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

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

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

so the answer would be A.

jdoe0001 jdoe0001 · Answer 2 · 2019-01-18T23:15:15+01:00

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.

[tex]\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[/tex]

[tex]\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[/tex]