Question

I need someone to walk me through this integral problem:

Answer 1

\[\int\limits_{0}^{3x} \sqrt{4+t^2 dt}\]

Answer 2

I am supposed to find f'(x) but I am not sure what to do.

Answer 3

since you only need f'(x) the actual integrating is irrelevant
assume the antiderivative of sqrt(4+t^2) exists, call it G(t)

\[f(x) = |_0^{3x} G(t) = G(3x) - G(0)\]
differentiate with respect to x
\[f'(x) = g(3x) - 0\]
\[f'(x) = \sqrt{4 + (3x)^2}\]

Answer 4

So I don't have to integrate it, I just have to put 3x and 0 into f'(x)=g(3x)-0

Answer 5

yes

Answer 6

Do I do that with all of the integral problems that I am taking the derivative of?

Answer 7

i was going to integrate it but no need
I just been to look up my notes as I had forgotten how to integrate this one

Answer 8

oh crap i forgot about the chain rule

you also have to multiply by derivative of "3x"

\[** f'(x) = 3 \sqrt{4 + (3x)^2}\]

Answer 9

Is finding the derivative of an integral the same thing as finding that ant-derivative?

Answer 10

no the anti derivative is same as the indefinite integral

so taking derivative of integral leaves you with original function
however when there are limits on the integral it may leave you with a slightly different function based on the limits

Answer 11

Okay so the derivative of the integral of f sin(x) is just f sin(x).

Answer 12

The integral is f -cos(x) and so the derivative of that is f sin(x)?

Answer 13

yes

Answer 14

What is the point to taking the derivative of the integral?

Answer 15

however
\[\frac{d}{dx} \int\limits_0^{3x} \sin t dt = 3 \sin (3x)\]

Answer 16

Thank you so much! You have been very helpful!

Answer 17

yw glad to help

Answer 18

I think I will be able to do the next few problems. :)