5E: Exercises

Exercise $\PageIndex{1}$

$\displaystyle\int_{-1}^{4} \frac{x}{\sqrt{x+4}} \, dx$

Answer

Let $u= x+4$, then $du=dx$ and $x=u-4$.

Now $\displaystyle\int_{-1}^{4} \frac{x}{\sqrt{x+4}} \, dx = \displaystyle \int_{-1}^{4} \frac{u-4}{\sqrt{u}} du$

$= \displaystyle \int_{-1}^{4} {(u-4)}u^{\frac{-1}{2}} du$

$=\displaystyle \int_{-1}^{4} u^{\frac{1}{2}} -4u^{\frac{-1}{2} }du$

$=\left(\displaystyle\frac{2}{3} u^{\frac{3}{2}} -4 \frac{2}{1}u^{\frac{1}{2}} \right) \left |_{x=-1}^{x=4} \right.$

$= \left(\displaystyle\frac{2}{3} (x+4)^{\frac{3}{2}} -8 (x+4)^{\frac{1}{2}} \right) \left |_{x=-1}^{x=4} \right.$

$=\frac{2}{3} ( 8^{\frac{3}{2} }- 3 ^{\frac{3}{2}} )-8( 8^{\frac{1}{2}}- 3^{\frac{1}{2}} )$

$=6\sqrt{3}-\frac{16}{3}\sqrt{2}.$

Exercise $\PageIndex{2}$

$\displaystyle\int_{0}^{1} x^3\sqrt{x^2+3}\, dx$

Answer

Let $u= x^2+3$, then $du=2xdx$ and $x^2=u-3$. Now,

$\displaystyle\int_{0}^{1} x^3\sqrt{x^2+3} \, dx = \displaystyle \frac{1}{2}\int_{0}^{1} x^2\sqrt{x^2+3}\, \,2xdx$

$= \displaystyle \frac{1}{2}\int_{0}^{1} (u-3) \sqrt{u}\, du)$

$= \displaystyle \frac{1}{2}\int_{0}^{1} (u^{\frac{3}{2}}-3 u^{\frac{-1}{2} }) \, du)$

$= \displaystyle \frac{1}{2} ( \frac{2}{5} u^{\frac{5}{2}}-3 \frac{2}{1} u^{\frac{1}{2} }) \left |_{x=0}^{x=1} \right.)$

$= \displaystyle \frac{1}{2} ( \frac{2}{5} (x^2+3)^{\frac{5}{2}}-3 \frac{2}{1} (x^2+3)^{\frac{1}{2} }) \left |_{x=0}^{x=1} \right.)$

$= \displaystyle \frac{1}{2} \left( \frac{2}{5} ( (1^2+3)^{\frac{5}{2}} -(0^2+3)^{\frac{5}{2}} \right)-6\left((1^2+3)^{\frac{1}{2} }-(0^2+3)^{\frac{1}{2} } \right)$

$= \displaystyle \frac{6\sqrt{3}}{5} -\frac{8}{5}$.

Exercise $\PageIndex{3}$

$\displaystyle\int_{1}^{\sqrt{2}} \frac{x}{x^4+3}\, dx$

Answer

Let $u= x^2$, then $du=2xdx$. Now,

$\displaystyle\int_{1}^{\sqrt{2}} \frac{x}{x^4+3}\, dx=\displaystyle \frac{1}{2}\int_{1}^{\sqrt{2}} \frac{1}{u^2+3}\, du$

$=\displaystyle \frac{1}{2 \sqrt{3}} \tan^{-1} \left( \frac{u}{\sqrt{3}} \right) \left|_{x=1}^{x=\sqrt{2}} \right.$

$=\displaystyle \frac{1}{2 \sqrt{3}} \tan^{-1} \left( \frac{x^2}{\sqrt{3}} \right) \left|_{x=1}^{x=\sqrt{2}} \right.$

$=\displaystyle \frac{1}{2 \sqrt{3}} \left(\tan^{-1} \left( \frac{(\sqrt{2})^2}{\sqrt{3}} \right)-\tan^{-1} \left( \frac{1^2}{\sqrt{3}}\right) \right)$

$=\displaystyle \frac{1}{2 \sqrt{3}} \left(\tan^{-1} \left( \frac{2}{\sqrt{3}} \right)-\tan^{-1} \left( \frac{1}{\sqrt{3}}\right) \right)$

$=\displaystyle \frac{1}{2 \sqrt{3}} \left(\tan^{-1} \left( \frac{2}{\sqrt{3}} \right)-\frac{\pi}{6} \right)$

$=\displaystyle \frac{\pi-6\tan^{-1} \left( \frac{2}{\sqrt{3}} \right)}{12\sqrt{3}}$.

Exercise $\PageIndex{4}$

$\displaystyle\int_{1}^{\sqrt{2}} xe^{-x^{2}}\, dx$

Answer

Let $u= -x^2$, then $du=-2xdx$. Now,

$\displaystyle\int_{1}^{\sqrt{2}} xe^{-x^{2}}\, dx= \displaystyle \frac{-1}{2}\int_{1}^{\sqrt{2} } e^u du$

$= \displaystyle \frac{-1}{2}e^u \left|_{x=1}^{x=\sqrt{2}} \right.$

$= \displaystyle \frac{-1}{2} e^{-x^{2}} \left|_{x=1}^{x=\sqrt{2}} \right.$

$= \displaystyle \frac{-1}{2} \left( e^{-(\sqrt{2})^{2}} - e^{-1}\right)$

$= \displaystyle \frac{-1}{2} \left( e^{-2} - e^{-1}\right)$

$= \displaystyle \frac{e-1}{2e^2}$.

Exercise $\PageIndex{5}$

$\displaystyle\int \frac{\sin(3x)}{e^{1+\cos(3x)} }\, dx$.

Answer

Let $u=e^{-(1+\cos(3x))}$, then $du= 3e^{-(1+\cos(3x))} \sin(3x) dx$. Now,

$\displaystyle\int \frac{\sin(3x)}{e^{1+\cos(3x)} }\, dx= \frac{1}{3} \int du= \frac{u}{3}+C = \frac{e^{-(1+\cos(3x))}}{6} +C.$