Question

FTC part 2 question

Answer 1

If I don't use FTC part 2 to do the following question and just integrate then take the derivative, will I always get the correct solution

Answer 2

I need to type the question, sorry, I thought I could upload it on my ipad

Answer 3

d/dx integral x^3 to 3 function is (2-t^2)dt

Answer 4

Yes, integrating and differentiated will get the correct answer.

Answer 5

upper limit is x^3 and lower limit is 3

Answer 6

Cool thanks. I will do it in that fashion

Answer 7

For example: \[
\frac{d}{dx}
\int\limits_{a}^{g(x)} f'(x)~dx= \frac{d}{dx} \bigg[f(g(x))-f(a)\bigg] = f'(g(x))g'(x)

\]

Answer 8

Well, putting the variable of integration in the limits is bad, so here is the correct way to write it: \[
\frac{d}{dt} \int\limits_{a}^{g(t)} f'(x)~dx= \frac{d}{dt} \bigg[f(g(t))-f(a)\bigg] = f'(g(t))g'(t)
\]

Answer 9

Ok Thanks, I am studying for a final exam and I am trying to do the problems in the way that makes sense to me. It makes sense to me to integrate then take the derivative

Answer 10

\int\limits_{1}^{x^3}e^{t^2}dt

Answer 11

Is this going to fail?

Answer 12

\[\frac{d}{dx}\int\limits_{1}^{x^3}e^{t^2}dt\]

Answer 13

Integrating it is difficult because the anti-derivative is not an elementary function.

Answer 14

true and I would need to use u sub and I have u but not du

Answer 15

You would have to do a Taylor series to integrate it, when in reality you don't have to integrate it to begin with.

Answer 16

learning the theorem is the best thing to do

Answer 17

I don't think my exam is on that level

Answer 18

here I found another problem

d/dx integral 0 to 3x^2 function is (t^2+4t ) dt

Answer 19

just looking for the duct tape approach

Answer 20

this one should work as well

Answer 21

Thanks for your help

Answer 22

The duct tape approach is to have some general anti-derivative and call it \(F\) because once you differentiate it, it will go away.