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^{(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.