
# 4E: Exercises


## Exercise $$\PageIndex{1}$$

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

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 $$\PageIndex{2}$$

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

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 $$\PageIndex{3}$$

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

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 $$\PageIndex{4}$$

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

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 $$\PageIndex{5}$$

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

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