Integration

Unit 4 Integration

Key unit competence: Apply integration as the inverse of differentiation and as the limit of a sum in determining areas, volumes and other quantities.

Key ideas

• Differentials: $dy=f^{\prime }(x)dx$, so $\Delta y\approx f^{\prime }(x)\Delta x$ for small $\Delta x$.
• $F$ is an antiderivative of $f$ if $F^{\prime }=f$; $\int f(x)dx=F(x)+c$. The integral is linear: $\int (af+bg)=a\int f+b\int g$.
• Basic integrals: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$ ($n\ne -1$); $\int \frac{dx}{x}=\ln |x|+c$; $\int e^{ax}dx=\frac{e^{ax}}{a}+c$; $\int \sin xdx=-\cos x+c$; $\int \cos xdx=\sin x+c$; $\int \frac{dx}{1+x^2}=\arctan x+c$; $\int \frac{dx}{\sqrt{1-x^2}}=\arcsin x+c$.
• Techniques: substitution; $\int \frac{u^{\prime }}{u}=\ln |u|+c$; long division then partial fractions for rational functions; trigonometric powers (use ${\sin }^{2}x=\frac{1-\cos 2x}{2}$, or save one factor); $t=\tan \frac{x}{2}$; integration by parts $\int udv=uv-\int vdu$; reduction formulae.
• Definite integral: ${\int }_a^{b}f(x)dx=F(b)-F(a)$. Properties: ${\int }_a^b=-{\int }_b^a$; Chasles ${\int }_a^c+{\int }_c^b={\int }_a^b$; if $f\ge 0$ on $[a,b]$ then ${\int }_a^{b}f\ge 0$.
• Area between curves: ${\int }_a^b|f(x)-g(x)|dx$. Volumes of revolution about the $x$-axis: discs $V=\pi {\int }_a^b{f}^{2}dx$; washers $V=\pi {\int }_a^b(R^2-r^2)dx$; shells about the $y$-axis $V=2\pi {\int }_a^{b}xh(x)dx$.
• Arc length $L={\int }_a^b\sqrt{1+(f^{\prime }(x){)}^2}dx$. Work by a variable force $W={\int }_a^{b}F(x)dx$ (spring: $F=kx$).

Formulas to know

Common misconceptions

• Forgetting the constant $c$ in indefinite integrals.
• "$\int fg=\int f\cdot \int g$." No: use integration by parts or substitution.
• Counting area below the $x$-axis as negative: split the interval where $f$ changes sign.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Find $\int (3x^2-4x+5)dx$.

   Show answer

   $x^3-2x^2+5x+c$.

   Common mistake: Forgetting $+c$.

2. The region under $y=x^2$, $0\le x\le 2$, is rotated about the $y$-axis. Use shells to find the volume.

   Show answer

   $V=2\pi {\int }_0^{2}x\cdot x^{2}dx=2\pi {\left[\frac{x^4}{4}\right]}_0^2=8\pi$.

   Common mistake: Using the disc formula about the wrong axis.

3. Find $\int x\sqrt{x^2+1}dx$.

   Show answer

   $u=x^2+1$, $du=2xdx$: $\frac{1}{2}\int \sqrt{u}du=\frac{1}{3}(x^2+1{)}^{3/2}+c$.

   Common mistake: Not adjusting for the factor 2 in $du$.

4. The region between $y=\sqrt{x}$ and $y=1$ for $1\le x\le 4$ is rotated about the $x$-axis. Find the volume.

   Show answer

   Washers: $\pi {\int }_1^4(x-1)dx=\pi {\left[\frac{x^2}{2}-x\right]}_1^4=\pi \left(4+\frac{1}{2}\right)=\frac{9\pi }{2}$.

   Common mistake: Using $(R-r{)}^2$ instead of $R^2-r^2$.

5. Find $\int \frac{3x+5}{(x+1)(x+2)}dx$.

   Show answer

   $\frac{3x+5}{(x+1)(x+2)}=\frac{2}{x+1}+\frac{1}{x+2}$, so the integral is $2\ln |x+1|+\ln |x+2|+c$.

   Common mistake: Errors when substituting $x=-1$ and $x=-2$ to find the constants.

6. Find the area enclosed by $y=x^2-4$ and the $x$-axis.

   Show answer

   Roots $\pm 2$; the curve is below the axis: ${\int }_{-2}^2(4-x^2)dx=16-\frac{16}{3}=\frac{32}{3}$.

   Common mistake: Giving the negative value $-\frac{32}{3}$.

7. Find $\int x^2{e}^{x}dx$.

   Show answer

   Parts twice: $x^2{e}^x-2\int x{e}^{x}dx=x^2{e}^x-2(x-1)e^x+c=e^x(x^2-2x+2)+c$.

   Common mistake: Sign error in the second integration by parts.

8. Find $\int {\sin }^{3}x{\cos }^{2}xdx$.

   Show answer

   ${\sin }^{3}x{\cos }^{2}x=(1-{\cos }^{2}x){\cos }^{2}x\sin x$. With $u=\cos x$: $-\int (u^2-u^4)du=-\frac{{\cos }^{3}x}{3}+\frac{{\cos }^{5}x}{5}+c$.

   Common mistake: Forgetting the minus sign from $du=-\sin xdx$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find $\int \tan xdx$.	a. $\frac{1}{2}\arctan \frac{x}{2}+c$
   2. Find the area enclosed by $y=x^2$ and $y=x+2$.	b. $\frac{1}{2}{e}^{x^2}+c$
   3. Find $\int \frac{dx}{x^2+4}$.	c. $-\ln |\cos x|+c$
   4. Find $\int x{e}^{x^2}dx$.	d. $\frac{9}{2}$

   Show answer

   1 → c ($-\ln |\cos x|+c$) · 2 → d ($\frac{9}{2}$) · 3 → a ($\frac{1}{2}\arctan \frac{x}{2}+c$) · 4 → b ($\frac{1}{2}{e}^{x^2}+c$)

   • Step 1. $\tan x=\frac{\sin x}{\cos x}$; let $u=\cos x$, $du=-\sin xdx$.
     Step 2. $-\int \frac{du}{u}=-\ln |\cos x|+c$.
   • Step 1. $x^2=x+2\Rightarrow x=-1$ or $x=2$.
     Step 2. ${\int }_{-1}^2(x+2-x^2)dx={\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}_{-1}^2=\frac{10}{3}+\frac{7}{6}=\frac{9}{2}$.
   • Step 1. $\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan \frac{x}{a}+c$.
     Step 2. $a=2$.
   • Step 1. Let $u=x^2$, $du=2xdx$.
     Step 2. $\int \frac{1}{2}{e}^{u}du=\frac{1}{2}{e}^{x^2}+c$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $4\pi$  ·  1 J  ·  $8\pi$  ·  $\frac{14}{3}$
   1. The length of $y=\frac{2}{3}{x}^{3/2}$ from $x=0$ to $x=3$ is .
   2. The region under $y=\sqrt{x}$, $0\le x\le 4$, is rotated about the $x$-axis. The volume is .
   3. A spring has $k=200$ N/m. The work to stretch it 0.1 m from its natural length is .
   Show answer

   1: $\frac{14}{3}$ · 2: $8\pi$ · 3: 1 J

   • Step 1. $y^{\prime }=\sqrt{x}$, so $L={\int }_0^3\sqrt{1+x}dx$.
     Step 2. ${\left[\frac{2}{3}(1+x{)}^{3/2}\right]}_0^3=\frac{2}{3}(8-1)=\frac{14}{3}$.
   • Step 1. $V=\pi {\int }_0^4(\sqrt{x}{)}^{2}dx=\pi {\int }_0^{4}xdx$.
     Step 2. $\pi {\left[\frac{x^2}{2}\right]}_0^4=8\pi$.
   • Step 1. $W={\int }_0^{0.1}200xdx={\left[100x^2\right]}_0^{0.1}$.
     Step 2. $100\times 0.01=1$ J.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Evaluate ${\int }_0^{\pi /2}\sin x\cos xdx$.	a. $\arcsin \frac{x}{3}+c$
   2. Find $\int x{e}^{x}dx$.	b. $\frac{1}{2}$
   3. Find $\int {\cos }^{2}xdx$.	c. $\frac{x}{2}+\frac{\sin 2x}{4}+c$
   4. Find $\int \frac{dx}{\sqrt{9-x^2}}$.	d. $(x-1)e^x+c$

   Show answer

   1 → b ($\frac{1}{2}$) · 2 → d ($(x-1)e^x+c$) · 3 → c ($\frac{x}{2}+\frac{\sin 2x}{4}+c$) · 4 → a ($\arcsin \frac{x}{3}+c$)

   • Step 1. Let $u=\sin x$, $du=\cos xdx$; $u$ goes from 0 to 1.
     Step 2. ${\int }_0^{1}udu=\frac{1}{2}$.
   • Step 1. Parts: $u=x$, $dv=e^{x}dx$, so $du=dx$, $v=e^x$.
     Step 2. $x{e}^x-\int e^{x}dx=x{e}^x-e^x+c$.
   • Step 1. ${\cos }^{2}x=\frac{1+\cos 2x}{2}$.
     Step 2. Integrate: $\frac{x}{2}+\frac{\sin 2x}{4}+c$.
   • Step 1. $\int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin \frac{x}{a}+c$.
     Step 2. $a=3$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $1$  ·  $\sqrt{2}-1$  ·  $\ln 2$  ·  $\frac{4}{3}$
   1. The area enclosed by $y=x^2$ and $y=2x$ is .
   2. ${\int }_0^1\frac{x}{\sqrt{1+x^2}}dx=$ →
   3. Evaluate ${\int }_1^e\ln xdx$. →
   Show answer

   1: $\frac{4}{3}$ · 2: $\sqrt{2}-1$ · 3: $1$

   • Step 1. $x^2=2x$ at $x=0$ and $x=2$; between them $2x\ge x^2$.
     Step 2. ${\int }_0^2(2x-x^2)dx={\left[x^2-\frac{x^3}{3}\right]}_0^2=4-\frac{8}{3}=\frac{4}{3}$.
   • Step 1. Let $u=1+x^2$, $du=2xdx$: antiderivative $\sqrt{1+x^2}$.
     Step 2. $\sqrt{2}-\sqrt{1}=\sqrt{2}-1$.
   • Step 1. Parts: $\int \ln xdx=x\ln x-x$.
     Step 2. $(e-e)-(0-1)=1$.

Take the Unit 4 quiz → · Back to contents