Added homeworks from Real Analysis II
Added the first half of the homeworks from Real Analysis II for Spring 2019
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Alex Tusa
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\documentclass[12pt,letterpaper]{article}
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\usepackage[utf8]{inputenc}
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\usepackage{pgfplots}
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\usepackage[english]{babel}
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\usepackage{amsmath}
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\usepackage{graphicx}
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\usepackage{array}
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\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
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\usepackage{enumitem}
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\usepackage{mathrsfs}
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\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}}
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\newcommand{\st}{\ \text{s.t.}\ }
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\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
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\DeclareMathOperator{\sign}{sgn}
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\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
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\theoremstyle{case}
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\newtheorem{case}{Case}
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\newtheorem{case*}{Case}
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\theoremstyle{definition}
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\newtheorem{definition}{Definition}[section]
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem*{theorem*}{Theorem}
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\newtheorem{corollary}{Corollary}[section]
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\newtheorem*{corollary*}{Corollary}
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem*{lemma*}{Lemma}
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\newtheorem*{remark}{Remark}
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\setlist[enumerate]{font=\bfseries}
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\renewcommand{\qedsymbol}{$\blacksquare$}
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\author{Alexander J. Tusa}
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\title{Real Analysis II Homework 4}
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\begin{document}
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\maketitle
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\begin{enumerate}
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\item Evaluate
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\begin{enumerate}
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\item $\displaystyle\lim\limits_{h \to 0} \frac{1}{h} \int_{x}^{x+h} \sqrt{t+1}\cos t\ dt$ \\
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\begin{align*}
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\limx{h}{0}\frac{1}{h} \int_{x}^{x+h} \sqrt{t+1}\cos t\ dt &= \limx{h}{0} \frac{\displaystyle\int_{0}^{x+h}\sqrt{t+1} \cos t\ dt - \int_{0}^{x} \sqrt{t+1}\cos t\ dt}{h} \\
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&=\limx{h}{0} \frac{F(x+h)-F(x)}{h},\ \text{where $F(x):=\int_{0}^{x}\sqrt{t+1}\cos t\ dt$} \\
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F'(x) &= \frac{d}{dx} \int_{0}^{x} \sqrt{t+1}\cos t\ dt \\
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&\text{let } h(x)=x,\ g(x)=0,\text{ and } f(x):=\sqrt{x+1}\cos x\text{ then} \\
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F'(x) &= f(h(x))\cdot h'(x) - f(g(x))\cdot g'(x) \\
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&= \sqrt{x+1}\cos x \cdot 1 -\sqrt{0+1}\cos 0 \cdot 0 \\
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&= \sqrt{x+1}\cos x
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\end{align*}
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Thus $F'(x)=\sqrt{x+1}\cos x$.\\
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\item $\displaystyle\lim\limits_{x \to a} \frac{x}{x-a} \int_{a}^{x} t^2\ dt$ \\
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\begin{align*}
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\limx{x}{a} \frac{x}{x-a} \int_{a}^{x} t^2\ dt &= \limx{x}{a} \frac{x \cdot \displaystyle\int_{a}^{x} t^2\ dt}{x-a} \\
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&= \lim\limits_{x \to a} \frac{x(x^2\cdot 1 -a^2\cdot 0)}{1-0} \\
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&= a^3
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\end{align*}
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\item $\displaystyle\lim\limits_{x \to 0} \frac{1}{x} \int_{0}^{x} \sqrt{9 + t^2}\ dt$
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\begin{align*}
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\limx{x}{0} \frac{1}{x} \int_{0}^{x} \sqrt{9+t^2}\ dt &= \limx{x}{0} \frac{\displaystyle\int_{0}^{x} \sqrt{9+t^2}\ dt}{x} \\
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&= \limx{x}{0} \frac{\sqrt{9+x^2}\cdot 1 - \sqrt{9+0^2}\cdot 0}{1}, &\text{by L'Hospital's Rule and Leibniz's Rule} \\
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&= \limx{x}{0} \sqrt{9+x^2} \\
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&= \sqrt{9} \\
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&= 3
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\end{align*}
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\end{enumerate}
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\item
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\begin{enumerate}
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\item Show that $(x^2\sin x) /2$ is not an antiderivative of $x \cos x$.
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\begin{proof}
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We want to show that $\frac{x^2\sin x}{2} \neq \int x\cos x $ So, we note that we can use \textit{Theorem 7.3.17 Integration by Parts}. So, let $u=x$ and $dv=\cos x\ dx$. Then
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\[du = u'dx=1\cdot dx = dx\]
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and
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\[v=\int \cos x\ dx = \sin x + C\]
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for some arbitrary constant $C$. Then,
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\begin{align*}
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\int x \cos x\ dx &= uv - \int v\ du \\
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&= x\sin x - \int \sin x\ dx \\
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&= x\sin x - (-\cos x + C) \\
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&= x\sin x + \cos x + C
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\end{align*}
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for some arbitrary constant $C$. And thus we have that the antiderivative of $x \cos x=x\sin x +\cos x$, which we note is \textit{not} equal to $\frac{x^2\sin x}{2}$. That is,
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\[\int x \cos x = x \sin x +\cos x \neq \frac{x^2\sin x}{2}\]
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\end{proof}
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\item If $x^2 \cos x= \displaystyle\int_{0}^{x} f(t)\ dt$, find $f(x)$.
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\\\\Since $\displaystyle\int_{0}^{x} f(t)\ dt = x^2\cos x$, we know that $x^2\cos x=F(x)$. Thus, in order to find $f(x)$, we must find $F'(x)=f(x)$. So,
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\[\frac{d}{dx} x^2\cos x = 2x\cos(x)-x^2\sin(x)\]
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And thus we have that $f(x)=2x\cos(x)-x^2\sin(x)$.\\
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\item Let $F(x)=\displaystyle\int_{0}^{x} xe^{t^2}\ dt$ for $x \in [0,1]$. find $F''(x)$ for $x \in [0,1]$. (Note: $F'(x) \neq xe^{x^2}$)
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\begin{align*}
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F(x) &= x \int_{0}^{x} e^{t^2}\ dt \\
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F'(x) &= x \cdot e^{x^2} + \int_{0}^{x} e^{t^2}\ dt \\
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F''(x) &= x \cdot e^{x^2} \cdot 2x + e^{x^2}+e^{x^2} \\
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&= 2(x^2+1)e^{x^2}
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\end{align*}
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So $F''(x)=2(x^2+1)e^{x^2}$.\\
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\end{enumerate}
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\item Suppose $f$ is nonnegative and continuous on $[1,2]$ and that $\displaystyle\int_{1}^{2} x^k f(x)\ dx=5+k^2$ for $k=0,1,2$.
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\\Prove each of the following:
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\begin{enumerate}
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\item $\displaystyle\int_{1}^{4} f(\sqrt{x})\ dx \leq 20$.\\\\
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Let $u=\sqrt{x}$, then $du = u'dx = \frac{1}{2\sqrt{x}}\ dx$ and thus $dx = 2\sqrt{x}\ du$. So,
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\[\int_{1}^{4} f(\sqrt{x})\ dx = \int_{1}^{2} f(u)\ 2\ \sqrt{x}\ du = 2 \int_{1}^{2} \sqrt{x} f(u)\ du = 2 \int_{1}^{2} u f(u)\ du=2(5+1)=12\]
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Thus $12 \leq 20$.\\
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\item $\displaystyle\int_{1/\sqrt{2}}^{1} f(1/x^2)\ dx \leq 5/2$.\\\\
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Substitute $u=\frac{1}{x^2}$ and $du=\frac{-2}{x^3}\ dx$. $\frac{1}{x^3}\ dx = \frac{-1}{2}\ du$.\\
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We know $\frac{1}{\sqrt{2}} \leq x \leq 1$ and $\frac{1}{2\sqrt{2}} \leq x^3 \leq 1$. So $2\sqrt{2} \geq \frac{1}{x^3} \geq 1$. Thus $1 \leq \frac{1}{x^3}$ and $\displaystyle\int_{\frac{1}{\sqrt{2}}}^{1} 1 f \left(\frac{1}{x^2}\right) \leq \displaystyle\int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{x^3} f\left(\frac{1}{x^2}\right)\ dx$. So
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\[\int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{x^3} f \left(\frac{1}{x^2}\right)\ dx = \int_{2}^{1} \frac{-1}{2} f(u)\ du = \frac{1}{2} \int_{1}^{2} u^0 f(u)\ du = \frac{1}{2} \cdot 5 = \frac{5}{2}\]
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Thus $\displaystyle\int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{x^3} f \left(\frac{1}{x^2}\right)\ dx \leq \frac{5}{2}$.
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\item $\displaystyle\int_{0}^{1} x^2 f(x+1)\ dx =2$. \\\\
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Let $u=x+1$. Then $du=u'dx=1dx=dx$. Then,
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\[\int_{0}^{1} x^2 f(x+1)\ dx = \int_{1}^{2} x^2 f(u)\ du\]
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\begin{align*}
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\int_{0}^{1} x^2 f(x+1)\ dx &= \int_{1}^{2} x^2 f(u)\ du \\
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&= \int_{1}^{2} (u-1)^2f(u)\ du \\
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&= \int_{1}^{2} (u^2-2u+1) f(u)\ du \\
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&= \int_{1}^{2} u^2 f(u)\ du -2uf(u)+f(u)\ du \\
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&= \int_{1}^{2} u^2f(u)\ du -2\int_{1}^{2} uf(u)\ du +\int_{1}^{2} f(u)\ du \\
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&= (5+2^2)-2(5+1^2)+(5+0^2) \\
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&= 9-12+5 \\
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&= 2
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\end{align*}
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$\therefore\ \displaystyle\int_{0}^{1} x^2f(x+1)\ dx = 2$.\\
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\end{enumerate}
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\item Suppose that $f \in \mathcal{R}[1/2, 2]$ and that $\displaystyle\int_{1/2}^{1} x^k f(x)\ dx = \displaystyle\int_{1}^{2} x^kf(x)\ dx + 2k^2 = 3+k^2$ for $k=0,1,2$. Compute the exact values of the following integrals:
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\begin{enumerate}
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\item $\displaystyle\int_{0}^{1} x^3 f(x^2+1)\ dx$\\\\
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Let $u=x^2+1$. Then $du=u'dx = 2x\ dx$. Thus $dx=\frac{du}{2x}$. Note that $x^2=u-1$. So, we have
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\begin{align*}
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\int_{0}^{1} x^3 f(x^2+1)\ dx &= \int_{0}^{1} x^3 f(u)\ \frac{du}{2x} \\
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&= \int_{1}^{2} x^2 f(u)\ \frac{du}{2} \\
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&= \frac{1}{2}\ \int_{1}^{2} (u-1) f(u)\ du \\
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&= \frac{1}{2}\ \int_{1}^{2} uf(u)-f(u)\ du \\
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&= \frac{1}{2}\ \int_{1}^{2} uf(u)\ du - \frac{1}{2}\ \int_{1}^{2} f(u)\ du \\
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&= \frac{1}{2}\ \int_{1}^{2} uf(u)\ du - \frac{1}{2} \cdot (3+0) \\
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&= \frac{1}{2} \left[\int_{1}^{2} uf(u)\ du + 3\right] \\
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&= \frac{1}{2} \left[\int_{1}^{2} u f(u)\ du + 2 - 5\right] \\
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&= \frac{1}{2} [3+1-5] \\
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&= -\frac{1}{2}
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\end{align*}
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$\therefore\ \displaystyle\int_{0}^{1} x^3f(x^2+1)\ dx = -\frac{1}{2}$.\\
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\item $\displaystyle\int_{0}^{\sqrt{3}/2} \frac{x^3}{\sqrt{1-x^2}} f(\sqrt{1-x^2})\ dx$\\\\
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Let $u=\sqrt{1-x^2}$. Then $du=u'dx = \frac{-2x}{2\sqrt{1-x^2}}\ dx$. Thus
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$dx=\frac{-2\sqrt{1-x^2}}{2x}\ du$. So,
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\begin{align*}
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\int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^3}{\sqrt{1-x^2}}f(\sqrt{1-x^2})\ dx &= \int_{1}^{\frac{1}{2}} \frac{x^3}{\cancel{u}} f(u) \cdot \frac{-2\cancel{u}}{2x}\ du \\
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&= \int_{1}^{\frac{1}{2}} \frac{-x^3 f(u)}{x}\ du \\
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&= \int_{1}^{\frac{1}{2}} -x^2f(u)\ du \\
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&= -\int_{\frac{1}{2}}^{1} (u^2-1) f(u)\ du \\
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&= \int_{\frac{1}{2}}^{1} (1-u^2) f(u)\ du \\
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&= \int_{\frac{1}{2}}^{1} f(u)\ du - \int_{\frac{1}{2}}^{1} u^2 f(u)\ du \\
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&= (3+0^2)-(3+2^2) \\
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&= 3-7 \\
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&= -4
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\end{align*}
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$\therefore\ \displaystyle\int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^3}{\sqrt{1-x^2}} f(\sqrt{1-x^2})\ dx = -4$.
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\end{enumerate}
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\item Suppose that $f,g$ are differentiable on $[0,e]$ and that $f', g' \in \mathcal{R}[0,e]$.
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\begin{enumerate}
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\item If $\displaystyle\int_{1}^{e} \frac{f(x)}{x}\ dx < f(e)$, prove that $\displaystyle\int_{1}^{e} f'(x) \ln x\ dx > 0$.
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\\\[\int_{1}^{e} f'(x)\cdot \ln (x)\ dx = \left.f(x)\ln(x)\right|_1^e - \int_{1}^{e} f(x) \cdot \frac{1}{x}\ dx\]
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where $u=\ln (x)$, $dv=f'(x)\ dx$, $du=\frac{1}{x}\ dx$, $v=f(x)$. So
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\[\left.\int_{1}^{e} f'(x)\cdot \ln (x)\ dx = f(x)\ln(x)\right|_1^e - \int_{1}^{e} f(x) \cdot \frac{1}{x}\ dx=f(e)-\int_{1}^{e} \frac{1}{x}\ dx > 0\]
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Since $\int_{1}^{e} \frac{f(x)}{x}\ dx < f(e)$.\\
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\item If $f(0)=f(1)=0$, prove that $\displaystyle\int_{0}^{1} e^x [f(x)+f'(x)]\ dx =0$.
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\begin{proof}
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\[\int_{0}^{1}e^x \left[f(x)+f'(x)\right]\ dx = \int_{0}^{1} e^xf(x)\ dx +\int_{0}^{1} e^xf'(x)\ dx\]
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Let us use \textit{Integration by Parts} on the second integral containing $f'(x)$. Let $u=e^x$. Then $du=e^x\ dx$, $dv=f'(x)\ dx$, and $v=f(x)$. Then we have the following:
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\begin{align*}
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\int_{0}^{1} e^xf(x)\ dx + \int_{0}^{1} f'(x)\ dx &= \left.\cancel{\int_{0}^{1} e^xf(x)\ dx} + e^xf(x)\right|_0^1 - \cancel{\int_{0}^{1} f(x)e^x\ dx} \\
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&= e^1f(1)-e^0f(0) \\
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&= e\cdot 0 - 1 \cdot 0 \\
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&= 0 - 0 \\
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&= 0
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\end{align*}
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$\therefore\ \displaystyle\int_{0}^{1} e^x\left[f(x)+f'(x)\right]\ dx = 0$.
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\end{proof}
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\end{enumerate}
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\item
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\begin{enumerate}
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\item Let $f:[0,b] \to \R,\ b>0$ be continuous and $f(x) \neq 0$ for all $x \in (0,b)$. Further, suppose $[f(x)]^2 = 2 \displaystyle\int_{0}^{x} f(t)\ dt$ for all $x \in [0,b]$. Prove that $f(x)=x$ for all $x \in [0,b]$.\\\\We have
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\[2f(x)f'(x)=2f(x)\]
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which implies $f(x)[f'(x)-1]=0$. Since $f(x) \neq 0$, then $f'(x)-1=0$. So $f'(x)=1$ and $f(x)=x+C$ for some arbitrary constant $C$. But $f(0)=0$ since $[f(0)]^2=0\implies f(0)=0$. So $f(x)=x$.\\
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\item Suppose that $f$ is defined on $[0,1]$ with $f(0)=0$ and $0<f'(x) \leq 1$. Prove that $\left[\displaystyle\int_{0}^{1} f(x)\ dx\right]^2 \geq \displaystyle\int_{0}^{1} [f(x)]^3\ dx$.
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\begin{proof}
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Let $x \in [0,1]$. Then $F(x):=\big[\int_{0}^{x} f\big] ^2 - \int_{0}^{3} f^3$. So $f(0)=0$. Thus $F'(x)=2\left[\int_{0}^{x} f\right] \cdot f(x) - f^3(x)=f(x)\left[2\int_{0}^{x} f-f^2\right]$ since $f(0)=0$ and $0 \leq f'(x) \leq 1$ which implies that $f$ is strictly increasing. So $f(x)\geq 0$.
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\end{proof}
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\end{enumerate}
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\end{enumerate}
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\end{document}
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