Homework 5

HW 5.pdf

Compute the derivative of the following functions:
1. \(F(x) = \displaystyle\int_{0}^{\sqrt{x}}\sqrt{3t^{2}- 4t + 1}dt\).
  
  Let \(\sqrt{x}=u\), then \(\left({{\displaystyle\int_0^{\sqrt{x}}\sqrt{3t^{2}-4t+1}dt}}\right)^{\prime}={{\displaystyle\left(\int_0^{u}\sqrt{3t^{2}-4t+1}dt\right)}} ^{\prime}=\sqrt{3u^{2}-4u+1}\cdot\frac{1}{2\sqrt{x}}=\sqrt{3x-4\sqrt{x}+1}\cdot\frac{1}{2\sqrt{x}} =\sqrt{\frac{3x}{4x}-\frac{4\sqrt{x}}{4x}+\frac{1}{4x}}=\sqrt{3-\frac{1}{\sqrt{x}}+\frac{1}{4x}}\) 2. \(F(x) = \left(\displaystyle\int_{0}^{x}(6t^{2}- 1) dt\right)^{2}\)
  
  Let \(\displaystyle\int_{0}^{x}(6t^{2}-1)dt=u\), then \(\left\lbrack\left({{{{{{\displaystyle\int_0^{x}(6t^2-1)dt}}}}}}\right)^{2}\right \rbrack^{\prime}=\left\lbrack u^{2}\right\rbrack^{\prime}\cdot u^{\prime}=2u=2{{{\displaystyle\int_0^{x}(6t^2-1)dt\cdot\left(\int_0^{x}(6t^2-1)dt\right)^{\prime}}}} =2\cdot\left(2x^{3}-x\right)\left(6x^{2}-1\right)=24x^{5}-16x^{3}+2x\) 3. \(F(x) = \displaystyle\int^{0}_{x^2}\sin(\cos t) dt\).
  
  \({{\displaystyle\left(\int_{x^2}^0\sin(\cos t)dt\right)^{\prime}=\left(-\int_0^{x^2}\sin(\cos t)dt\right)^{\prime}}} =-\left(\displaystyle\int_{0}^{x^2}\sin(\cos t)dt\right)^{\prime}\)
  
  Let \(u=x^2\), then \(=-\left({{\displaystyle\int_0^{x^2}\sin(\cos t)dt}}\right)^{\prime}=-\left({{\displaystyle\int_0^{u}\sin(\cos t)dt}} \right)^{\prime}=-\sin(\cos u)\cdot u^{\prime}=-\sin(\cos x^{2})\cdot2x=-2x\sin(\cos x^{2})\)
Find the Taylor series at 0 of the function \(F(x) = \displaystyle\int_{0}^{x}e^{-t^2}dt\)

Since \(T(x)=F\left(0\right)+\frac{F^{\prime}\left(0\right)}{1!}x+\frac{F''\left(0\right)}{2!} x^{2}+...\), then

Since \(e^{x}=\sum_{n=0}^{\infty}\left(\frac{x^{n}}{n!}\right)\), then \(e^{-t^2}=\sum_{n=0}^{\infty}\frac{\left(-t^{2}\right)^{n}}{n!}=\sum_{n=0}^{\infty} \left(-1\right)^{n}\frac{t^{2n}}{n!}\)

Thus \(F(x)=\displaystyle\int_{0}^{x}\sum_{n=0}^{\infty}\left(-1\right)^{n}\frac{t^{2n}}{n!} dt=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\int_{0}^{x}t^{2n}dt=\sum_{n=0} ^{\infty}\frac{\left(-1\right)^{n}}{n!}\frac{t^{2n+1}}{2n+1}\big|^{x}_{0}=\sum_{n=0} ^{\infty}\frac{\left(-1\right)^{n}}{n!}\frac{x^{2n+1}}{2n+1}\)

Thus \(T(x)=x+\frac{\left(-1\right)}{1!\cdot3}x^{3}+\frac{\left(-1\right)^{2}}{2!\cdot5} x^{5}+\cdots+\frac{\left(-1\right)^{n}\cdot x^{2n+1}}{n!\cdot\left(2n+1\right)}+\cdots =x-\frac{x^{3}}{3}+\frac{x^{5}}{10}+\cdots+\frac{\left(-1\right)^{n}\cdot x^{2n+1}}{n!\cdot\left(2n+1\right)} +\cdots\)
Let \(f\) be a continuous function on \([a, b]\) and \(\phi : [\alpha, \beta] \rightarrow \mathbb{R}\) be continuously differentiable such that \(\phi(\alpha) = a\) and \(\phi(\beta) = b\). Show that \(\displaystyle\int_{a}^{b}f(x) dx = \displaystyle\int_{\alpha}^{\beta}f(\phi(t)) \phi'(t) dt\).

By theorem, \({\displaystyle\int_{a}^{b}f(x)dx}=F\left(b\right)-F\left(a\right)\) where \(F'(x)=f(x)\)

Similarly \(\displaystyle\int_{\alpha}^{\beta}f(\phi(t)) \phi'(t) dt=G(\phi(\beta))-G(\phi(\alpha ))=G(b)-G(a)\) where \(G'(\phi(t))=f(\phi(t))\phi'(t)\), then let \(x=\phi(t)\), \(G'(x)=f(x)\cdot x'=f(x)\)

Thus \(G'(x)=F'(x)\), then \(G(x)=F(x)+c\)

Thus \({\displaystyle\int_{\alpha}^{\beta}f(\phi(t))\phi^{\prime}(t)dt=G(b)-G(a)}=F\left (b\right)+c-F\left(a\right)-c=F\left(b\right)-F\left(a\right)\)

Thus \(\displaystyle\int_{a}^{b}f(x) dx = \displaystyle\int_{\alpha}^{\beta}f(\phi(t)) \phi'(t) dt\)
Determine the area of the region enclosed by:
1. \(y = \sin(x)\), \(y = \cos(x)\), \(x = \frac{\pi}{2}\) and the \(y\)-axis.
  
  We consider two part: 1. \([0,\frac{\pi}{4}]\) 2. \([\frac{\pi}{4},\frac{\pi}{2}]\)
  
  For the first part, \(\cos(x)\geq \sin(x)\), thus the area \(A_{1}=\displaystyle\int_{0}^{\frac{\pi}{4}}(\cos x-\sin x)dx=\int_{0}^{\frac{\pi}{4}} \cos xdx-\int_{0}^{\frac{\pi}{4}}\sin xdx=\sin(\frac{\pi}{4})-\sin(0)-(-\cos(\frac{\pi}{4} )-(-\cos(0)))=\sqrt{2}-1\)
  
  For the second part, \(\cos(x)\leq \sin(x)\), thus the area \(A_{2}=\displaystyle\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}(\sin x-\cos x)dx=\int^{\frac{\pi}{2}} _{\frac{\pi}{4}}\sin xdx-\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\cos xdx=-\sin(\frac{\pi}{2} )+\sin(\frac{\pi}{4})+(-\cos\frac{\pi}{2}-(-\cos\frac{\pi}{4}))=\sqrt{2}-1\)
  
  Thus \(A=A_1+A_2=2\sqrt2-2\) 2. \(x = -y^2 + 10\) and \(x = (y - 2)^2\).
  
  \(-y^{2}+10=(y-2)^{2}\Rightarrow -y^{2}+10=y^{2}-4y+4\Rightarrow 2y^{2}-4y-6=0\Rightarrow y^{2}-2y-3=0\Rightarrow y=3,-1\)
  
  Thus the intersection point is \((1,3)\) and \((9,-1)\), we calculate in terms of \(y\)
  
  Thus \(A=\displaystyle \int^{3}_{-1}-y^{2}+10-(y-2)^{2}dy= \int^{3}_{-1}-2y^{2}+4y+6dy= \int^{3}_{-1}-2y^{2}dy+\int^{3}_{-1}4ydy+\int^{3}_{-1}6dy= \frac{-2}{3}3^{3}-\frac{-2}{3} (-1)^{3}+2\cdot3^{2}-2\cdot(-1)^{2}+6\cdot 3-6\cdot(-1)=-18-\frac{2}{3}+18-2+18+6= 21\frac{1}{3}=\frac{64}{3}\)