Question

Ignore the chicken scratch... but how do you do #10

Answer 1

to the risk of sounding redundant, bearing in mind the posting above by @LammettHash is terrific.

so the region is that in the picture below.

$\bf \displaystyle\int\limits_{0}^{2}~e^{\frac{x}{2}}\cdot dx \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using some substitution}}{u=\cfrac{x}{2}\implies u=\cfrac{1}{2}x\implies} \cfrac{du}{dx}=\cfrac{1}{2}\implies 2du=dx \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int\limits_{0}^{2}~e^{u}\cdot 2du\implies 2\int\limits_{0}^{2}~e^u\cdot du \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{changing the bounds}}{u(0)=\cfrac{0}{2}}\implies u(0)=\boxed{0}~\hspace{7em} u(2)=\cfrac{2}{2}\implies u(2)=\boxed{1} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle 2\int\limits_{0}^{1}~e^u\cdot du\implies \left. 2e^u\cfrac{}{} \right]_{0}^{1}\implies 2e-2$

Answer 2

The region bounded by $y=e^{x/2}$ and $x=2$ in the first quadrant is the set

$\{(x,y)\mid0\le x\le2,0\le y\le e^{x/2}\}$

The area would be given by the integral

$\displaystyle\int_{x=0}^{x=2}e^{x/2}\,\mathrm dx$

You can substitute $x=2w$, so that $\mathrm dx=2\,\mathrm dw$. Then when $x=0$, you also have $w=0$; when $x=2$, you have $w=1$. Then the integral is equivalent to

$\displaystyle2\int_{w=0}^{w=1}e^w\,\mathrm dw=2e^w\bigg|_{w=0}^{w=1}=2(e^1-e^0)=2e-2$

so the answer would be A.