Unit 01 - Essential problems

Shells

01

Shells volume - offset graph, $y$-axis

Consider the region in the first quadrant bounded by the lines $x=0$, $x=2$, $y=0$, and the curve $y=\frac{1}{\sqrt{x^2+1}}$. Revolve this about the $y$-axis.

Find the volume of the resulting solid.

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Solution

Solutions - 0030-01

(1) Set up integral.

$$\begin{array}{cc}V& ={\int }_a^{b}2\pi Rhdr\\ & \\ & \gg \gg {\int }_0^{2}2\pi x\left(\frac{1}{\sqrt{x^2+1}}\right)dx\end{array}$$

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(2) Perform $u$-substitution with $u=x^2+1$ and $du=2x$:

$$\begin{array}{c}\gg \gg V={\int }_0^5\frac{\pi }{\sqrt{u}}du\end{array}$$

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(3) Integrate with power rule:

$$\begin{array}{c}\gg \gg \pi {\frac{u^{1/2}}{1/2}|}_0^5\\ \\ \gg \gg 2\pi (\sqrt{5}-1)\end{array}$$Link to original

IBP

02

Integration by parts - A and T

Compute the integral:

$$\int x^2\sin xdx$$

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Solution

Solutions - 0010-02

(1) Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u=x^2& v^{\prime }=\sin x\\ u^{\prime }=2x& v=-\cos x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }\text{original}\\ ⟶\int u^{\prime }\cdot v\text{final}\end{array}$$

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(2) Apply IBP formula $uv-\int u^{\prime }\cdot v$:

$$\begin{array}{cccc}& uv-\int u^{\prime }\cdot v& \gg \gg x^2(-\cos x)-\int 2x(-\cos x)dx& \\ & & & \\ & & \gg \gg -x^2\cos x+2\int x\cos xdx& \text{(Exp. A)}\end{array}$$

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(3) Select another $u=x$ and $v^{\prime }=\cos x$ and do IBP again:

$$\begin{array}{cc}\int x\cos x& \gg \gg x\sin x-\int \sin xdx\\ & \\ & \gg \gg x\sin x+\cos x+C\end{array}$$

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(4) Put all together in (A):

$$\begin{array}{ccc}& \gg \gg -x^2\cos x+2(x\sin x+\cos x+C)& \\ & & \\ & \gg \gg -x^2\cos x+2x\sin x+2\cos x+C& \text{(Note B)}\end{array}$$

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Note B: We can change notation $2C$ to $C$ because the value of $C$ is arbitrary.

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03

Integration by parts - A and L

Compute the integral:

$$\int x^3\ln xdx$$

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Solution

Solutions - 0010-03

(1) Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u=\ln x& v^{\prime }=x^3\\ u^{\prime }={\displaystyle \frac{1}{x}}& v={\displaystyle \frac{x^4}{4}}\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }=\int x^3\ln x\text{original}\\ ⟶\int u^{\prime }\cdot v=\int {\displaystyle \frac{x^3}{4}}\text{final}\end{array}$$

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(2) Apply IBP formula $uv-\int u^{\prime }\cdot v$:

$$\begin{array}{cc}uv-\int u^{\prime }\cdot v& \gg \gg \frac{x^4}{4}\ln x-\int \frac{x^3}{4}dx\\ & \\ & \gg \gg \frac{x^4}{4}\ln x-\frac{x^4}{16}+C\end{array}$$Link to original

05

Integration by parts - A and I

Compute the integral:

$$\int {\tan }^{-1}(x)dx$$

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Solution

Solutions - 0010-05

(1) Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u={\tan }^{-1}x& v^{\prime }=1\\ u^{\prime }={\displaystyle \frac{1}{1+x^2}}& v=x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }=\int {\tan }^{-1}x\text{original}\\ ⟶\int u^{\prime }\cdot v=\int {\displaystyle \frac{x}{1+x^2}}\text{final}\end{array}$$

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(2) Apply IBP formula $uv-\int u^{\prime }\cdot v$ and compute integral:

$$\begin{array}{cccc}& uv-\int u^{\prime }\cdot v& \gg \gg x{\tan }^{-1}x-\int \frac{x}{1+x^2}dx& \text{(Exp. A)}\end{array}$$

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(3) Perform $u$-sub with $u=1+x^2$ and $du=2xdx$:

$$\begin{array}{cc}\int \frac{x}{1+x^2}dx& \gg \gg \frac{1}{2}\int \frac{2xdx}{1+x^2}\\ & \\ & \gg \gg \frac{1}{2}\int \frac{du}{u}\\ & \\ & \gg \gg \frac{1}{2}\ln |u|+C\\ & \\ & \gg \gg \frac{1}{2}\ln |1+x^2|+C\end{array}$$

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(4) Insert result in Exp. (A):

$$\begin{array}{ccc}& (A)\gg \gg x{\tan }^{-1}x-\frac{1}{2}\ln (1+x^2)+C& \text{(Note B)}\end{array}$$

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Note B: We can change $|1+x^2|$ to $(1+x^2)$ because the inner expression is never negative.

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Trig power products

01

Somewhat odd power product

Compute the integral:

$$\int {\sin }^{2}x\cdot {\cos }^{3}xdx$$

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Solution

Solutions - 0040-01

(1) Notice odd power on $\cos x$. Swap the even bunch:

$${\sin }^{2}x\cdot {\cos }^{3}x\gg \gg {\sin }^{2}x(1-{\sin }^{2}x)\cos x$$

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(2) Integrate with $u$-sub setting $u=\sin x$ and thus $du=\cos xdx$:

$$\begin{array}{cc}\int {\sin }^{2}x(1-{\sin }^{2}x)\cos xdx& \gg \gg \int u^2(1-u^2)du\\ & \\ & \gg \gg \frac{u^3}{3}-\frac{u^5}{5}+C\\ & \\ & \gg \gg \frac{{\sin }^{3}x}{3}-\frac{{\sin }^{5}x}{5}+C\end{array}$$Link to original

02

Tangent and secant both even

Compute the integral:

$$\int {\tan }^{2}x\cdot {\sec }^{2}xdx$$

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Solution

Solutions - 0040-02

Notice ${\sec }^{2}xdx$. Therefore integrate with $u$-sub setting $u=\tan x$ and $du={\sec }^{2}xdx$:

$$\begin{array}{cc}\int {\tan }^{2}x\cdot {\sec }^{2}xdx& \gg \gg \int u^{2}du\\ & \\ & \gg \gg \frac{u^3}{3}+C\\ & \\ & \gg \gg \frac{{\tan }^{3}x}{3}+C\end{array}$$Link to original

05

Tangent and secant mixed parity

Compute the integral:

$$\int {\tan }^{3}x{\sec }^{2}xdx$$

(a) Using $du={\sec }^{2}xdx$.

(b) Using $du=\sec x\tan xdx$.

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Solution

Solutions - 0040-05

(a) Select $u=\tan x$ and thus $du={\sec }^{2}xdx$:

$$\begin{array}{cc}\int {\tan }^{3}x{\sec }^{2}xdx& \gg \gg \int u^{3}du\\ & \\ & \gg \gg \frac{u^4}{4}+C\\ & \\ & \gg \gg \frac{{\tan }^{4}x}{4}+C\end{array}$$

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(b)

(1) Select $u=\sec x$ and thus $du=\sec x\tan xdx$:

$$\int {\tan }^{3}x{\sec }^{2}xdx\gg \gg \int {\tan }^{2}x\sec x(\sec x\tan x)dx$$

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(3) Swap even bunch using $1+{\tan }^{2}x={\sec }^{2}x$:

$$\gg \gg \int ({\sec }^{2}x-1)\sec x(\sec x\tan x)dx$$

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(4) Perform $u$-sub with $u=\sec x$ and integrate:

$$\begin{array}{cc}& \gg \gg \int (u^2-1)udu\\ & \\ & \gg \gg \frac{u^4}{4}-\frac{u^2}{2}+C\\ & \\ & \gg \gg \frac{{\sec }^{4}x}{4}-\frac{{\sec }^{2}x}{2}+C\end{array}$$Link to original

Trig subs

01

Trig sub

Compute the definite integral:

$${\int }_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}dx$$

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Solution

Solutions - 0050-01

(1) Substitute $x=\sin \theta$ and thus $dx=\cos \theta d\theta$. Adjust the bounds as follows:

$$\begin{array}{cc}x& =0\gg \gg \theta =0\\ x& =\frac{1}{2}\gg \gg \theta =\frac{\pi }{6}\end{array}$$

Rewrite the integral:

$$\begin{array}{cccc}& {\int }_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}dx& \gg \gg {\int }_0^{\pi /6}\frac{{\sin }^2\theta \cos \theta }{\sqrt{1-{\sin }^2\theta }}d\theta & \\ & & & \\ & & \gg \gg {\int }_0^{\pi /6}{\sin }^2\theta d\theta & \text{(Note A)}\end{array}$$

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(2) Use power-to-frequency conversion:

$$\begin{array}{cc}{\int }_0^{\pi /6}{\sin }^2\theta d\theta & \gg \gg \frac{1}{2}{\int }_0^{\pi /6}1-\cos 2\theta d\theta \\ & \\ & \gg \gg \frac{1}{2}{(\theta -\frac{1}{2}\sin 2\theta )|}_0^{\pi /6}\\ & \\ & \gg \gg \frac{\pi }{12}-\frac{\sin (\frac{\pi }{3})}{4}-\frac{1}{2}(0-0)\\ & \\ & \gg \gg \frac{\pi }{12}-\frac{\sqrt{3}}{8}\end{array}$$

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Note A: Use $1-{\sin }^2\theta ={\cos }^2\theta$, then $\sqrt{{\cos }^2\theta }=|\cos \theta |$ and this equals $\cos \theta$ for $0\le \theta \le \pi /6$.

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03

Trig sub

Compute the integral:

$$\int \frac{dx}{x^3\sqrt{x^2-4}}$$

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Solution

Solutions - 0050-03

(1) Notice $x^2-4$ pattern, so we should make use of the identity ${\sec }^{2}x-1={\tan }^{2}x$.

Select $x=2\sec \theta$ and thus $dx=2\sec \theta \tan \theta d\theta$. Then:

$$x^2-4\gg \gg 4{\sec }^2\theta -4\gg \gg 4{\tan }^{2}x$$

Plug in and simplify:

$$\begin{array}{c}\int \frac{dx}{x^3\sqrt{x^2-4}}\gg \gg \int \frac{2\sec \theta \tan \theta }{8{\sec }^3\theta |2\tan \theta |}d\theta \\ \\ \gg \gg \frac{1}{8}\int {\cos }^2\theta d\theta \end{array}$$

(We must assume that $\tan \theta >0$ for the relevant values of $\theta$ here.)

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(2) Use power-to-frequency conversion:

$$\begin{array}{c}\gg \gg \frac{1}{8}\int \frac{1}{2}(1+\cos 2\theta )d\theta \gg \gg \frac{1}{16}(\theta +\frac{1}{2}\sin 2\theta )+C\end{array}$$

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(3) Convert back to terms of $x$:

First draw a triangle expressing $\sec \theta ={\displaystyle \frac{x}{2}}$:

Therefore:

$$\sin \theta =\frac{2}{x},\cos \theta =\frac{\sqrt{x^2-4}}{x}$$

For $\sin 2\theta$, use the double-angle identity:

$$\sin 2\theta =2\sin \theta \cos \theta \gg \gg 2\frac{\sqrt{x^2-4}}{x}\frac{2}{x}$$

Therefore:

$$\begin{array}{c}\frac{1}{16}(\theta +\frac{1}{2}\sin 2\theta )+C\\ \\ \gg \gg \frac{1}{16}({\sec }^{-1}\frac{x}{2}+\frac{2\sqrt{x^2-4}}{x^2})+C\end{array}$$Link to original

05

Trig sub

Compute the integral:

$$\int \frac{x^2}{{(x^2+1)}^{3/2}}dx$$

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Solution

Solutions - 0050-05

(1) Notice $x^2+1$ pattern, so we should make use of the identity ${\tan }^2\theta +1={\sec }^2\theta$.

Select $x=\tan \theta$ and thus $dx={\sec }^2\theta d\theta$. Then:

$${(x^2+1)}^{3/2}\gg \gg {({\sec }^2\theta )}^{3/2}\gg \gg {\sec }^3\theta$$

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(2) Convert to $\theta$ and integrate:

$$\begin{array}{c}\int \frac{x^2}{{(x^2+1)}^{3/2}}dx\gg \gg \int \frac{{\tan }^2\theta {\sec }^2\theta }{{\sec }^3\theta }d\theta \\ \\ \gg \gg \int \frac{{\tan }^2\theta }{\sec \theta }d\theta \gg \gg \int \frac{{\sec }^2\theta -1}{\sec \theta }d\theta \\ \\ \gg \gg \int \sec \theta -\cos \theta d\theta \gg \gg \ln |\tan \theta +\sec \theta |-\sin \theta +C\end{array}$$

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(3) Convert back to terms of $x$:

Draw a triangle expressing $\tan \theta ={\displaystyle \frac{x}{1}}$:

Therefore $\sec \theta =\sqrt{x^2+1}$ and $\sin \theta ={\displaystyle \frac{x}{\sqrt{x^2+1}}}$. Then:

$$\begin{array}{c}\ln |\tan \theta +\sec \theta |-\sin \theta +C\\ \\ \gg \gg \ln |x+\sqrt{x^2+1}|-\frac{x}{\sqrt{x^2+1}}+C\end{array}$$Link to original

Partial fractions

01

Distinct linear factors

Compute the integral:

$$\int \frac{1}{(x+2)(x-3)}dx$$

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Solution

Solutions - 0060-01

(1) Write the partial fractions general form equation:

$$\frac{1}{(x+2)(x-3)}=\frac{A}{x+2}+\frac{B}{x-3}$$

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(2) Solve for constants.

Cross multiply:

$$\gg \gg 1=A(x-3)+B(x+2)$$

Plug in $x=3$, obtain $1=B\cdot 5$ so $B=1/5$.

Plug in $x=-2$, obtain $1=A(-5)$ so $A=-1/5$.

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(3) Integrate each term:

$$\begin{array}{cc}\int \frac{1}{(x+2)(x-3)}dx& \gg \gg \int \frac{-1/5}{x+2}dx+\int \frac{1/5}{x-3}dx\\ & \\ & \gg \gg -\frac{1}{5}\ln |x+2|+\frac{1}{5}\ln |x-3|+C\end{array}$$Link to original

02

Long division first

Compute the integral:

$$\int \frac{2x^3+3x^2+7x+4}{x+1}dx$$

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Solution

Solutions - 0060-02

$$\int \frac{2x^3+3x^2+7x+4}{x+1}dx$$

(1) Numerator degree is not smaller! Long division first:

$$\frac{2x^3+3x^2+7x+4}{x+1}\gg \gg 2x^2+x+6-\frac{2}{x+1}$$

Now this already has the form of a partial fraction decomposition, so we proceed directly to integration.

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(2) Integrate using power rule (with log):

$$\begin{array}{c}\int 2x^2+x+6-\frac{2}{x+1}dx\\ \\ \gg \gg \frac{2}{3}{x}^3+\frac{1}{2}{x}^2+6x-2\ln |x+1|+C\end{array}$$Link to original

07

Partial fractions - linear and quadratic

Compute the integral:

$$\int \frac{5x^2-5x+14}{(x-2)(x^2+4)}dx$$

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Solution

Solutions - 0060-07

(1) Denominator has degree 3, numerator has degree 2, therefore long division is not necessary.

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(2) Write the partial fractions general form equation:

Notice that $x^2+4$ is an irreducible quadratic (cannot be factored). So we have:

$$\frac{5x^2-5x+14}{(x-2)(x^2+4)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+4}$$

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(3) Solve for constants:

Cross multiply:

$$\gg \gg 5x^2-5x+14=A(x^2+4)+(Bx+C)(x-2)$$

Plug in $x=2$, obtain:

$$20-10+14=A(8)\gg \gg 24=8A\gg \gg A=3$$

Expand RHS:

$$\begin{array}{cc}5x^2-5x+14& =3x^2+12+B{x}^2-2Bx+Cx-2C\\ & \\ & =(3+B)x^2+(-2B+C)x+(12-2C)\end{array}$$

Comparing $x^2$ terms, obtain: $5=3+B$ and thus $B=2$.

Comparing constant terms, $C=-1$.

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(4) Integrate by terms:

$$\begin{array}{ccc}& \int \frac{5x^2-5x+14}{(x-2)(x^2+4)}dx\gg \gg \int \frac{3}{x-2}dx+\int \frac{2x-1}{x^2+4}dx& \\ & & \\ & \gg \gg 3\ln |x-2|+\int \frac{2x}{x^2+4}dx-\int \frac{1}{x^2+4}dx& \\ & & \\ & \gg \gg 3\ln |x-2|+\ln |x^2+4|-\frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C& \text{(Note A)}\end{array}$$

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Note A: For the last term, use the formula:

$$\int \frac{dx}{x^2+h^2}=\frac{1}{h}{\tan }^{-1}\left(\frac{x}{h}\right)+C$$Link to original

08

Partial fractions - repeated factor

Compute the integral:

$$\int \frac{1}{x{(x-1)}^3}dx$$

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Solution

Solutions - 0060-08

(1) Write the partial fractions general form equation:

$$\frac{1}{x{(x-1)}^3}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{{(x-1)}^2}+\frac{D}{{(x-1)}^3}$$

Observe that $(x-1)$ appears in degree 3 in the integrand, so we have one term for each power up to 3 in the partial fraction decomposition.

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(2) Solve for constants:

Cross multiply:

$$1=A{(x-1)}^3+Bx{(x-1)}^2+Cx(x-1)+Dx$$

Plug in $x=0$, obtain $1=A(-1)$ so $A=-1$.

Plug in $x=1$, obtain $1=D$.

Plug in $x=2$, obtain:

$$1=(-1)\cdot 1+B\cdot 2+C\cdot 2+1\cdot 2\gg \gg 0=B+C$$

Plug in $x=-1$, obtain:

$$\begin{array}{c}1=(-1)(-8)+B(-1)\cdot 4+C(-1)(-2)+1\cdot (-1)\\ \\ \gg \gg -6=-4B+2C\\ \\ \gg \gg -6=+4C+2C\\ \\ \gg \gg C=-1,B=+1\end{array}$$

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(3) Integrate each term:

$$\begin{array}{c}\int \frac{1}{x{(x-1)}^3}dx\\ \\ \gg \gg \int \frac{-1}{x}dx+\int \frac{1}{x-1}dx+\int \frac{-1}{{(x-1)}^2}dx+\int \frac{1}{{(x-1)}^3}dx\\ \\ \gg \gg -\ln |x|+\ln |x-1|+\frac{1}{x-1}-\frac{1}{2{(x-1)}^2}+C\end{array}$$

Optional simplification:

$$\gg \gg \ln \left|\frac{x-1}{x}\right|+\frac{2x-3}{2{(x-1)}^2}+C$$Link to original

Simpson’s Rule

02

Simpson’s Rule for volume by shells

Use Simpson’s Rule with $n=6$ to compute the volume of the solid obtained by revolving the pictured region about the $y$-axis. Can you do it without using a calculator?

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Solution

Solutions - 0070-02

(1) Recall shells formula:

$$V={\int }_a^{b}2\pi Rhdr$$

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(2) Interpret:

$$R=xh=f(x)dr=dx$$

Bounded above by $f(x)$. Bounded below by $x$-axis.

Bounded left by $x=2$. Bounded right by $x=8$.

Obtain:

$$V=2\pi {\int }_2^{8}xf(x)dx$$

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(3) Create table of values to apply Simpson’s Rule:

$x_i$	$f(x_i)$	$x_{i}f(x_i)$
$x_0=2$	$f(x_0)=0$	$x_{0}f(x_0)=0$
$x_1=3$	$f(x_1)=1$	$x_{1}f(x_1)=3$
$x_2=4$	$f(x_2)=3$	$x_{2}f(x_2)=12$
$x_3=5$	$f(x_3)=3$	$x_{3}f(x_3)=15$
$x_4=6$	$f(x_4)=2$	$x_{4}f(x_4)=12$
$x_5=7$	$f(x_5)=3$	$x_{5}f(x_5)=21$
$x_6=8$	$f(x_6)=0$	$x_{6}f(x_6)=0$

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(4) Recall Simpson’s Rule formula:

$$S_n=\frac{1}{3}\mathrm{\Delta }x(y_0+4y_1+2y_2+\cdots +2y_{n-2}+4y_{n-1}+y_n)$$

Here $y_i=x_{i}f(x_i)$ since $y_i$ in this formula represents the integrand values.

Note that $\mathrm{\Delta }x=1$.

Plug in:

$$\begin{array}{cc}{S}_n& =\frac{1}{3}(x_{0}f(x_0)+4x_{1}f(x_1)+2x_{2}f(x_2)+\cdots +4x_{5}f(x_5)+x_{6}f(x_6))\\ & \\ & =\frac{1}{3}(0+4(3)+2(12)+4(15)+2(12)+4(21)+0)=68\end{array}$$

Therefore:

$${\int }_2^{8}xf(x)dx\approx 68$$

Therefore:

$$V=2\pi {\int }_2^{8}xf(x)dx\approx 2\pi \cdot 68\approx 427.26$$Link to original

03

Area of a garden bed

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Solution

Solutions - 0070-03

Note: you can also start from $y_0=4.5$ and $y_8=4.0$. This gives a different answer, $\approx 1.3$.

(1) Set up integration:

Set $(x,y)=(\mathrm{0,0})$ at the left upper corner, with $+x$ extending to the right, $+y$ extending downwards. Then:

$$A={\int }_0^{20}f(x)dx$$

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(2) Create table of values:

$i$	$x_i$	$f(x_i)$
$0$	$0$	$0$
$1$	$2$	$4.5$
$2$	$4$	$3.3$
$3$	$6$	$3.3$
$4$	$8$	$5$
$5$	$10$	$6$
$6$	$12$	$5$
$7$	$14$	$4$
$8$	$16$	$5$
$9$	$18$	$4$
$10$	$20$	$0$

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(3) Recall Simpson’s Rule formula:

$$S_n=\frac{1}{3}\mathrm{\Delta }x(y_0+4y_1+2y_2+\cdots +2y_{n-2}+4y_{n-1}+y_n)$$

Here $y_i=f(x_i)$ and $\mathrm{\Delta }x=2$.

Thus:

$$\begin{array}{cc}A\approx S_{10}& =\frac{1}{3}\cdot 2\cdot (0+4\cdot 4.5+2\cdot 3.3+4\cdot 3.3+\cdots +4\cdot 4+0)\\ & \\ & \approx 82.53\end{array}$$

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(4) Compute cubic yards from known surface area:

Mulch is $0.5\mathrm{ft}$ deep, so the volume is:

$$\begin{array}{cc}82.53{\mathrm{ft}}^2\times 0.5\mathrm{ft}& \gg \gg 41.265{\mathrm{ft}}^3\\ & \\ & \gg \gg 1.53{\mathrm{yd}}^3\end{array}$$Link to original