Could someone help me with this problem?
It involves u-substitution I think

Question

Could someone help me with this problem?
It involves u-substitution I think

anonymous · Answer

$$\frac{ d }{ dx } \int\limits_{2}^{x^2}e ^{x^3}$$

anonymous · Answer

*dx at the end

freckles · Answer

$$\int\limits_{a(x)}^{b(x)}f(t) dt=F(t)|_{a(x)}^{b(x)}=F(b(x))-F(a(x))$$
we know this right?

freckles · Answer

where F'=f

anonymous · Answer

Yeah thats the fundamental theorem correct?

freckles · Answer

now if we differentiate this we have:
$$\frac{d}{dx}(F(b(x))-F(a(x))=b'(x) \cdot f(b(x))-a'(x) \cdot f(a(x))$$

freckles · Answer

by chain rule

anonymous · Answer

so no U-substitution?

freckles · Answer

your $$f(x)=e^{x^3} \ a(x)=2 \ b(x)=x^2 $$

freckles · Answer

I think you can find the following:
$$f(b(x))=? \ f(a(x))=? \ a'(x)=? \ b'(x)=?$$

freckles · Answer

I didn't use u substitution

anonymous · Answer

okay, I was just checking, thanks :) 
so f(b(x))= F(x^2)=e^(x^6)
f(a(x))= e^8
a'(x)=0
b'(x)=2x

anonymous · Answer

Okay, so for my answer, I get 2xe^6?

freckles · Answer

sorta

freckles · Answer

you are missing a thingy

anonymous · Answer

which?

anonymous · Answer

oh, + C

freckles · Answer

$$2x e^{x^6}$$

freckles · Answer

oh no definitely do not put +C

freckles · Answer

that was a definite integral
and we just took derivative of it so no C for sure

freckles · Answer

I was just talking about your exponent

anonymous · Answer

right, i see that now. What i missed was the variable in the power for the exponent

freckles · Answer

i'm sure it was a type-o though

anonymous · Answer

yeah, thanks so much for the help, I've been getting a lot of questions like that lately and didn't know where to start.

freckles · Answer

well does it make sense now?

anonymous · Answer

yes, because I was so caught up on whether I should've been using u-sub, I didn't think of other ways to go about doing the question

freckles · Answer

we didn't need a u sub because we didn't need integrate
but also I'm pretty sure the integral of e^(x^3) is a non-elementary integral

freckles · Answer

which means a simple u-sub wouldn't have worked to integrate it

anonymous · Answer

Well that explains why I was having difficulties in the first place, because I didn't realize I didn't need to sub anything, so i was trying to make it work and it couldn't

freckles · Answer

yeah you would have been trying all day and had no success with the integral :p

anonymous · Answer

Thank you!

freckles · Answer

np :)