From dd1124bde4f8614fc6f24dbe118556203f85ed49 Mon Sep 17 00:00:00 2001 From: Camil Staps Date: Tue, 6 Oct 2015 20:54:12 +0200 Subject: Assignment 5 --- assignment5.tex | 138 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 138 insertions(+) create mode 100644 assignment5.tex diff --git a/assignment5.tex b/assignment5.tex new file mode 100644 index 0000000..26eb79c --- /dev/null +++ b/assignment5.tex @@ -0,0 +1,138 @@ +\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 +\DeclareTextCommandDefault{\textregistered}{\textcircled{\check@mathfonts\fontsize\sf@size\z@\math@fontsfalse\selectfont R}} + +\usepackage{fancyhdr} +\renewcommand{\headrulewidth}{0pt} +\renewcommand{\footrulewidth}{0pt} +\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 \begin{align*} + \int_0^\infty e^{-x^2}\diff x &= \lim\limits_{t\to\infty} \int_0^t e^{-x^2}\diff x \qquad\text{where $x=z$} \\ + &= \lim\limits_{t\to\infty} \int_0^t 1\cdot e^{-x^2}\diff x \\ + &= \lim\limits_{t\to\infty} \left(x\cdot e^{-x^2} - \int_0^t x\cdot-2\cdot e^{-x^2} \diff x\right) \\ + &= \lim\limits_{t\to\infty} \left(\left.x\cdot e^{-x^2}\right|_0^t + 2\int_0^t xe^{-x^2}\diff x\right) \\ + &= \lim\limits_{t\to\infty} \left.xe^{-x^2}\right|_0^t + 1 \qquad\text{($-x^2$ is dominant over $x$, and see 4.a)} \\ + &= 0 + 1 = 1. + \end{align*} + \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} + -- cgit v1.2.3