Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

solving for g'(x) when g(x)=integral from sqrt(x) to x^2, (sin t)(sqrt t) dt

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

derivative of the integral is the integrand, plus chain rule replace \(t\) by \(x^2\) and multiply by the derivative of \(x^2\) get \[\sin(x^2)\sqrt{x^2}\times 2x\] for the top part bottom part is the same, but make it negative
I think I am suppose to break it up. integral from sqrt x to 0 plus 0 to x^2. But i have no idea why this is the way to do it
yes but you can break up anyway you like

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

the top part gives what i wrote above, the bottom part is analogous, just make it negative
the bottom can be any number??
I know I'm suppose to use the second fundamental theorem
yes \[\int_{\sqrt{x}}^{x^2}f(t)dt=\int_{\sqrt{x}}^af(t)dt+\int_a^{x^2}f(t)dt\]
My teacher chose 0 though. Is there a reason for that? should I just choose 0 everytime?
take the derivative of the second part, get \(f(x^2)\times 2x\)
the derivative of the first part is \(-f(\sqrt{x})\times \frac{1}{2\sqrt{x}}\)
if you like to pick zero, go ahead a pick zero makes no difference
Ok I understand it now. Thanks!!!! :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question