Question

Find the derivative of the integral from x to x^4.

Answer 1

\[f(x)=\int\limits_{x}^{x^4}t^3dt\]

Answer 2

The Second Fundamental Theorem of Calculus states:
if \[ f(x)=\int\limits_{a}^{x}g(t)dt\]
then \[f'(x)= g(x)\]

Answer 3

I separated the integral :
\[\huge f(x)= -\int\limits_{a}^{x}t^3dt+\int\limits_{a}^{x^4}t^3dt\]
But now I'm stuck.

Answer 4

\[\int\limits\limits_{x}^{x^4}t^3dt=\left[ t^4/4 \right]= {(x^4)^4-x^4}/4\]

Answer 5

That isn't explicitly using the Second Fundamental Theorem though. And it isn't correct.

Answer 6

y,

Answer 7

you aren't accounting for the fact that the bottom boundary is x and not a.

Answer 8

a is constant whereas x isn't.

Answer 9

but wat difference does it make, if u integrate where u stuck , u will ge the same expression

Answer 10

The integral will be different from 2-16 than it will be from 3-81.

Answer 11

f'(x) = -t^3 +4x^3t^3

Answer 12

Notice your variables don't match...?

Answer 13

oh, t should be x

Answer 14

How did you get that answer? It doesn't make sense

Answer 15

Can you use the equation tool? I cant read that.

Answer 16

sr, I dont know that tool =.=

Answer 17

It's right beside the draw tool T.T

Answer 18

just substitute the endpoint of x with t and reverse. then use the theorem to get answer

Answer 19

ps: I'm a member for 1 day, so, sorry about that

Answer 20

I am too. Just a minute let me try that.

Answer 21

No. That isn't right.

Answer 22

what you mean, the final answer is -x^3+4x^6 ?

Answer 23

No, it isnt.

Answer 24

I'll just ask my prof tomorrow if nobody can help me.

Answer 25

i referred to a formula, it's called Leibniz's rule