## ta123 4 years ago need help on the attachment @Calculus1

1. ta123

2. across

$f(x)=2x^2$

3. across

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

4. across

$y=4x-2.$

5. across

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

6. across

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

7. across

There it is. Can you evaluate that integral?

8. ta123

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

9. across

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

10. across

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

11. across

There has got to be an easier method.

12. ta123

thanks again across, like the new pic of you

13. across

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

14. ta123

"professional",lol