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\documentclass[10pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[margin=2cm]{geometry}
\usepackage{enumitem}
\setenumerate[1]{label=\arabic*.}
\setenumerate[2]{label=(\alph*)}
% textcomp package is not available everywhere, and we only need the Copyright symbol
% taken from http://tex.stackexchange.com/a/1677/23992
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\usepackage{fancyhdr}
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\fancyhead{}
\fancyfoot[C]{Copyright {\textcopyright} 2015 Camil Staps}
\pagestyle{fancy}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\newcommand*\diff{\mathop{}\!\mathrm{d}}
\parindent0pt
\title{Calculus en Kansrekenen - assignment 5}
\author{Camil Staps\\\small{s4498062, group Bram}}
\begin{document}
\maketitle
\thispagestyle{fancy}
\begin{enumerate}
\item \begin{enumerate}
\item $\int\sin x\cdot\cos x\diff x = -\frac14\cos2x+c$ since $\left(-\frac14\cos2x\right)' = -\frac14\cdot2\cdot-\sin2x = \sin x\cdot\cos x$.
\item $\int\ln(ax)\diff x = \int\ln a+\ln x\diff x = \frac1x + x\cdot\ln a + c$ since $\left(\frac1x + x\cdot\ln a\right)' = \ln x+\ln a = \ln(ax)$.
\item $\int\cos^2x\diff x = \int\cos x\cdot\cos x\diff x = \sin x\cdot\cos x - \int\sin x\cos'x\diff x = \sin x\cdot\cos x + \int\sin^2x \diff x = \sin x\cdot\cos x + \int1-\cos^2x \diff x = \sin x\cdot\cos x+x+c-\int\cos^2x\diff x$.
But then also: $2\cdot\int\cos^2x\diff x = \sin x\cdot\cos x + x + c$, so $\int\cos^2x\diff x = \frac12\left(\sin x\cdot\cos x + x\right) + c'$.
\item $\int\frac1{\sqrt{1-4x^2}}\diff x = \int\frac1{\sqrt{1-u^2}}\cdot\frac{\diff\frac12u}{\diff u}\diff u = \int\frac1{\sqrt{1-u^2}}\cdot\frac12\diff u = \frac12\arcsin(2x) + c$.
\item %todo
\end{enumerate}
\item \begin{enumerate}
\item \begin{align*}
\int_{-1}^1\sqrt{1+f'(x)^2}\diff x &= \int_{-1}^1\sqrt{1+\left(\frac{-x}{\sqrt{1-x^2}}\right)^2}\diff x \qquad\text{since $f'(x)=\tfrac12\cdot(1-x^2)^{-\frac12}\cdot-2x = -\frac{x}{\sqrt{1-x^2}}$}\\
&= \int_{-1}^1 \sqrt{1+\frac{x^2}{1-x^2}} \diff x \\
&= \int_{-1}^1 \sqrt{\frac1{1-x^2}} \diff x \\
&= \int_{-1}^1 \frac1{\sqrt{1-x^2}} \diff x \\
&= \left.\arcsin x+c\right|_{-1}^1 \\
&= \arcsin1 - \arcsin-1 = \pi.
\end{align*}
\item On the unit circle we have $x^2+y^2=1$, so $y=\sqrt{1-x^2}$. This is then one half of the length of the unit circle, because $0=y=\sqrt{1-x^2} \vDash x=-1 \lor x=1$. The whole unit circle has length $2\pi$, so one half has length $\pi$.
\end{enumerate}
\item \begin{enumerate}
\item \begin{align*}
\int_{-1}^1\sqrt{1-x^2}\diff x &= \int_{\arcsin-1}^{\arcsin1} \sqrt{1-\sin^2u}\cdot\frac{\diff\sin u}{\diff u}\diff u \\
&= \int_{-\frac\pi 2}^{\frac\pi 2} \cos^2u\diff u \\
&= \left.\tfrac12(\sin x\cos x+x)\right|_{-\frac\pi 2}^{\frac\pi 2} \qquad\text{(see 1.c)} \\
&= \tfrac\pi 2.
\end{align*}
\item This is the area of one half of the unit circle, whose area is $\pi\cdot1^2=\pi$. Therefore, this must be $\frac\pi 2$.
\end{enumerate}
\item \begin{enumerate}
\item \begin{align*}
\int_0^\infty re^{-r^2} \diff r &= \lim\limits_{t\to\infty} \int_0^t re^{-r^2}\diff r \\
&= \lim\limits_{t\to\infty} \int_0^{-t^2} \sqrt{-u}\cdot e^u \frac{\diff\sqrt{-u}}{\diff u}\diff u \qquad\text{with $u=-r^2, r=\sqrt{-u}$} \\
&= \lim\limits_{t\to\infty} -\tfrac12\int_0^{t^2} e^u\diff u \\
&= \lim\limits_{t\to\infty} -\tfrac12\left(\left.e^u\right|_0^{-t^2}\right) \\
&= -\tfrac12 \lim\limits_{t\to\infty} \left(e^{-t^2} - e^0\right) \\
&= -\tfrac12 \cdot(0-1) = \tfrac12.
\end{align*}
\item $\int_0^{2\pi} \left(\int_0^\infty re^{-r^2}\diff r\right)\diff t = 2\pi\cdot\int_0^\infty re^{-r^2}\diff r = 2\pi\cdot\frac12 = \pi$ (see 4.a).
\setcounter{enumii}{3}
\item In 4.c we saw that $\int_{-\infty}^\infty e^{-z^2}\diff z = \sqrt\pi$. We know that $f(x)=e^{-x^2}$ is even because of the power of two, therefore $\int_0^\infty e^{-x^2}\diff x = \frac12 \int_{-\infty}^\infty e^{-z^2}\diff z = \frac12\sqrt\pi$.
\end{enumerate}
\item \begin{enumerate}
\item $\int_0^\infty e^{-x}\diff x = \lim_{t\to\infty} \int_0^t e^{-x}\diff x = \lim_{t\to\infty}\left.-e^{-x}\right|_0^t = \lim_{t\to\infty}-e^{-t}+1 = 0 + 1 = 1.$
\item \begin{align*}
\int_0^\infty xe^{-x}\diff x &= \lim_{t\to\infty} \int_0^t xe^{-x}\diff x \\
&= \lim_{t\to\infty} \left(\left.-xe^{-x}\right|_0^t - \int_0^t-e^{-x}\diff x\right) \\
&= \lim_{t\to\infty} \left(\left.-xe^{-x}\right|_0^t - \left.e^{-x}\right|_0^t\right) \\
&= \lim_{t\to\infty} \left((-t-1)e^{-t} + e^0\right) \\
&= 0 + 1 = 1.
\end{align*}
\setcounter{enumii}{3}
\item $\int_0^\infty x^{-\frac12}e^{-x}\diff x = \int_0^\infty\frac1ue^{-u^2}\frac{\diff u^2}{\diff u}\diff u = 2\int_0^\infty e^{-u^2}\diff u = 2\cdot1 = 2$ (see 4.d).
\end{enumerate}
\item \begin{enumerate}
\item These vertices are the intersections of the lines.
$x+2=-x+6$ gives $x=2$ and $y=4$.\\
$x+2=2x-3$ gives $x=5$ and $y=7$.\\
$-x+6=2x-3$ gives $x=3$ and $y=3$.
That gives us $(2,4)$, $(5,7)$ and $(3,3)$.
We then see that on $(2,3)$ we have $x+2>-x+6$ and on $(3,5)$ we have $x+2>2x-3$, so as the area we take:
\begin{align*}
\int_2^3(x+2)-(-x+6)\diff x + \int_3^5(x+2)-(2x-3)\diff x &= \int_2^3 2x-4\diff x + \int_3^5 5-x \diff x\\
&= \left.x^2-4x\right|_2^3 + \left.5x-\tfrac12x^2\right|_3^5 \\
&= 3.
\end{align*}
\item $(x-1)^3=(x-1)^2$ gives $x-1=0,x=1$ or $x-1=1, x=2$. On $(1,2)$ we have $(x-1)^2>(x-1)^3$ (take e.g. $x=\frac12$ and note that there are no intersections).
Therefore we take
\begin{align*}
\int_1^2(x-1)^2-(x-1)^3\diff x &= \int_1^2(2-x)(x^2-2x+1) \diff x \\
&= \int_1^2 -x^3 + 4x^2 - 5x + 2 \diff x \\
&= \left. -\tfrac14x^4 + \tfrac43x^3 - \tfrac52x^2 + 2x\right|_1^2 \\
&= \tfrac1{12}.
\end{align*}
\end{enumerate}
\end{enumerate}
\end{document}
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