Question

how can i find the derivative of
r(x) = \int\limits_{\sqrt{x}}^{x^3}\sqrt{t} \sin(t) dt

Answer 1

\[r(x) = \int\limits\limits_{\sqrt{x}}^{x^3}\sqrt{t} \sin(t) dt\]

Answer 2

the derivative of an integral is simply that function.

Answer 3

and dont forget wolfram alpha integrator is your friend!
http://www.wolframalpha.com/input/?i=integral

Answer 4

Yeah, thank! But i didn't understood this question. So the derivative of an integral is...?

Answer 5

you have a function f(x)
when you integrate it you get g(x)
so the derivative of g(x) is just f(x)
in this problem f(x) is that t sin(t) crap insid ethe integral

Answer 6

fundamental theorem its called or somethign i think.

Answer 7

yes, unsual question, the question actually shows an integral(and a definite integral at that, although it does use x's as the upper and lower bounds), so you would normally solve this to find the original function then find derivative of that function

Answer 8

\[\frac{d}{dx}\int\limits_{a(x)}^{b(x)}g(t)dt=g(b(x))b'(x)-g(a(x))a'(x)\]

Answer 9

Integration by parts is used for products.

Answer 10

Thanks a lot for your answers! You all are true heroes!

Answer 11

yes im sorry i made a mistake you have to solve integral using integratino by parts then plug in the lower and upper bounds for the definite integral then find the derivative of teh final function

Answer 12

just use the formula I provided...no need to integrate

Answer 13

\[\int\limits u.dv = uv - \int\limits v.du\]

Answer 14

zarkon does that formula use the chain rule or somethign? I'm trying to remember all this stuff for a differential equation class.

Answer 15

chain rule with the fundamental theorem of calculus

Answer 16

ty haha imma keep an eye out for you later this year

Answer 17

Zarkon what is the name of that equation?

Answer 18

I don't know that it has a name

Answer 19

it is a generalization of

\[\frac{d}{dx}\int\limits_{a}^{x}f(t)dt=f(x)\]

Answer 20

ok! thanks!