Question

need help on the attachment @Calculus1

Answer 1

\[f(x)=2x^2\]

Answer 2

\[f'(x)=4x\implies f(1)=4,\]\[f(1)=2.\]

Answer 3

\[y=4x-2.\]

Answer 4

\[f_{1}^{-1}(x)=\sqrt{\frac{1}{2}x}\]\[f_{2}^{-1}(x)=\frac{x+2}{4}\]

Answer 5

\[\int_{0}^{2}\int_{\sqrt{\frac{y}{2}}}^{\frac{y+2}{4}}dxdy\]

Answer 6

There it is. Can you evaluate that integral?

Answer 7

I believe so, you just have to separate both them right?

Answer 8

\[\frac{1}{4}\int_{0}^{2}ydy+\frac{1}{2}\int_{0}^{2}dy-\sqrt{\frac{1}{2}}\int_{0}^{2}\sqrt{y}dy\]yep

Answer 9

\[\frac{1}{8}\left[y^2\right]_{0}^{2}+\frac{1}{2}\left[y\right]_{0}^{2}-\frac{1}{3}\sqrt{2}\left[y^{\frac{3}{2}}\right]_{0}^{2}\]seems a bit tedious though, but it gives the right answer :/

Answer 10

There has got to be an easier method.

Answer 11

thanks again across, like the new pic of you

Answer 12

Thought it'd look more "professional." n_O Thank you.

Answer 13

"professional",lol