Question

How would you find the derivative of the integral if the lower limit is pi?

Answer 1

@satellite73

Answer 2

@ganeshie8

Answer 3

@oldrin.bataku

Answer 4

Would you view it differently if the pi were exchanged from some constant like 2?
\( \displaystyle -\int_{2}^{x^2} \dfrac{d}{du} \left[ u^2 + 4u \right] \ du \)

Answer 5

Yes.

Answer 6

You just take the derivative and then integrate (which undo each other), and then plug into that integrand: f(x^2) - f(2). That is clear?

Answer 7

Yep.

Answer 8

But why don't you use f(g(x) * g'(x) here?

Answer 9

Unless something is not written correctly, taking the derivative of the inner function is just power rule. d/du ( u^2 + 4u ) = 2u + 4. Then integrating we get the same function we started with, but we evaluate from u=2 to u=x^2. So we have never needed a chain rule part here.

Answer 10

I mean the 2nd part of the calculus theorem. What do you use that for?

Answer 11

I just use it in evaluating that definite integral. Our sequential derivative and antiderivative undo each other, leaving us with an antiderivative of f(u) = u^2 + 4u. Our limits of integration were u=2 and u=x^2, so f(x^2) - f(2) is the evaluated definite integral.

Answer 12

Alright. So would you always use that method?

Answer 13

With a definite integral after we had found the antiderivative, I would do so. Or at least see if that is useful for the question.

I am suspicious of what the question was asking though, as I read "derivative of an integral" in the first post and the problem I viewed was the integral of a derivative. If we were taking the derivative with respect to x afterwards, we would need some more work.

Answer 14

I don't get the integral of a derivative or a derivative of an integral.

Answer 15

i think the derivative sign should be outside of the integral
it is just a guess, but that is my feeling
maybe it doesn't make any difference
are you looking for the derivative of the integral?

Answer 16

\[\frac{d}{dx}\int _a^xf(t)dt=f(x)\] for example
\[\frac{d}{dx}\int_0^x\sin(t)dt=\sin(x)\] and by the chain rule
\[\frac{d}{dx}\int_0^{x^2}\sin(t)dt=\sin(x^2)\times 2x\]

Answer 17

what @AccessDenied said