Problems

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Solution: \begin{align*} && x &= \frac{1}{t^2-1} \\ && t &= \sqrt{\frac{x+1}{x}}\\ \Rightarrow && \frac{\d x}{\d t} &= \frac{-2t}{(t^2-1)^2} \\ \Rightarrow && I &= \int \frac{1}{\sqrt{x(x+1)}} \d x \\ &&&= \int \frac{1}{\sqrt{\frac1{t^2-1} \frac{t^2}{t^2-1}}} \cdot \frac{-2 t}{(t^2-1)^2} \d t \\ &&&= \int \frac{t^2-1}{t} \frac{-2t}{(t^2-1)^2} \d t \\ &&&= -\int \frac{2}{t^2-1} \d t \\ &&&= - \frac{2}{2 \cdot 1} \ln \left | \frac{t-1}{t+1} \right| +C \\ &&&= \ln \left | \frac{t+1}{t-1} \right| + C \\ &&&= \ln \left | \frac{\sqrt{\frac{x+1}{x}}+1}{\sqrt{\frac{x+1}{x}}-1} \right| + C \\ &&&= \ln \left | \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}-\sqrt{x}} \right| + C \\ &&&= 2\ln \left | \sqrt{x+1}+\sqrt{x}\right| + C \\ &&&= 2\ln \left ( \sqrt{x+1}+\sqrt{x}\right) + C \\ \end{align*} \begin{align*} && V&= \pi \int_{1/8}^{9/16} y^2 \d x \\ &&&= \pi \int_{1/8}^{9/16} \left ( \frac1x + \frac{1}{x+1} - \frac{2}{\sqrt{x(x+1)}}\right) \d x \\ &&&= \pi \left [ \ln x + \ln (x+1) - 4 \ln(\sqrt{x+1} + \sqrt{x}) \right]_{1/8}^{9/16} \\ &&&= \pi \left ( \ln \frac{9}{16} + \ln \frac{25}{16} - 4 \ln \left ( \frac54 + \frac34\right) \right) +\\ &&&\quad -\pi \left ( \ln \frac{1}{8} + \ln \frac{9}{8} - 4 \ln \left ( \frac1{2\sqrt{2}} + \frac3{2\sqrt{2}}\right) \right) \\ &&&= \pi \left ( 2 \ln 3 - 8 \ln 2 + 2 \ln 5 - 4\ln2 \right) - \pi \left ( -6 \ln 2 + 2\ln 3 - 2\ln 2\right) \\ &&&= \pi (2 \ln 5 - 4 \ln 2 ) \\ &&&= 2 \pi \ln \tfrac54 \end{align*}

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Solution:

1. Notice that \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \text{Evaluate at }x = 1: && 2^n &= \sum_{i=0}^n \binom{n}{i} \end{align*}
2. \(\,\) \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i-1} \\ \text{Evaluate at }x = 1: && n2^{n-1} &= \sum_{i=1}^n i\binom{n}{i} \end{align*}
3. \(\,\) \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \Rightarrow && \int_0^1(1+x)^n \d x &= \int_0^1 \sum_{i=0}^n \binom{n}{i} x^i \d x \\ \Rightarrow && \frac{1}{n+1}(2^{n+1}-1) &= \sum_{i=0}^n \binom{n}{i}\int_0^1 x^i \d x\\ &&& = \sum_{i=0}^n \frac{1}{i+1}\binom{n}{i} \\ \end{align*}
4. \(\,\) \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i-1} \\ \times x: && nx(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i} \\ \frac{\d}{\d x}: && n(1+x)^{n-1}+n(n-1)x(1+x)^{n-2} &= \sum_{i=1}^n i^2\binom{n}{i} x^{i-1} \\ \text{Evaluate at }x = 1: && \sum_{i=1}^n i^2\binom{n}{i} &= n(1+1)^{n-1}+n(n-1)x(1+1)^{n-2} \\ &&&= 2^{n-2} \left (n(n-1) + 2n \right) \\ &&&= n(n+1)2^{n-2} \end{align*}

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Solution: \begin{align*} \cos 3x &\equiv \cos (2x + x) \\ &\equiv \cos 2x \cos x - \sin 2x \sin x \\ &\equiv (2\cos^2 x - 1) \cos x - 2 \sin x \cos x \sin x \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (\sin^2 x) \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (1- \cos^2 x) \\ &\equiv 4\cos^3 x - 3\cos x \end{align*} Similarly, \begin{align*} \sin 3x &\equiv \sin (2x + x) \\ &\equiv \sin 2x \cos x + \cos 2x \sin x \\ &\equiv 2 \sin x \cos x \cos x + (1-2\sin^2 x) \sin x \\ &\equiv 2 \sin x (1-\sin^2 x) + \sin x - 2 \sin^3 x \\ &\equiv 3 \sin x -4 \sin ^3 x \end{align*}

1. \begin{align*} I(\alpha) &= \int_0^{\alpha} (7 \sin x - 8 \sin^3 x) \d x \\ &= \int_0^{\alpha} (7 \sin x - (6\sin x-2 \sin 3x) ) \d x \\ &= \int_0^{\alpha} (\sin x +2 \sin 3x ) \d x \\ &= -\cos \alpha - \frac23 \cos 3\alpha +1+\frac23 \\ &= -c - \frac23 (4c^3-3c) + \frac53 \\ &= -\frac83 c^3 +c + \frac53 \end{align*} as required. When \(c = -1\) this value is \(0\). Eustace will obtain the value \(\frac{7}{2} \sin^2 \beta - 2 \sin^4 \beta = \frac72 (1-\cos^2 \beta) - 2(1-\cos^2 \beta)^2 = \frac32 + \frac12\cos^2 \beta -2\cos^4 \beta\) So if \(\cos \beta = -\frac16\) he will obtain \(\frac32 + \frac{1}{2\cdot36} - \frac{2}{6^4}\) and he should obtain \(\frac{8}{3} \frac{1}{6^3} - \frac{1}{6} + \frac{5}{3}\) which are equal. We want to find all roots of: \begin{align*} && \frac32 + \frac12 c^2 - 2c^4 &= -\frac83 c^3+ c + \frac53 \\ \Rightarrow && 0 &=2c^4-\frac83c^3-\frac12 c^2+c +\frac{1}{6} \\ &&&= 12c^4-16c^3-3c^2+6c+1\\ &&&= (6c+1)(2c^3-3c^2+1) \\ &&&= (6c+1)(2c+1)(c-1)^2 \end{align*} Therefore \(\cos \alpha = - \frac16, -\frac12, 1\) will give the correct answers.

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Solution:

1. \(\,\) \begin{align*} && I &= \int_0^a \frac{f(x)}{f(x)+f(a-x)} \d x \\ u =a-x, \d u = - \d x: &&& \int_{u=a}^{u=0} \frac{f(a-u)}{f(a-u)+f(u)} (-1) \d u \\ &&&= \int_0^a \frac{f(a-u)}{f(u)+f(a-u)} \d u \\ &&&= \int_0^a \frac{f(a-x)}{f(x)+f(a-x)} \d x \\ \Rightarrow && 2 I &= \int_0^a \left ( \frac{f(x)}{f(x)+f(a-x)} + \frac{f(a-x)}{f(x)+f(a-x)} \right) \d x \\ &&&= \int_0^a 1 \d x \\ &&&= a \\ \Rightarrow && I &= \frac{a}{2} \end{align*} \begin{align*} && J &= \int_0^1 \frac{\ln (x+1)}{\ln (2+x-x^2)}\, \d x \\ &&&= \int_0^1 \frac{\ln (x+1)}{\ln((x+1)(2-x))} \d x \\ &&&= \int_0^1 \frac{\ln (x+1)}{\ln(x+1) + \ln ((1-x)+1)} \d x \\ &&&= \frac{1}{2} \tag{\(f(x) = \ln (x+1)\)} \\ \\ && K &= \int_0^{\frac\pi 2} \frac{\sin x } {\sin(x+\frac \pi 4 )} \, \d x \\ &&&= \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}} \\ &&&= \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x + \sin (\frac{\pi}{2}-x)} \d x\\ &&&= \frac{\pi}{2\sqrt{2}} \end{align*}
2. \(\,\) \begin{align*} &&I &= \int_{\frac12}^2 \frac{\sin x }{x(\sin x + \sin \frac1x)} \d x \\ u = 1/x, \d u = -1/x^2 \d x : &&&= \int_{u = 2}^{u=\frac12} \frac{\sin \frac1u}{\frac{1}{u}(\sin \frac1u + \sin u)} (-\frac{1}{u^2} ) \d u \\ &&&= \int_{\frac12}^2 \frac{\sin \frac1u}{u (\sin u + \sin \frac1u)} \d u \\ \Rightarrow && 2I &= \int_{\frac12}^2 \left ( \frac{\sin x }{x(\sin x + \sin \frac1x)} + \frac{\sin \frac1x }{x(\sin x + \sin \frac1x)}\right) \d x \\ &&&= \int_{\frac12}^2 \frac{1}{x} \d x\\ &&&= 2\ln2 \\ \Rightarrow && I &= \ln 2 \end{align*}

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Solution:

The curves touch when \(\sin x = 1\) ie \(x = \frac{4n+1}{2} \pi\). therefore \begin{align*} && I &= \int e^{-x} \sin x \d x \\ &&&= -e^{-x} \sin x + \int e^{-x} \cos x \d x \\ &&&= -e^{-x} \sin x -e^{-x} \cos x - I \\ && I &= -\frac{e^{-x}}2 ( \sin x + \cos x)\\ \\ && A_n &= \int_{\frac{4n+3}{2}\pi}^{\frac{4n-1}{2}\pi} e^{-x} (1-\sin x) \d x \\ &&&= \left [ -e^{-x} - \frac{e^{-x}}2(\sin x + \cos x) \right]_{\frac{4n+1}{2}\pi}^{\frac{4n+5}{2}\pi} \\ &&&= -\frac12\exp \left ( -\frac{4n+1}{2}\pi\right)+ \frac12\exp \left ( -\frac{4n-3}{2}\pi\right) \\ &&&= \frac12 (e^{2 \pi}-1) \left ( -\frac{4n+1}{2}\pi\right) \\ \\ && \sum_{n=1}^\infty A_n &= \frac12(e^{2\pi}-1)e^{-\pi/2} \sum_{n=1}^\infty e^{-2n\pi} \\ &&&= \frac12(e^{2\pi}-1)e^{-\pi/2} \frac{e^{-2\pi}}{1-e^{-2\pi}} \\ &&&= \frac12e^{-\pi/2} \end{align*}

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Solution:

1. \(\cosh a = \frac12 (e^a + e^{-a})\) \begin{align*} \int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x &= \int_0^1 \frac{1}{x^2+(e^a+e^{-a})x+e^ae^{-a}} \d x \\ &= \int_0^1 \frac{1}{e^a-e^{-a}}\left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a} \int_0^1 \left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a}\left [\ln(x+e^{-a})-\ln(x+e^a) \right]_0^1 \\ &= \frac{1}{2 \sinh a} \left (\ln(1+e^a)-\ln(1+e^{-a}) - (\ln e^{-a}-\ln e^a) \right) \\ &= \frac{1}{2\sinh a}\left (2a + \ln \frac{1+e^a}{1+e^{-a}}\right) \\ &= \frac1{2\sinh a} \left ( 2a -a \right) \\ &= \frac{a}{2 \sinh a} \end{align*}
2. \begin{align*} \int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x &= \int_1^{\infty} \frac{1}{(x+e^a)(x-e^{-a})} \d x \\ &= \int_1^{\infty} \frac{1}{e^a+e^{-a}} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \int_1^{\infty} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \left [\ln(x-e^{-a}) - \ln (x + e^{a} ) \right]_1^{\infty} \\ &= \frac1{2\cosh a} \left [ \ln \frac{x-e^{-a}}{x+e^{a}} \right]_1^{\infty} \\ &= \frac{1}{2\cosh a} \left ( 0 - \ln \frac{1-e^{-a}}{1+e^a}{}\right) \\ &= \frac{1}{2\cosh a} \ln \frac{1+e^a}{1-e^{-a}}\\ &= \frac{1}{2\cosh a} \left ( a + \ln \coth \frac{a}{2} \right) \end{align*} and \begin{align*} \int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x &= \int_0^\infty\frac{1}{(x^2+e^a)(x^2+e^{-a})} \d x \\ &= \int_0^\infty \frac{1}{e^a-e^{-a}} \left ( \frac{1}{x^2+e^{-a}} - \frac{1}{x^2+e^{a}} \right) \d x \\ &= \frac{1}{2\sinh a} \left [ \frac{1}{e^{-a/2}} \tan^{-1} \frac{x}{e^{-a/2}} - \frac{1}{e^{a/2}}\tan^{-1} \frac{x}{e^{a/2}} \right]_0^{\infty} \\ &= \frac{1}{2\sinh a} \left (e^{a/2}\frac{\pi}{2}-e^{-a/2}\frac{\pi}{2} - 0 \right) \\ &= \frac{1}{2\sinh a} \pi \sinh \frac{a}{2} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{2\sinh a} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{4\sinh \tfrac{a}{2} \cosh \tfrac{a}{2}} \\ &= \frac{\pi}{4\cosh \tfrac{a}{2}} \end{align*}

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Solution: \begin{align*} && \int \frac{{\rm P} ( x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x &= \int \frac{{\rm Q} (x){\rm R}'(x) - {\rm Q}'(x){\rm R}(x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x \\ &&&= \int \frac{\d}{\d x} \left ( \frac{R(x)}{Q(x)} \right) \d x \\ &&&= \frac{R(x)}{Q(x)} + C \end{align*}

1. Suppose \(Q(x) = 1 + 2x + 3x^2, P(x) = 5x^2-4x-3\), and \(R(x) = a +bx + cx^2\), then \begin{align*} && 5x^2-4x-3 &= (1 + 2x + 3x^2)(2cx+b) - (6x+2)(a+bx+cx^2) \\ &&&= (6c-6c)x^3 + (3b+4c-6b-2c)x^2 + \\ &&&\quad+(2c+2b-6a-2b)x + (b-2a) \\ \Rightarrow && 2c-3b &= 5 \\ && 2c-6a &= -4 \\ && b - 2a &= -3 \\ \Rightarrow && b &= 2a - 3\\ && c &= 3a-2 \end{align*} So say \(a = 0, b = -3, c = -2\) we will have \begin{align*} \int \frac{5x^2 - 4x - 3} {(1 + 2x + 3x^2 )^2 } \, \d x &= \frac{-3x-2x^2}{1+2x+3x^2} + C \end{align*} Suppose we have a different value of \(a\), then we end up with: \begin{align*} \frac{a+(2a-3)x+(3a-2)x^2}{1+2x+3x^2} = a +\frac{-3x-2x^2}{1+2x+3x^2} \end{align*} So the different antiderivatives differ by a constant.
2. \(\,\) \begin{align*} && \frac{\d }{\d x} \left ( \frac{1}{1+\cos x + 2 \sin x } y\right) &= \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \\ \end{align*} We want to find \(R(x) = A \cos x + B\sin x + C\) such that \begin{align*} &&5-3\cos x + 4 \sin x &= (1+\cos x + 2 \sin x)R'(x) - R(x)(-\sin x + 2 \cos x) \\ &&&= (1+\cos x + 2 \sin x)(-A\sin x + B \cos x) \\ &&&\quad- (A\cos x + B \sin x + C)(-\sin x + 2 \cos x) \\ &&&=(-A+C) \sin x + (B-2C)\cos x +\\ &&&\quad\quad (2B-A+A-2B)\sin x \cos x \\ &&&\quad\quad (-2A+B)\sin^2x+(B-2A)\cos^2x \\ &&&= (-A+C)\sin x + (B-2C)\cos x +(B-2A) \\ \Rightarrow && B-2A &= 5\\ && C-A &= 4 \\ && B-2C &= -3 \\ \end{align*} There are many solutions so WLOG \(C=4, A = 0, B = 5\) and so \begin{align*} && \int \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \d x &= \frac{5\sin x +4}{1+\cos x + 2 \sin x} + K \\ \Rightarrow && y &= 5\sin x + 4 + K(1 + \cos x + 2 \sin x) \end{align*}

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Solution:

1. \(\,\) \begin{align*} && \int_{1/m}^m \frac{x^2}{x+1} \d x &= \int_{1/m}^m \left ( x- 1 + \frac{1}{x+1} \right) \d x \\ &&&= \left [ \frac{x^2}{2} - x + \ln (x+1) \right]_{1/m}^m \\ &&&= \left ( m^2/2 - m + \ln(m+1) \right)- \left ( \frac{1}{2m^2} - \frac{1}{m} + \ln\left(\frac1m+1\right) \right) \\ &&&= \frac{m^4-2m^3-1+2m}{2m^2} + \ln (m+1) - \ln(m+1) + \ln m \\ &&&= \frac{(m-1)^3(m+1)}{2m^2} + \ln m \end{align*}
2. \(\,\) \begin{align*} u = \frac{1}x, \d x = -\frac{1}{u^2} \d u:&& \int_{1/m}^m \frac1 {x^n(x+1)}\,\d x &= \int_{u=m}^{u=1/m} \frac{1}{u^{-n}(u^{-1}+1)} \frac{-1}{u^2} \d u \\ &&&= \int_{1/m}^m \frac{u^{n-1}}{u+1} \d u \end{align*}
3. • \(\bf (a)\) \(\,\) \begin{align*} && I &= \int_{1/2}^2 \frac {x^5+3}{x^3(x+1)}\,\d x \\ &&&= \int_{1/2}^2 \left ( \frac{x^2}{x+1} + \frac{3}{x^3(x+1)} \right) \d x \\ &&&= \int_{1/2}^2 \frac{x^2}{x+1} \d x + 3 \int_{1/2}^2 \frac{x^2}{x+1} \d x \\ &&&= 4 \left ( \frac{(2-1)^3(2+1)}{2 \cdot 2^2} + \ln 2 \right) \\ &&&= \frac32+4 \ln 2 \end{align*}
   • \(\bf (b)\) \(\,\) \begin{align*} && J &= \int_1^2 \frac{x^5+x^3 +1}{x^3(x+1)}\, \d x \\ && K &= \int_1^2 \frac{x^5 +1}{x^3(x+1)}\, \d x\\ u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=1}^{u=1/2} \frac{u^{-5}+1}{u^{-3}(u^{-1}+1)} \frac{-1}{u^2} \d u \\ &&&= \int_{1/2}^1 \frac{1 + u^5}{u^3(u+1)} \d u \\ \Rightarrow && K &= \frac12 \int_{1/2}^2 \frac{x^5+1}{x^3(x+1)} \d x \\ &&&= \frac{(2-1)^3(2+1)}{2 \cdot 2^2} + \ln 2 \\ &&&= \frac38 + \ln 2 \\ && L &= \int_1^2 \frac{x^3}{x^3(x+1)} \d x \\ &&&= \ln (3) - \ln 2 \\ \Rightarrow && J &= \frac38 + \ln 3 \end{align*}

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Solution: \begin{align*} && I &= \int_0^{2 \pi} e^{x} \cos m x \d x \\ &&&= \left [e^x \cos m x \right]_0^{2 \pi}-\int_0^{2 \pi} e^x m (-\sin mx) \d x\\ &&&= e^{2\pi}-1 + m\int_0^{2\pi}e^x \sin m x \d x \\ &&&= e^{2\pi}-1 + m\left [e^x \sin m x \right]_0^{2\pi} - m \int_0^{2\pi} e^x m \cos x \d x \\ &&&= e^{2\pi}-1+0 - m^2 I\\ \Rightarrow && (m^2+1)I &= e^{2\pi}-1 \\ \Rightarrow && I &= \frac{1}{m^2+1} (e^{2\pi}-1) \end{align*}

1. \(\,\) \begin{align*} && \cos(A+B) + \cos(A-B) &= \cos A\cos B - \sin A \sin B + \cos A \cos B + \sin A \sin B \\ &&&= 2 \cos A \cos B \end{align*} Therefore \begin{align*} && I &= \int_0^{2\pi} e^x \cos x \cos 6x \d x \\ &&&= \int_0^{2\pi} e^x \frac12\left (\cos 7x + \cos 5x \right) \d x\\ &&&= \left ( \frac{1}{2(1+7^2)} + \frac1{2(1+5^2)}\right)(e^{2\pi}-1) \\ &&&= \left (\frac{1}{100}+\frac{1}{52} \right) (e^{2\pi}-1) \\ &&&= \frac{19}{650}(e^{2\pi}-1) \end{align*}
2. \(\,\) \begin{align*} && I &= \int_0^{2\pi} e^x \sin 2x \sin 4x \cos x \d x\\ &&&= \int_0^{2\pi} e^x \tfrac12(\cos2x-\cos 6x) \cos x \d x\\ &&&= \frac12 \int_0^{2\pi} e^x \left (\cos 2x \cos x -\cos 6x \cos x \right) \d x \\ &&&= \frac14 \int_0^{2\pi} e^x \left (\cos 3x + \cos x-\cos 7x-\cos 5x \right) \d x \\ &&&= \frac14 \left (\frac{1}{1+3^2}+\frac{1}{1+1^2}-\frac{1}{1+7^2} - \frac{1}{1+5^2} \right)(e^{2\pi}-1) \\ &&&= \frac14 \left (\frac{1}{10}+\frac{1}{2}-\frac{1}{50} - \frac{1}{26} \right)(e^{2\pi}-1) \\ &&&= \frac{44}{325}(e^{2\pi}-1) \end{align*}

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Solution:

1. \par
   

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2. Let \(Y \sim N(0,\frac14)\), then: \begin{align*} &&\int_0^\infty \frac{1}{\sqrt{2 \pi \cdot \frac14}} e^{-2x^2} \, dx &= \frac12\\ \Rightarrow && \int_0^\infty e^{-2x^2} &= \frac{\sqrt{\pi}}{2 \sqrt{2}} \\ \Rightarrow && k &= \boxed{\frac{2\sqrt{2}}{\sqrt{\pi}}} \end{align*}
3. \begin{align*} \mathbb{E}[X] &= \int_0^\infty x f(x) \, dx \\ &= \frac{2\sqrt{2}}{\sqrt{\pi}}\int_0^\infty x e^{-2x^2}\, dx \\ &= \frac{2\sqrt{2}}{\sqrt{\pi}} \left [-\frac{1}{4}e^{-2x^2} \right]_0^\infty \\ &= \frac{1}{\sqrt{2\pi}} \\ \end{align*} In order to calculate \(\mathbb{E}(X^2)\) it is useful to consider the related computation \(\mathbb{E}(Y^2)\). In fact, by symmetry, these will be the same values. Therefore \(\mathbb{E}(X^2) = \mathbb{E}(Y^2) = \mathrm{Var}(Y) = \frac{1}{4}\) (since \(\mathbb{E}(Y) = 0\)). Therefore \(\mathrm{Var}(Y) = \mathbb{E}(Y^2) - \mathbb{E}(Y)^2 = \frac14 - \frac{1}{2\pi}\)
4. \begin{align*} && \mathbb{P}(X < x) &= \frac12 \\ \Leftrightarrow && 2\mathbb{P}(0 \leq Y < x) &= \frac12 \\ \Leftrightarrow && 2\l \mathbb{P}(Y < x) - \frac12 \r &= \frac12 \\ \Leftrightarrow && \mathbb{P}(Y < x)&= \frac34 \\ \Leftrightarrow && \mathbb{P}(\frac{Y-0}{1/2} < \frac{x}{1/2})&= \frac34 \\ \Leftrightarrow && \mathbb{P}(Z < \frac{x}{1/2})&= \frac34 \\ \Leftrightarrow && \Phi(2x)&= \frac34 \\ \Leftrightarrow && 2x &= 0.6744895\cdots \\ \Leftrightarrow && x &= 0.3372\cdots \\ \Leftrightarrow && &= 0.337 \, (3 \text{sf}) \\ \end{align*}

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Solution:

1. \begin{align*} \mathcal{L}\{e^{-bt}f(t)\}(s) &= \int_0^{\infty}e^{-st}\{ e^{-bt}f(t) \} \d t \\ &= \int_0^{\infty} e^{-(s+b)t}f(t) \d t \\ &= F(s+b) \end{align*}
2. \begin{align*} \mathcal{L}\{f(at)\}(s) &= \int_0^{\infty} e^{-st}f(at) \d t \\ &= \int_{u=0}^{\infty}e^{-s \frac{u}{a}} f\left(a \tfrac{u}{a}\right)\frac{1}{a} \d u \\ &= \int_0^{\infty}e^{-su/a}f(u) a^{-1} \d u \\ &= a^{-1} \int_0^{\infty} e^{-(s/a)u}f(u) \d u \\ &= a^{-1} F\left (\frac{s}{a} \right) \end{align*}
3. \begin{align*} \mathcal{L}\{f'(t)\}(s) &= \int_0^{\infty} e^{-st}f'(t) \d t \\ &= \left [e^{-st} f(t) \right]_0^{\infty} - \int_0^{\infty} -s e^{-st} f(t) \d t\\ &= -f(0)+sF(s) \\ &= sF(s) - f(0) \end{align*}
4. Since \(f''(t) = -f(t)\) we must have: \begin{align*} && -\mathcal{L}(f)&= \mathcal{L}(f'') \\ &&&= s\mathcal{L}(f') -f'(0) \\ &&&= s(s\mathcal{L}(f)-f(0)) - f'(0) \\ &&&= s^2\mathcal{L}(f) - 1 \\ \Rightarrow && (1+s^2) \mathcal{L}(f) &= 1 \\ \Rightarrow && F(s) &= \frac{1}{1+s^2} \end{align*}

\begin{align*} \mathcal{L}\{e^{-pt}\cos qt\}(s) &= \mathcal{L}\{\cos qt\}(s+p) \\ &= q^{-1}\mathcal{L}\{\cos t\}\left (\frac{s+p}{q} \right) \\ &= q^{-1}\mathcal{L}\{\sin'\}\left (\frac{s+p}{q} \right) \\ &= q^{-1} \left (\frac{s+p}{q} \right) \mathcal{L}\{\sin\} \left (\frac{s+p}{q} \right) - q^{-1}\sin \left (0\right) \\ &= \frac{s+p}{q^2} \frac{1}{1+\left (\frac{s+p}{q} \right)^2 } \\ &= \frac{s+p}{q^2+(s+p)^2} \end{align*}

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Solution:

1. \(\,\) \begin{align*} && \lim_{x \to 0} x^m(\ln x)^n &= \lim_{t \to \infty} (e^{-t})^m (\ln e^{-t})^n \\ &&&= \lim_{t \to \infty} e^{-mt} (-t)^n \\ &&&= (-1)^n \lim_{t \to \infty} e^{-mt} t^n = 0 \\ \\ && \lim_{x \to 0} x^x &= \lim_{x \to 0} e^{x \ln x} \\ &&&= \exp \left (\lim_{x \to 0} x \ln x \right) \\ &&&= \exp \left (0 \right) = 1 \end{align*}
2. \(\,\) \begin{align*} && I_{n} &= \int_0^1 x^m (\ln x)^n \d x \\ && I_{n+1} &= \int_0^1 x^m (\ln x)^{n+1} \d x \\ &&&= \left [\frac{x^{m+1}}{m+1} (\ln x)^{n+1} \right]_0^1 - \frac{1}{m+1} \int_0^1 x^{m+1} (n+1) (\ln x)^n \cdot x^{-1} \d x \\ &&&= 0 - \frac{1}{m+1} \lim_{x \to 0} \left (x^{m+1} (\ln x)^{n+1} \right) - \frac{n+1}{m+1} \int_0^1 x^m (\ln x)^n \d x \\ &&&= - \frac{n+1}{m+1} I_n \\ \\ && I_0 &= \int_0^1 x^m \d x = \frac{1}{m+1} \\ && I_1 &= -\frac{1}{(m+1)^2} \\ && I_2 &= \frac{2}{(m+1)^3} \\ && I_n &= (-1)^n \frac{n!}{(m+1)^{n+1}} \end{align*}
3. \(\,\) \begin{align*} && \int_0^1 x^x \d x &= \int_0^1 e^{x \ln x} \d x \\ &&&= \int_0^1 \sum_{k=0}^{\infty} \frac{x^k(\ln x)^k}{k!} \d x \\ &&&= \sum_{k=0}^{\infty} \int_0^1 \frac{x^k(\ln x)^k}{k!} \d x\\ &&&= \sum_{k=0}^{\infty} (-1)^k \frac{k!}{k!(k+1)^{k+1}}\\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^{k+1}}\\ &&&= 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \cdots \end{align*}