \cos\left(\frac{3i}{n}\right)\) as a definite integral.
24. (✳).
Let \(\displaystyle R_n= \sum_{i=1}^{n}
\frac{i e^{i/n}}{n^2}\text{.}\) Express \(\displaystyle\lim_{n\to\infty}R_n\) as a definite integral. Do not evaluate this integral.
25. (✳).
Express \(\displaystyle\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2i/n}\cdot \frac{2}{n} \bigg)\) as an integral in three different ways.
26.
Evaluate the sum \(1+r^3+r^6+r^9+\cdots+r^{3n}\text{.}\)
27.
Evaluate the sum \(r^5+r^6+r^7+\cdots+r^{100}\text{.}\)
28. (✳).
Evaluate \({\displaystyle\int_{-1}^2 |2x|\ \,\dee{x}}\text{.}\)
29.
Evaluate the following integral by interpreting it as a signed area, and using geometry:
\begin{equation*}
\int_{-3}^5 |t-1| \,\dee{t}
\end{equation*}
30.
Evaluate the following integral by interpreting it as a signed area, and using geometry:
\begin{equation*}
\int_a^b x \,\dee{x}
\end{equation*}
where \(0 \leq a \leq b\text{.}\)
31.
Evaluate the following integral by interpreting it as a signed area, and using geometry:
\begin{equation*}
\int_a^b x\, \dee{x}
\end{equation*}
where \(a \leq b \leq 0\text{.}\)
32.
Evaluate the following integral by interpreting it as a signed area, and using geometry:
\begin{equation*}
\int_0^4 \sqrt{16-x^2} \dee{x}
\end{equation*}
33. (✳).
Use elementary geometry to calculate \(\displaystyle \int_0^3 f(x)\,\,\dee{x}\text{,}\) where
\begin{align*}
f(x) = \begin{cases}
x, & \text{if } x \le 1,\\
1, & \text{if } x \gt 1.
\end{cases}
\end{align*}
34. (✳).
A car's gas pedal is applied at \(t=0\) seconds and the car accelerates continuously until \(t=2\) seconds. The car's speed at half-second intervals is given in the table below. Find the best possible upper estimate for the distance that the car traveled during these two seconds.
| \(t\) (s) | \(0\) | \(0.5\) | \(1.0\) | \(1.5\) | \(2\) |
| \(v\) (m/s) | 0 | 14 | 22 | 30 | 40 |
35.
True or false: the answer you gave for Question 34 is definitely greater than or equal to the distance the car travelled during the two seconds in question.
36.
An airplane's speed at one-hour intervals is given in the table below. Approximate the distance travelled by the airplane from noon to 4pm using a midpoint Riemann sum.
| time | 12:00 pm | 1:00 pm | 2:00 pm | 3:00 pm | 4:00 pm |
| speed (km/hr) | 800 | 700 | 850 | 900 | 750 |
Exercises — Stage 3 .
37. (✳).
(a) Express
\begin{equation*}
\lim_{n\rightarrow\infty}
\sum_{i=1}^n\frac{2}{n}\sqrt{4-\left(-2+\frac{2i}{n}\right)^2}
\end{equation*}
as a definite integal.
(b) Evaluate the integral of part (a).
38. (✳).
Consider the integral:
\begin{gather*}
\int_0^3 (7 + x^3) \,\,\dee{x} .\qquad\qquad (*)
\end{gather*}
- Approximate this integral using the left Riemann sum with \(n=3\) intervals.
- Write down the expression for the right Riemann sum with \(n\) intervals and calculate the sum. Now take the limit \(n \to \infty\) in your expression for the Riemann sum, to evaluate the integral (\(*\)) exactly.
You may use the identity
\begin{gather*}
\sum_{i=1}^{n} i^3 = \frac{n^4 +2n^3 + n^2}{4}
\end{gather*}
39. (✳).
Using a limit of right--endpoint Riemann sums, evaluate \(\displaystyle\int_2^4 x^2\ \,\dee{x}\text{.}\) You may use the formulas \(\sum\limits_{i=1}^n i = \frac{n(n + 1)}{2}\) and \(\sum\limits_{i=1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}\text{.}\)
40. (✳).
Find \(\displaystyle\int_0^2 (x^3+x)\,\,\dee{x}\) using the definition of the definite integral. You may use the summation formulas \(\sum\limits_{i=1}^{n}i^3 = \frac{n^4+2n^3+n^2}4\) and \(\sum\limits_{i=1}^{n} i = \frac{n^2+n}{2}\text{.}\)
41. (✳).
Using a limit of right-endpoint Riemann sums, evaluate \(\displaystyle\int_1^4 (2x-1)\,\,\dee{x}\text{.}\) Do not use anti-differentiation, except to check your answer. 25 You may use the formula \(\sum\limits_{i=1}^{n} i = \frac{n(n+1)}{2}\text{.}\)
42.
Give a function \(f(x)\) that has the following expression as a right Riemann sum when \(n=10\text{,}\) \(\Delta(x)=10\) and \(a=-5\text{:}\)
\begin{equation*}
\sum_{i=1}^{10} 3(7+2i)^2\sin(4i)\,.
\end{equation*}
43.
Using the method of Example 1.1.2, evaluate
\begin{equation*}
\int_0^1 2^x \dee{x}
\end{equation*}
44.
- Using the method of Example 1.1.2, evaluate\begin{equation*} \int_a^b 10^x \dee{x} \end{equation*}
Using your answer from above, make a guess for
\begin{equation*}
\int_a^b c^x \dee{x}
\end{equation*}
where \(c\) is a positive constant. Does this agree with Question 43?
45.
Evaluate \(\displaystyle\int_0^a \sqrt{1-x^2}\dee{x}\) using geometry, if \(0 \leq a \leq 1\text{.}\)
46.
Suppose \(f(x)\) is a positive, decreasing function from \(x=a\) to \(x=b\text{.}\) You give an upper and lower bound on the area under the curve \(y=f(x)\) using \(n\) rectangles and a left and right Riemann sum, respectively, as in the picture below.
- What is the difference between the lower bound and the upper bound? (That is, if we subtract the smaller estimate from the larger estimate, what do we get?) Give your answer in terms of \(f\text{,}\) \(a\text{,}\) \(b\text{,}\) and \(n\text{.}\)
- If you want to approximate the area under the curve to within 0.01 square units using this method, how many rectangles should you use? That is, what should \(n\) be?
47.
Let \(f(x)\) be a linear function, let \(a \lt b\) be integers, and let \(n\) be a whole number. True or false: if we average the left and right Riemann sums for \(\displaystyle\int_a^b f(x)\dee{x}\) using \(n\) rectangles, we get the same value as the midpoint Riemann sum using \(n\) rectangles.
This should remind the reader of the approach taken to compute the slope of a tangent line way way back at the start of differential calculus.
Approximating the area in this way leads to a definition of integration that is called Riemann integration. This is the most commonly used approach to integration. However we could also approximate the area by using long thin horizontal strips. This leads to a definition of integration that is called Lebesgue integration. We will not be covering Lebesgue integration in these notes.
If we want to be more careful here, we should construct two approximations, one that is always a little smaller than the desired area and one that is a little larger. We can then take a limit using the Squeeze Theorem and arrive at the exact area. More on this later.
Notice that since \(e^x\) is an increasing function, this choice of heights means that each of our rectangles is smaller than the strip it came from.
We haven't proved that this will give us the exact area, but it should be clear that taking this limit will give us a lower bound on the area. To complete things rigorously we also need an upper bound and the squeeze theorem. We do this in the next optional subsection.
If you do not recall L'Hôpital's rule and indeterminate forms then we recommend you skim over your differential calculus notes on the topic.
Say if you don't recall l'Hôpital's rule and have not had time to revise it.
gcc.gnu.org/ml/gcc/2001-10/msg00610.htmlwww.economist.com/news/books-and-arts/21723088-hyphens-can-be-tricky-they-need-not-drive-you-crazy-hysteria-over-hyphensRecall that if we have 3 functions \(f(x), g(x), h(x)\) that satisfy \(f(x) \leq g(x) \leq h(x)\) and we know that \(\lim_{x \to a} f(x) = \lim_{x\to a} h(x) = L\) exists and is finite, then the squeeze theorem tells us that \(\lim_{x\to a} g(x) = L\text{.}\)
Some careful addition shows it is \(\frac{46181}{176400}\text{.}\)
Since all the sums are finite, this isn't too hard. More care must be taken when the sums involve an infinite number of terms. We will examine this in Chapter 3.
Of course, any finite sum can be computed exactly — just sum together the terms. What we mean by “computed exactly” in this context, is that we can rewrite the sum as a simple, and easily evaluated, formula involving the terminals of the sum. For example \(\sum_{k=m}^n r^k = \frac{r^{n+1}-r^m}{r-1}\) provided \(r\neq1\text{.}\) No matter what finite integers we choose for \(m\) and \(n\text{,}\) we can quickly compute the sum in just a few arithmetic operations. On the other hand, the sums, \(\sum_{k=m}^n \frac{1}{k}\) and \(\sum_{k=m}^n \frac{1}{k^2}\text{,}\) cannot be expressed in such clean formulas (though you can rewrite them quite cleanly using integrals). To explain more clearly we would need to go into a more detailed and careful discussion that is beyond the scope of this course.
Generating functions are frequently used in mathematics to analyse sequences and series, but are beyond the scope of the course. The interested reader should take a look at “Generatingfunctionology” by Herb Wilf. It is an excellent book and is also free to download.
\(a\) and \(b\) are also called the bounds of integration.
We'll eventually allow \(a\) and \(b\) to be any two real numbers, not even requiring \(a \lt b\text{.}\) But it is easier to start off assuming \(a \lt b\text{,}\) and that's what we'll do.
Bernhard Riemann was a 19th century German mathematician who made extremely important contributions to many different areas of mathematics — far too many to list here. Arguably two of the most important (after Riemann sums) are now called Riemann surfaces and the Riemann hypothesis (he didn't name them after himself).
When a function is decreasing the situation is reversed — the left Riemann sum is always larger than the integral while the right Riemann sum is smaller than the integral. For more general functions that both increase and decrease it is perhaps easiest to study each increasing (or decreasing) interval separately.
If this were the only interpretation then integrals would be a nice mathematical curiousity and unlikely to be the core topic of a large first year mathematics course.
We'll be more precise about what “reasonable” means shortly.
Again, we'll explain this “reasonable” shortly
Here we are using the extreme value theorem — its proof is beyond the scope of this course. The theorem says that any continuous function on a closed interval must attain a minimum and maximum at least once. In this situation this implies that for any continuous function \(f(x)\text{,}\) there are \(x_{j-1}\le {\overline x}_j, {\underline x}_j\le x_j\) such that \(f({\underline x}_j)\le f(x) \le f({\overline x}_j)\) for all \(x_{j-1}\le x\le x_j\text{.}\)
Recall that the mean value theorem states that for a function continuous on \([a,b]\) and differentiable on \((a,b)\text{,}\) there exists a number \(c\) between \(a\) and \(b\) so that \(f'(c) = \frac{f(b)-f(a)}{b-a}.\)
You'll learn about this method starting in Section 1.3. You can also check this answer using geometry.
