# Definate Integral Help

• Aug 26th 2009, 01:35 PM
JayGatsby
Definate Integral Help
I feel like the solution to this is painfully obviously, but I just can't seem to get it right... Any help would be greatly appreciated.

$\displaystyle f(x)=\int_{3}^{x^3}{t^2}{dt}$

It wants to know $\displaystyle f'(x)$. Wouldn't that just be $\displaystyle x^2$ then? Since the derivative cancels out the integral? What is messing me up is the limits, since I am pretty sure they make it something other than $\displaystyle x^2$.
• Aug 26th 2009, 01:58 PM
Random Variable
let $\displaystyle F(x) = \int^{x}_{3} t^{2}dt$

then $\displaystyle \frac{d}{dx} \Big( \int^{x^{3}}_{3} t^{2} dt \Big) = F'(x^{3}) \frac{d}{dx} x^{3} = x^{6} \cdot 3x^{2} = 3x^{8}$
• Aug 26th 2009, 02:08 PM
Random Variable
In general, if you have $\displaystyle \int^{v(x)}_{u(x)}f(t)dt$

$\displaystyle \frac{d}{dx} \int^{v(x)}_{u(x)}f(t)dt = F'\big(v(x)\big) \frac{d}{dx} v(x) - F'\big(u(x)\big) \frac{d}{dx} u(x)$

where F(x) is an antiderivative of f(t)
• Aug 26th 2009, 02:09 PM
eXist
Random Variable is correct. Sometimes it helps to actually integrate the function, then take the derivative if you need help understanding.

Ex:

$\displaystyle f(x) = \int_3^{x^3} t^2 \, dt = \frac{x^9}{3} - \frac{27}{3}$

From there, take the derivative:

$\displaystyle \frac{d}{dx}\bigg(\frac{x^9}{3} - \frac{27}{3}\bigg) = \frac{9x^8}{3} - 0 = 3x^8$
• Aug 26th 2009, 05:41 PM
HallsofIvy
Quote:

Originally Posted by JayGatsby
I feel like the solution to this is painfully obviously, but I just can't seem to get it right... Any help would be greatly appreciated.

$\displaystyle f(x)=\int_{3}^{x^3}{t^2}{dt}$

It wants to know $\displaystyle f'(x)$. Wouldn't that just be $\displaystyle x^2$ then? Since the derivative cancels out the integral? What is messing me up is the limits, since I am pretty sure they make it something other than $\displaystyle x^2$.

What is true is that the derivative of $\displaystyle \int_a^x f(t) dt$ is f(x). If you have $\displaystyle F(x)= \int_a^{g(x)} f(t)dt$, for some function g(x), make a "change of variable". Let u= g(x) so that this becomes $\displaystyle F(u)= \int_a^u f(t)dt$. Now we have $\displaystyle \frac{dF}{dx}= \frac{dF}{du}\frac{du}{dx}$. Since u= g(x), $\displaystyle \frac{du}{dx}= \frac{dg}{dx}$ and so $\displaystyle \frac{dF}{dx}= f(g(x))\frac{dg}{dx}$

Here, "f(t)" is $\displaystyle t^2$ and g(x) is $\displaystyle x^3$ so the derivative is $\displaystyle (x^3)^2(3x^2)= 3x^8$.