Homework 7

% Differentials d[something]/d[something] \gdef\diff#1#2{\frac{\mathrm{d}#1}{\mathrm{d}#2}} % Shortcut for dy/dx \gdef\dydx{\diff{y}{x}} % Differential letter "d" with a thin space before it \gdef\dd{\mathop{}\!\mathrm{d}} % Shortcut for not implies \gdef\nimplies{\;\;\;\not\nobreak\!\!\!\!\implies\;} % Shortcuts for extended brackets \gdef\({\left(} \gdef\){\right)} \gdef\[{\left[} \gdef\]{\right]} % Shortcut for real number symbol \gdef\R{\mathbb{R}} % More spacing between lines in arrays (override by using \[5em]) \gdef\arraystretch{2em}

Mos Kullathon
921425216

Section 3.4

Express the rational function as a sum or difference of two simpler rational expressions.

(183) x2+1x(x+1)(x+2)\displaystyle\frac{x^2+1}{x(x+1)(x+2)}

x2+1x(x+1)(x+2)=Ax+Bx+1+Cx+2;A,B,C∈Rx2+1=A(x+1)(x+2)+Bx(x+2)+Cx(x+1)\begin{align*} \frac{x^2+1}{x(x+1)(x+2)} &= \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2};\quad A,B,C\in\R \\ x^2 + 1 &= A(x+1)(x+2) + Bx(x+2) + Cx(x+1) \end{align*}

Let x=0x=0.

02+1=A(0+1)(0+2)1=2A∴A=120^2 + 1 = A(0+1)(0+2) \\ 1 = 2A \\ \therefore A = \frac{1}{2}

Let x=−1x = -1.

(−1)2+1=B(−1)(−1+2)2=−B∴B=−2(-1)^2 + 1 = B(-1)(-1+2) \\ 2 = -B \\ \therefore B = -2

Let x=−2x = -2.

(−2)2+1=C(−2)(−2+1)5=2C∴C=52(-2)^2 + 1 = C(-2)(-2+1) \\ 5 = 2C \\ \therefore C = \frac{5}{2}

∴x2+1x(x+1)(x+2)=12x−2x+1+52(x+2)\therefore \frac{x^2+1}{x(x+1)(x+2)} = \frac{1}{2x} -\frac{2}{x+1} +\frac{5}{2(x+2)}

(185) 3x+1x2\displaystyle\frac{3x+1}{x^2}

3x+1x2=Ax+Bx2;A,B∈R3x+1=Ax+B\begin{align*} \frac{3x+1}{x^2} &= \frac{A}{x} + \frac{B}{x^2};\quad A,B\in\R \\ 3x+1 &= Ax + B \end{align*}

By comparing coefficients, A=3,B=1A = 3, B= 1.

∴3x+1x2=3x+1x2\therefore\frac{3x+1}{x^2} = \frac{3}{x} + \frac{1}{x^2}

(187) 2x4x2−2x\displaystyle\frac{2x^4}{x^2-2x}

x2−2x      2x2+4x+8(x2−2x)(2x4‾x2−2x      2x4−4x3‾x2−2x      2x4−4x3x2−2x      2x4−4x3−8x2‾x2−2x      2x4−4x3−8x2x2−2x      2x4−4x3−8x2−16x‾x2−2x      2x4−4x3−8x2−16x\gdef\longdiv#1#2{ \mathstrut#1\kern.25em\smash{\raisebox.3ex\hbox{)\kern.1em}} \mkern-8mu\overline{\enspace\mathstrut#2} } \def\arraystretch{1em} \begin{array}{l} \phantom{x^2-2x\;\;\;} 2x^2+4x+8 \\ \longdiv{x^2-2x}{2x^4} \\ \phantom{x^2-2x\;\;\;} \underline{2x^4-4x^3} \\ \phantom{x^2-2x\;\;\;2x^4-} 4x^3 \\ \phantom{x^2-2x\;\;\;2x^4-} \underline{4x^3-8x^2} \\ \phantom{x^2-2x\;\;\;2x^4-4x^3-} 8x^2 \\ \phantom{x^2-2x\;\;\;2x^4-4x^3-} \underline{8x^2-16x} \\ \phantom{x^2-2x\;\;\;2x^4-4x^3-8x^2-} 16x \\ \end{array}

∴2x4x2−2x=2x2+4x+8+16xx2−2x=2x2+4x+8+16xx(x−2)=2x2+4x+8+16x−2\begin{align*} \therefore \frac{2x^4}{x^2-2x} &= 2x^2+4x+8+\frac{16x}{x^2-2x} \\ &= 2x^2+4x+8+\frac{16x}{x(x-2)} \\ &= 2x^2+4x+8+\frac{16}{x-2} \end{align*}

(189) 1x2(x−1)\displaystyle\frac{1}{x^2(x-1)}

1x2(x−1)=Ax+Bx2+Cx−1;A,B,C∈R1=Ax(x−1)+B(x−1)+Cx2\begin{align*} \frac{1}{x^2(x-1)} &= \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1};\quad A,B,C\in\R \\ 1 &= Ax(x-1) + B(x-1) + Cx^2 \end{align*}

Let x=0x=0.

1=B(0−1)1=−B∴B=−11 = B(0-1) \\ 1 = -B \\ \therefore B = -1

Let x=1x=1.

1=C(12)∴C=11 = C(1^2) \\ \therefore C = 1

Let x=2x=2 and substitute B=−1,C=1B=-1, C=1.

1=A(2)(2−1)−(2−1)+(22)1=2A−1+4∴A=−11 = A(2)(2-1) -(2-1) + (2^2) \\ 1 = 2A-1+4 \\ \therefore A = -1

∴1x2(x−1)=−1x−1x2+1x−1\therefore\frac{1}{x^2(x-1)} = -\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x-1}

(192) 1x4−1=1(x+1)(x−1)(x2+1)\displaystyle\frac{1}{x^4-1} = \frac{1}{(x+1)(x-1)(x^2+1)}

1(x+1)(x−1)(x2+1)=Ax+1+Bx−1+Cx2+1;A,B,C∈R1=A(x−1)(x2+1)+B(x+1)(x2+1)+C(x+1)(x−1)\begin{align*} \frac{1}{(x+1)(x-1)(x^2+1)} &= \frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x^2+1};\quad A,B,C\in\R \\ 1 &= A(x-1)(x^2+1) + B(x+1)(x^2+1) + C(x+1)(x-1) \end{align*}

Let x=−1x=-1.

1=A(−1−1)(−12+1)1=−4A∴A=−141 = A(-1-1)(-1^2+1) \\ 1 = -4A \\ \therefore A =-\frac{1}{4}

Let x=1x=1.

1=B(1+1)(12+1)1=4B∴B=141 = B(1+1)(1^2+1) \\ 1 = 4B \\ \therefore B = \frac{1}{4}

Let x=0x=0 and substitute A=−14,B=14A=-\frac{1}{4},B=\frac{1}{4}.

1=A(0−1)(02+1)+B(0+1)(02+1)+C(0+1)(0−1)1=−A+B−C1=14+14−C12=−C∴C=−121 = A(0-1)(0^2+1) + B(0+1)(0^2+1) + C(0+1)(0-1) \\ 1 = -A+B-C \\ 1 = \frac{1}{4}+\frac{1}{4}-C \\ \frac{1}{2} = -C \\ \therefore C = -\frac{1}{2}

∴1x4−1=1(x+1)(x−1)(x2+1)=−14(x+1)+14(x−1)−12(x2+1)\begin{align*} \therefore \frac{1}{x^4-1} &= \frac{1}{(x+1)(x-1)(x^2+1)} \\ &= -\frac{1}{4(x+1)} + \frac{1}{4(x-1)} - \frac{1}{2(x^2 + 1)} \end{align*}

Section 3.7

Determine whether the improper integrals converge or diverge. If possible, determine the value of the integrals that converge.

(359) ∫−∞∞1x2+1 ⁣dx\displaystyle\int_{-\infty}^\infty\frac{1}{x^2+1}\dd x

∫−∞∞1x2+1 ⁣dx=∫−∞01x2+1 ⁣dx+∫0∞1x2+1 ⁣dx\int_{-\infty}^\infty\frac{1}{x^2+1}\dd x = \int_{-\infty}^0\frac{1}{x^2+1}\dd x + \int_{0}^\infty\frac{1}{x^2+1}\dd x

∫−∞01x2+1 ⁣dx=lim⁡n→−∞∫n01x2+1 ⁣dx=lim⁡n→−∞[tan⁡−1x]n0=lim⁡n→−∞−tan⁡−1n=π2∫0∞1x2+1 ⁣dx=lim⁡n→∞∫0n1x2+1 ⁣dx=lim⁡n→∞[tan⁡−1x]0n=lim⁡n→∞tan⁡−1n=π2∴∫−∞∞1x2+1 ⁣dx=∫−∞01x2+1 ⁣dx+∫0∞1x2+1 ⁣dx=π2+π2=π\begin{align*} \int_{-\infty}^0\frac{1}{x^2+1}\dd x &= \lim_{n\to-\infty}\int_{n}^0\frac{1}{x^2+1}\dd x \\ &= \lim_{n\to-\infty}\bigg[\tan^{-1}x\bigg]_n^0 \\ &= \lim_{n\to-\infty}-\tan^{-1}n \\ &= \frac{\pi}{2} \\ \int_{0}^\infty\frac{1}{x^2+1}\dd x &= \lim_{n\to\infty}\int_0^n\frac{1}{x^2+1}\dd x \\ &= \lim_{n\to\infty}\bigg[\tan^{-1}x\bigg]_0^n \\ &= \lim_{n\to\infty}\tan^{-1}n \\ &= \frac{\pi}{2} \end{align*} \\ \begin{align*} \therefore \int_{-\infty}^\infty\frac{1}{x^2+1}\dd x &= \int_{-\infty}^0\frac{1}{x^2+1}\dd x + \int_{0}^\infty\frac{1}{x^2+1}\dd x \\ &= \frac{\pi}{2} + \frac{\pi}{2} \\ &= \pi \end{align*}

(362) ∫0∞e−x ⁣dx\displaystyle\int_0^\infty e^{-x}\dd x

∫0∞e−x ⁣dx=lim⁡n→∞∫0ne−x ⁣dx=lim⁡n→∞[−e−x]0n=lim⁡n→∞−e−n−(−e0)=lim⁡n→∞−1en+lim⁡n→∞1=0+1=1\begin{align*} \int_0^\infty e^{-x}\dd x &= \lim_{n\to\infty} \int_0^n e^{-x}\dd x \\ &= \lim_{n\to\infty} \bigg[-e^{-x}\bigg]_0^n \\ &= \lim_{n\to\infty} -e^{-n}-(-e^0) \\ &= \lim_{n\to\infty} -\frac{1}{e^n} + \lim_{n\to\infty} 1 \\ &= 0+1 \\ &= 1 \end{align*}

(365) ∫01 ⁣dxx3\displaystyle\int_0^1\frac{\dd x}{\sqrt[3]{x}}

∫01 ⁣dxx3=lim⁡n→0∫n1x−13 ⁣dx=lim⁡n→0[x232/3]n1=lim⁡n→0[3x2/32]n1=lim⁡n→03(1)2/32−3n2/32=32\begin{align*} \int_0^1\frac{\dd x}{\sqrt[3]{x}} &= \lim_{n\to0}\int_n^1x^{-\frac{1}{3}}\dd x \\ &= \lim_{n\to0}\bigg[\frac{x^\frac{2}{3}}{2/3}\bigg]_n^1 \\ &= \lim_{n\to0}\bigg[\frac{3x^{2/3}}{2}\bigg]_n^1 \\ &= \lim_{n\to0}\frac{3(1)^{2/3}}{2} - \frac{3n^{2/3}}{2} \\ &= \frac{3}{2} \end{align*}

Evaluate the integrals. If the integral diverges, answer “diverges.”

(374) ∫1∞ ⁣dxxe\displaystyle\int_1^\infty\frac{\dd x}{x^e}

∫1∞ ⁣dxxe=lim⁡n→∞∫1nx−e ⁣dx=lim⁡n→∞[x1−e1−e]1n=lim⁡n→∞n1−e1−e−11−e1−e=lim⁡n→∞11−e(ne−1)−11−e1−e=0−11−e=1e−1\begin{align*} \int_1^\infty\frac{\dd x}{x^e} &= \lim_{n\to\infty} \int_1^n x^{-e} \dd x \\ &= \lim_{n\to\infty} \bigg[\frac{x^{1-e}}{1-e}\bigg]_1^n \\ &= \lim_{n\to\infty} \frac{n^{1-e}}{1-e} - \frac{1^{1-e}}{1-e}\\ &= \lim_{n\to\infty} \frac{1}{1-e(n^{e-1})} - \frac{1^{1-e}}{1-e}\\ &= 0 - \frac{1}{1-e} \\ &= \frac{1}{e-1} \end{align*}

(375) ∫01 ⁣dxxπ\displaystyle\int_0^1\frac{\dd x}{x^\pi}

∫01 ⁣dxxπ=lim⁡n→0∫n1x−π ⁣dx=lim⁡n→0[x1−π1−π]n1=lim⁡n→011−π−01−π1−π=∞\begin{align*} \int_0^1\frac{\dd x}{x^\pi} &= \lim_{n\to0}\int_n^1 x^{-\pi} \dd x\\ &= \lim_{n\to0}\bigg[\frac{x^{1-\pi}}{1-\pi}\bigg]_n^1 \\ &= \lim_{n\to0}\frac{1}{1-\pi} - \frac{0^{1-\pi}}{1-\pi} \\ &= \infty \end{align*}

Therefore, ∫01 ⁣dxxπ\displaystyle\int_0^1\frac{\dd x}{x^\pi} diverges.

(382) ∫0∞xe−x ⁣dx\displaystyle\int_0^\infty xe^{-x}\dd x

∫0∞xe−x ⁣dx=lim⁡n→∞∫0nxe−x ⁣dxu=x  ⟹  u′=1v′=e−x  ⟹  v=−e−x∫0nxe−x ⁣dx=[uv]0n−∫0nvu′ ⁣dx=[−xe−x]0n+∫0ne−x ⁣dx=−ne−n+[−e−x]0n=−ne−n−e−n+1=−nen−1en+1\int_0^\infty xe^{-x}\dd x = \lim_{n\to\infty}\int_0^n xe^{-x}\dd x \\ \begin{darray}{cc} u = x \implies u' = 1 \\ v' = e^{-x} \implies v = -e^{-x} \end{darray} \\ \begin{align*} \int_0^n xe^{-x}\dd x &= \bigg[uv\bigg]_0^n - \int_0^n vu'\dd x \\ &= \bigg[-xe^{-x}\bigg]_0^n + \int_0^n e^{-x}\dd x \\ &= -ne^{-n} + \bigg[-e^{-x}\bigg]_0^n \\ &= -ne^{-n} -e^{-n} + 1 \\ &= -\frac{n}{e^n} - \frac{1}{e^n} + 1 \end{align*}

By L’Hôpital’s:

lim⁡n→∞−nen=lim⁡n→∞−1en=0.\lim_{n\to\infty}-\frac{n}{e^n} = \lim_{n\to\infty}-\frac{1}{e^n} = 0.

∴∫0∞xe−x ⁣dx=lim⁡n→∞−nen−1en+1=−0−0+1=1\begin{align*} \therefore \int_0^\infty xe^{-x}\dd x &= \lim_{n\to\infty} -\frac{n}{e^n} - \frac{1}{e^n} + 1 \\ &= -0-0+1 \\ &= 1 \end{align*}

(394) Find the area of the region in the first quadrant between the curve y=e−6xy=e^{-6x} and the x-axis.

A=∫0∞e−6x ⁣dx=lim⁡n→∞∫0ne−6x ⁣dx=lim⁡n→∞[−e−6x6]0n=lim⁡n→∞−e−6n6+16=−0+16=16\begin{align*} A &= \int_0^\infty e^{-6x}\dd x \\ &= \lim_{n\to\infty}\int_0^n e^{-6x}\dd x \\ &= \lim_{n\to\infty}\bigg[-\frac{e^{-6x}}{6}\bigg]_0^n \\ &= \lim_{n\to\infty} -\frac{e^{-6n}}{6} + \frac{1}{6} \\ &= -0+\frac{1}{6} \\ &= \frac{1}{6} \end{align*}

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