# Math Help - fairly simple integral, more interested in how to find it

1. ## fairly simple integral, more interested in how to find it

f(x)= int[1,x^3] t^3dt

f'(x)=

2. $\int_{4}^{x^{3}} t^3 dt$

differentiate x using FTC

$f '(x) = 3x^{11}$

3. Originally Posted by thedoge
f(x)= int[1,x^3] t^3dt

f'(x)=
If $u=x^3$, then by the FTC:

$\frac{df}{du}=u^3$,

and:

$\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}=u^3 (3 x^2)=3 x^{11}$

RonL