Answer to Question #115069 in Calculus for hah112

Question

[img]https://upload.cc/i1/2020/05/09/2hm3MA.jpg[/img]


whole question is in picture,R=4,you can sub R=4

Answer 1

(a) Let $R=4$

(i) $\int \frac{x+6}{x(x^2+6^2)}\,dx=\int\Big(\frac{6-x}{6(x^2+36)}+\frac{1}{6x}\Big)\,dx$

$=\int \frac{dx}{x^2+36}-\frac{1}{12}\int\frac{d(x^2+36)}{x^2+36}+\frac{1}{6}\int\frac{dx}{x}$

$=\frac{1}{6}\arctan(\frac{x}{6})-\frac{1}{12}\ln(x^2+36)+\frac{1}{6}\ln x+C$

(ii) $\int\sin^3x\cos^2x\,dx=\int (1-\cos^2x)\cos^2x\sin x\,dx$

$|\cos x=t, \,\sin x\,dx=-dt|$

$=\int(1-t^2)t^2(-dt)=\int(t^4-t^2)\,dt=\frac{t^5}{5}-\frac{t^3}{3}+C$

$=\frac{1}{5}\cos^5 x-\frac{1}{3}\cos^3 x+C$

(iii)$\int_0^3(x-2)\sqrt{x+1}\,dx=$

$\Big|u=\sqrt{x+1}, \, x=u^2-1,\, dx=2u\,du\Big|$

$=\int_1^2(u^2-3)2u^2\,du=\int_1^2(2u^4-6u^2)\,du=-\frac{8}{5}=-1.6$

(b) (i) $\int x^2e^{-x}\,dx=-(x^2e^{-x}-\int e^{-x} \,dx^2)$

$=-x^2e^{-x}+\int 2xe^{-x}\,dx$

$=-x^2e^{-x}-2xe^{-x}+2\int e^{-x}\,dx$

$=-x^2e^{-x}-2xe^{-x}-2e^{-x}+C$

$=-(x^2+2x+2)e^{-x}+C$

(ii) $L=\int_0^2\Big(\frac{\ln(t+1)}{t+1}\Big)^2\,dt$

$\Big|t=e^x-1,\, x=\ln(t+1),\, dx=\frac{dt}{t+1}\Big|$

$=\int_0^{\ln 3} x^2e^{-x}\,dx$

$=2-\frac{1}{3}(\ln^23+2\ln3+2)\approx 0.1986$