5 (✳)
Suppose \(\displaystyle\int_2^3 f(x)\,\dee{x} = -1\) and \(\displaystyle\int_2^3 g(x)\,\dee{x} = 5\text{.}\) Evaluate \(\displaystyle \int_2^3 \big( 6 f(x) - 3 g(x) \big)\,\dee{x}\text{.}\)
6 (✳)
If \(\displaystyle\int_0^2 f(x)\,\dee{x} = 3\) and \(\displaystyle\int_0^2 g(x)\,\dee{x} = -4\text{,}\) calculate \(\displaystyle \int_0^2 \big( 2 f(x) + 3 g(x) \big)\,\dee{x}\text{.}\)
7 (✳)
The functions \(f(x)\) and \(g(x)\) obey
\begin{align*}
\int_0^{-1} f(x)\,\dee{x} &= 1 &
\int_0^2 f(x)\,\dee{x} &= 2 \\
\int_{-1}^0 g(x)\,\dee{x} &= 3 &
\int_0^2 g(x)\,\dee{x} &= 4
\end{align*}
Find \(\int_{-1}^2 \big[3g(x)-f(x)\big]\,\dee{x}\text{.}\)
8
In Question 1.1.8.45, Section 1.1, we found that
\begin{equation*}
\int_0^a\sqrt{1-x^2}\dee{x}=\frac{\pi}{4} - \frac{1}{2}\arccos(a)+\frac{1}{2}a\sqrt{1-a^2}
\end{equation*}
when \(0\le a\le 1\text{.}\)
Using this fact, evaluate the following:
- \(\displaystyle\int_{a}^0 \sqrt{1-x^2}\dee{x}\text{,}\) where \(-1 \leq a \leq 0\)
- \(\displaystyle\int_{a}^1 \sqrt{1-x^2}\dee{x}\text{,}\) where \(0 \leq a \leq 1\)
9 (✳)
Evaluate \({\displaystyle\int_{-1}^2 |2x|\dee{x}}\text{.}\)
You may use the result from Example 1.2.6 that \(\int\limits_a^b x\dee{x}=\frac{b^2-a^2}{2}
\text{.}\)
10
Evaluate \(\displaystyle\int_{-5}^5 x|x|\dee{x}\,.\)
11
Suppose \(f(x)\) is an even function and \(\displaystyle\int_{-2}^2 f(x)\dee{x}=10\text{.}\) What is \(\displaystyle\int_{-2}^0 f(x)\dee{x}\text{?}\)