A community for students.
Here's the question you clicked on:
 0 viewing
ta123
 3 years ago
need help on the attachment @Calculus1
ta123
 3 years ago
need help on the attachment @Calculus1

This Question is Closed

across
 3 years ago
Best ResponseYou've already chosen the best response.1\[f'(x)=4x\implies f(1)=4,\]\[f(1)=2.\]

across
 3 years ago
Best ResponseYou've already chosen the best response.1\[f_{1}^{1}(x)=\sqrt{\frac{1}{2}x}\]\[f_{2}^{1}(x)=\frac{x+2}{4}\]

across
 3 years ago
Best ResponseYou've already chosen the best response.1\[\int_{0}^{2}\int_{\sqrt{\frac{y}{2}}}^{\frac{y+2}{4}}dxdy\]

across
 3 years ago
Best ResponseYou've already chosen the best response.1There it is. Can you evaluate that integral?

ta123
 3 years ago
Best ResponseYou've already chosen the best response.1I believe so, you just have to separate both them right?

across
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{4}\int_{0}^{2}ydy+\frac{1}{2}\int_{0}^{2}dy\sqrt{\frac{1}{2}}\int_{0}^{2}\sqrt{y}dy\]yep

across
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{8}\left[y^2\right]_{0}^{2}+\frac{1}{2}\left[y\right]_{0}^{2}\frac{1}{3}\sqrt{2}\left[y^{\frac{3}{2}}\right]_{0}^{2}\]seems a bit tedious though, but it gives the right answer :/

across
 3 years ago
Best ResponseYou've already chosen the best response.1There has got to be an easier method.

ta123
 3 years ago
Best ResponseYou've already chosen the best response.1thanks again across, like the new pic of you

across
 3 years ago
Best ResponseYou've already chosen the best response.1Thought it'd look more "professional." n_O Thank you.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.