Homework 7

HW 7.pdf

Compute the arc length of:
1. \(f(x) = \ln(\sin(x))\) on the interval \([ \frac{\pi}{4}, \frac{\pi}{3} ]\)
  
  We know \(\displaystyle L = \int_{a}^{b} \sqrt{1+f'(x)^{2}}\, dx\).
  
  \(f'(x) = \frac{\cos x}{\sin x}\), then \(1+f^{\prime}(x)^{2}=\frac{\sin x\cdot\sin x+\cos x\cdot\cos x}{\sin x^{2}}=\frac{1}{\sin x^{2}}\). Then \(\sqrt{1+f'(x)^2} = \left|\frac{1}{\sin x}\right|\).
  
  Since \(x\in\left[\frac{\pi}{4},\frac{\pi}{3}\right]\), then \(\sin x > 0\), then \(\displaystyle L=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{1}{\sin x}\,dx=\int_{\frac{\pi}{4}} ^{\frac{\pi}{3}}\csc x\,dx=\ln|\csc x - \cot x|\Big|^\frac{\pi}{3}_\frac{\pi}{4}\).
  
  \(=\ln\left|\csc\left(\frac{\pi}{3}\right) - \cot\left(\frac{\pi}{3}\right)\right| - \ln\left|\csc\left(\frac{\pi}{4}\right) - \cot\left(\frac{\pi}{4}\right)\right| =\ln\left|\frac{\sqrt{3}}{3}\right| - \ln|\sqrt{2}- 1|=\ln\left(\frac{\sqrt{3}}{3}\right) - \ln(\sqrt{2}- 1)\) 2. \(f(x) = \frac{1}{3}(x^2 + 2)^{3/2}\) on the interval \([0, a]\)
  
  \(f^{\prime}(x)=x\sqrt{x^{2}+2}\), then \(1+f^{\prime}(x)^{2}=x^{4}+2x^{2}+1\), then \(\sqrt{1+f^{\prime}(x)^{2}}=\sqrt{x^{4}+2x^{2}+1}=|x^{2}+1|=x^{2}+1\) since \(x^{2} \ge 0\)
  
  Then \(L={{\displaystyle\int_0^{a}\sqrt{1+f^{\prime}(x)^{2}}dx={{\displaystyle\int_0^{a}\left(x^2+1\right)dx=\frac{x^{3}}{3}+x\Big|_0^{a}=\left(\frac{a^{3}}{3}+a\right)-\left(\frac{0^{3}}{3}+0\right)=\frac{a^{3}}{3}+a}}}}\)
  
  Then \(L= \frac{a^3}{3}+a\)
Determine if the following integrals converge or diverge. If the integral converges, determine its value.
1. \(\displaystyle \int_{-\infty}^{0}xe^{3x}\, dx\)
  
  Let \(x = u\), \(e^{3x} = v'\), \(u' = 1\), \(v = \frac{1}{3} e^{3x}\). Then \({\displaystyle\int_{-\infty}^0xe^{3x}dx=\lim_{b\to-\infty}\int_{b}^0xe^{3x}dx}\)
  And \(\displaystyle\int_{b}^0xe^{3x}dx = \frac{x}{3}e^{3x}\Big|_{b}^{0}- \int_{b}^{0}\frac{1}{3}e^{3x}dx = -\frac{b}{3}e^{3b}- \frac{1}{3} \int_{b}^{0}e^{3x}dx\)
  
  Since \(\displaystyle \int_{b}^{0} e^{3x} dx = \frac{1}{3} e^{3x} \Big|_{b}^{0} = \frac{1}{3} - \frac{1}{3} e^{3b},\) then \(\displaystyle \int_{b}^{0} x e^{3x} dx = -\frac{b}{3} e^{3b} - \frac{1}{9} + \frac{1}{9} e^{3b}\)
  
  Then \(\lim_{b \to -\infty}\displaystyle \int_{b}^{0}x e^{3x}dx = -\frac{1}{9}= \displaystyle \int_{-\infty}^{0}x e^{3x}dx\) which is convergent. 2. \(\displaystyle \int_{e}^{\infty}\frac{1}{x \ln(x)}\, dx\)
  
  Let \(\ln(x) = t\), then \(\frac{1}{x} dx = dt\). Then \(\displaystyle \int_{1}^{\infty} \frac{1}{x} \cdot \frac{1}{t} dt \cdot x = \displaystyle \int_{1}^{\infty} \frac{1}{t} dt = \lim_{b \to \infty} \displaystyle \int_{1}^{b} \frac{1}{t} dt = \lim_{b \to \infty} \ln(b) - \ln(1) = \infty\)
  Thus this is divergent. 3. \(\displaystyle \int_{-7}^{\infty}\frac{1}{x^{2} + 4x + 29}\, dx\)
  
  Since \(x \geq 7\), then \(\frac{1}{x^2 + 4x + 29} \leq \frac{1}{x^2}.\) Since \(\displaystyle \int_{1}^{\infty} \frac{1}{x^2} dx\) is convergent, then \(\displaystyle \int_{1}^{\infty} \frac{1}{x^2 + 4x + 29} dx\) is convergent by comparison test.
  
  And \(\frac{1}{x^2 + 4x + 29}\) is continuous on \([-7, 1]\), then \(\displaystyle \int_{-7}^{1} \frac{1}{x^2 + 4x + 29} dx\) is convergent since it's a proper integral.
  
  Then \(\displaystyle \int_{-7}^{\infty} \frac{1}{x^2 + 4x + 29} dx = \displaystyle \int_{-7}^{1} \frac{1}{x^2 + 4x + 29} dx + \displaystyle \int_{1}^{\infty} \frac{1}{x^2 + 4x + 29} dx\) is convergent.
  
  Since \(x^2 + 4x + 29 = (x+2)^2 + 25,\) then \(\displaystyle \int_{-7}^{b} \frac{1}{x^2 + 4x + 29} dx = \displaystyle \int_{-7}^{b} \frac{1}{(x+2)^2 + 25} dx = \frac{1}{5} \displaystyle \int_{-7}^{b} \frac{1}{\left(\frac{x+2}{5}\right)^2 + 1} dx.\)
  
  Let \(t = \frac{x+2}{5}, \quad \frac{1}{5} dx = dt,\) then \(\displaystyle \int_{-7}^{b} \frac{1}{x^2 + 4x + 29} dx = \frac{1}{5} \displaystyle \int_{-1}^{\frac{b+2}{5}} \frac{1}{t^2 + 1} dt = \frac{1}{5} \arctan \left( \frac{x+2}{5} \right) \Bigg|_{-7}^{b}.\)
  
  Then \(\displaystyle\lim_{b \to \infty}\int_{-7}^{b}\frac{1}{x^{2}+ 4x + 29}dx = \lim_{b \to \infty}\left(\frac{1}{5}\arctan \left(\frac{b+2}{5}\right) - \frac{1}{5}\arctan (-1)\right) = \frac{\pi}{10}+ \frac{1}{5}\times \frac{\pi}{4}= \frac{3 \pi}{20}.\)
Determine if the following integrals converge or diverge. If the integral converges, determine its value.
1. \(\displaystyle \int_{1}^{5}xe^{3x}\, dx\)
  
  Since \(e^{3x} > 1\), then \(x \leq x e^{3x}\) when \(x \in [1, 5]\). And we know \(\displaystyle \int_1^5 x \, dx\) is convergent, then by comparison test \(\displaystyle \int_1^5 x e^{3x} \, dx\) is convergent.
  
  Let \(x = u\), \(e^{3x} = v'\), \(u' = 1\), \(v = \frac{1}{3} e^{3x}\). Then \(\displaystyle \int_1^5 x e^{3x} \, dx = \frac{x}{3} e^{3x} \Big|_1^5 - \displaystyle \int_1^5 \frac{1}{3} e^{3x} \, dx = \frac{5}{3} e^{15} - \frac{1}{3} e^3 - \frac{1}{3} \displaystyle \int_1^5 e^{3x} \, dx.\)
  
  And \(\displaystyle \int_1^5 e^{3x} \, dx = \frac{1}{3} e^{3x} \Big|_1^5 = \frac{1}{3} e^{15} - \frac{1}{3} e^3,\) then \(\displaystyle \int_1^5 x e^{3x} \, dx = \frac{15}{9} e^{15} - \frac{3}{9} e^3 - \frac{1}{9} e^{15} + \frac{1}{9} e^3 = \frac{14}{9} e^{15} - \frac{2}{9} e^3.\) 2. \(\displaystyle \int_{0}^{1}\frac{\ln(x)}{\sqrt[3]{x}}\, dx\)
  
  Let \(\ln(x) = u\), \(x^{-\frac{1}{3}}= v'\), \(u' = \frac{1}{x}\), \(v = \frac{3}{2} x^{\frac{2}{3}}\). Then \(\displaystyle \int_{0}^{1} \frac{\ln(x)}{\sqrt[3]{x}}\, dx = \frac{3}{2} x^{\frac{2}{3}} \ln(x) \Big|_{0}^{1} - \displaystyle \int_{0}^{1} \frac{3}{2} x^{\frac{2}{3}}\cdot \frac{1}{x} \, dx =- \frac{3}{2} \displaystyle \int_{0}^{1} x^{-\frac{1}{3}}\, dx = -\frac{3}{2} \left( \frac{3}{2} x^{\frac{2}{3}} \Big|_{0}^{1} \right) = -\frac{9}{4}.\) 3. \(\displaystyle \int_{0}^{1}\ln(x) \, dx\)
  
  Consider \(\ln(x) = u\), \(v' = 1\), then \(u' = \frac{1}{x}\), \(v = x\).
  
  Then \(\displaystyle \int_b^1 \ln(x) \, dx = x \ln(x) \Big|_b^1 - \displaystyle \int_b^1 x \cdot \frac{1}{x} \, dx = x \ln(x) \Big|_b^1 - (1 - b) = -b \ln b - 1 + b.\)
  
  Then \(\lim_{b \to 0^+} \displaystyle \int_b^1 \ln(x) \, dx = \displaystyle \int_0^1 \ln(x) \, dx = \lim_{b \to 0^+} b (1 - \ln b) - 1 = -1.\)