Pure Mathematics Practice: Integration — Answers

Questions: Integration Drills.

Question 1

(a) Let $u=x^2$, so $du=2xdx$.

$${\int }_0^{1}x{e}^{x^2}dx=\frac{1}{2}{\int }_0^1{e}^{u}du=\frac{1}{2}(e-1).$$

(b) By integration by parts,

$$\int x\ln xdx=\frac{x^2}{2}\ln x-\frac{x^2}{4}.$$

Therefore

$${\int }_1^{e}x\ln xdx={\left[\frac{x^2}{2}\ln x-\frac{x^2}{4}\right]}_1^e=\frac{e^2+1}{4}.$$

(c) Since

$$2x=(2x+4)-4$$

and

$$x^2+4x+5=(x+2{)}^2+1,$$

we obtain

$$\int \frac{2x}{x^2+4x+5}dx=\ln (x^2+4x+5)-4{\tan }^{-1}(x+2)+C.$$

Question 2

(a) At an intersection,

$$x(4-x)=x.$$

Thus $x(3-x)=0$, giving $x=0$ or $x=3$. The intersection points are

$$(0,0)\text{and}(3,3).$$

(b) For $0<x<3$,

$$x(4-x)-x=x(3-x)>0,$$

so $y=x(4-x)$ lies above $y=x$. The required area is

$${\int }_0^3[x(4-x)-x]dx.$$

(c)

$${\int }_0^3(3x-x^2)dx={\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]}_0^3=\frac{9}{2}.$$

Question 3

(a) Put $u=x+1$, so the limits become $u=1$ and $u=e$:

$${\int }_0^{e-1}(x+1)\ln (x+1)dx={\int }_1^{e}u\ln udu.$$

Using integration by parts,

$$\int u\ln udu=\frac{u^2}{2}\ln u-\frac{u^2}{4}.$$

Hence

$${\int }_1^{e}u\ln udu={\left[\frac{u^2}{2}\ln u-\frac{u^2}{4}\right]}_1^e=\frac{e^2+1}{4}.$$

(b) A cross-section perpendicular to the $x$-axis is a disc of radius $y$, so

$$V=\pi {\int }_0^{e-1}{y}^{2}dx=\pi {\int }_0^{e-1}(x+1)\ln (x+1)dx.$$

Using part (a),

$$V=\frac{\pi (e^2+1)}{4}.$$

(c) Rotating an ordinate of length $y$ about the $x$-axis forms a circular disc of radius $y$. Its cross-sectional area is $\pi y^2$, so the volume integrand uses $y^2$, not $y$.

Question 4

(a) Since

$$3x+5=\frac{3}{2}(2x+2)+2$$

and $x^2+2x+5=(x+1{)}^2+4$,

$$\int \frac{3x+5}{x^2+2x+5}dx=\frac{3}{2}\ln (x^2+2x+5)+{\tan }^{-1}\left(\frac{x+1}{2}\right)+C.$$

(b) Let $u=1+\tan x$, so $du={\sec }^{2}xdx$. The limits are $u=1$ and $u=2$:

$${\int }_0^{\pi /4}\frac{{\sec }^{2}x}{1+\tan x}dx={\int }_1^2\frac{1}{u}du=\ln 2.$$

(c) The signed integral is

$${\int }_0^3(x^2-2x)dx={\left[\frac{x^3}{3}-x^2\right]}_0^3=0.$$

For geometric area, split at $x=2$:

$$\begin{array}{rl}A& ={\int }_0^2(2x-x^2)dx+{\int }_2^3(x^2-2x)dx\\ & =\frac{4}{3}+\frac{4}{3}\\ & =\frac{8}{3}.\end{array}$$