Definate Integral Help

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August 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.

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

It wants to know $f'(x)$ . Wouldn't that just be $x^2$ 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 $x^2$ .
August 26th 2009, 01:58 PM
Random Variable

let $F(x) = \int^{x}_{3} t^{2}dt$

then $\frac{d}{dx} \Big( \int^{x^{3}}_{3} t^{2} dt \Big) = F'(x^{3}) \frac{d}{dx} x^{3} = x^{6} \cdot 3x^{2} = 3x^{8}$
August 26th 2009, 02:08 PM
Random Variable

In general, if you have $\int^{v(x)}_{u(x)}f(t)dt$

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

where F(x) is an antiderivative of f(t)
August 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:

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

From there, take the derivative:

$\frac{d}{dx}\bigg(\frac{x^9}{3} - \frac{27}{3}\bigg) = \frac{9x^8}{3} - 0 = 3x^8$
August 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.

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

It wants to know $f'(x)$ . Wouldn't that just be $x^2$ 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 $x^2$ .

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

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