W03 Regular

Due date: Sunday 2/1, 11:59pm

Partial fractions

01

04

Partial fractions - irreducible quadratic

Compute the integral:

$$\int \frac{x^2}{x^2+9}dx$$

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Solution

Solutions - 0060-04

(1) Perform long division:

$$\frac{x^2}{x^2+9}\gg \gg 1-\frac{9}{x^2+9}$$

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(2) Use $\arctan$ to integrate:

Recall formula:

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

Choose $h=3$. Then:

$$\int \frac{9}{x^2+9}dx\gg \gg 9\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+C$$

The final answer is therefore:

$$x-3{\tan }^{-1}\left(\frac{x}{3}\right)+C$$Link to original

02 $★$

05

Partial fractions - long division

Compute the integral:

$$\int \frac{x^2-x+1}{x^2+x}dx$$

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Solution

Solutions - 0060-05

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

$$\frac{x^2-x+1}{x^2+x}\gg \gg 1+\frac{-2x+1}{x^2+x}$$

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(2) Factor denominator:

$$x^2+x\gg \gg x(x+1)$$

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(3) Write the partial fractions general form equation (for the second term):

$$\frac{-2x+1}{x^2+x}=\frac{A}{x}+\frac{B}{x+1}$$

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(4) Solve for constants:

Cross multiply:

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

Plug in $x=0$, obtain $1=A\cdot 1$ so $A=1$.

Plug in $x=-1$, obtain $3=B(-1)$ so $B=-3$.

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(5) Integrate by terms:

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

03

06

Partial fractions - big generic

Give the generic partial fraction decomposition (no need to solve for the constants):

$$\frac{x+2}{(x^2+2){(x-1)}^3(x^2-9)}=\frac{A}{?}+?$$

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Solution

Solutions - 0060-06

(1) Observe that $x^2+9=(x+3)(x-3)$:

$$\gg \gg \frac{x+2}{(x^2+2){(x-1)}^3(x-3)(x+3)}$$

On the other hand, $x^2+2$ cannot be factored further. (Its zeros are imaginary.)

Now all denominator factors are either linear or irreducible quadratic.

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(2) Write the partial fractions general form equation:

$$\begin{array}{c}\frac{x+2}{(x^2+2){(x-1)}^3(x^2-9)}\\ \\ \gg \gg \frac{Ax+B}{x^2+2}+\frac{C}{x-1}+\frac{D}{{(x-1)}^2}+\frac{E}{{(x-1)}^3}+\frac{F}{x-3}+\frac{G}{x+3}\end{array}$$

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(3) Notice a few things:

• Quadratic $x^2+2$ acquires linear term $Ax+B$ on top
• Linear $x-1$ is to 3rd power so it has repetition up to 3rd power
• Linear $x-3$ and $x+3$ are only to 1st power.

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04 $★$

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

05 $★$

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

06

09

Partial fractions - rationalize first

For each of these integrals, make a $u$-substitution that changes the integrand into a rational function. Write the integral in terms of $u$ for your answer. You do not have to compute the $u$-integral.

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

(b) $\int \frac{\sqrt{x}}{x-1}dx$

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Solution

Solutions - 0060-09

(a) (1) $u$-substitution with $u=e^x$ and $du=e^{x}dx$:

Notice that $dx=u^{-1}du$. Then:

$$\begin{array}{c}\int \frac{1}{e^x-1}dx\gg \gg \int \frac{1}{u-1}{u}^{-1}du\\ \\ \gg \gg \int \frac{1}{u(u-1)}du\end{array}$$

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(2) Partial fractions:

$$\begin{array}{c}\frac{1}{u(u-1)}=\frac{A}{u}+\frac{B}{u-1}\\ \\ \gg \gg 1=A(u-1)+B(u)\\ \\ \text{set }u=1:\gg \gg 1=B\\ \\ \text{set }u=0:\gg \gg 1=A(-1)\gg \gg A=-1\end{array}$$

Therefore:

$$\begin{array}{c}\int \frac{1}{u(u-1)}du\gg \gg -\int \frac{1}{u}du+\int \frac{1}{u-1}du\\ \\ \gg \gg -\ln |u|+\ln |u-1|+C\\ \\ \gg \gg -x+\ln |e^x-1|+C\end{array}$$

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(b) (1) Set $u=\sqrt{x}$, so $x=u^2$ and $dx=2udu$:

$$\int \frac{\sqrt{x}}{x-1}dx\gg \gg \int \frac{2u^2}{u^2-1}du$$

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(2) Partial fractions:

$$\begin{array}{c}\frac{2u^2}{u^2-1}\gg \gg \frac{2u^2}{(u+1)(u-1)}=\frac{A}{u+1}+\frac{B}{u-1}\\ \\ \gg \gg 2u^2=A(u-1)+B(u+1)\\ \\ \text{set }u=1:\gg \gg 2=B(2)\gg \gg B=1\\ \\ \text{set }u=-1:\gg \gg 2=A(-2)\gg \gg A=-1\end{array}$$

Therefore:

$$\begin{array}{c}\int \frac{2u^2}{u^2-1}du\gg \gg \int \frac{1}{u+1}du+\int \frac{-1}{u-1}du\\ \\ \gg \gg \ln |u+1|-\ln |u-1|+C\\ \\ \gg \gg \ln \left|\frac{u+1}{u-1}\right|+C\\ \\ \gg \gg \ln \left|\frac{\sqrt{x}+1}{\sqrt{x}-1}\right|+C\end{array}$$Link to original

Simpson’s Rule

07

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

08 $★$

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