# integral problem

• Nov 30th 2006, 05:40 PM
viet
integral problem
$\displaystyle G(x) = \int_{1}^{x} xt \; dt$
find
$\displaystyle G '(x) =$

using the first fundamental theorem of calc, i got it to be x(x) = 2x.
the answer is wrong. any ideas?
• Nov 30th 2006, 05:54 PM
ThePerfectHacker
Quote:

Originally Posted by viet
$\displaystyle G(x) = \int_{1}^{x} xt \; dt$
find
$\displaystyle G '(x) =$

using the first fundamental theorem of calc, i got it to be x(x) = 2x.
the answer is wrong. any ideas?

Note that,
$\displaystyle G(x)=x\int_1^x t dt$
Then, by product rule,
$\displaystyle G'(x)=(x)\left( \int_1^x tdt \right)'+(x)' \left( \int_1^x tdt\right)$
Use fundamental theorem,
$\displaystyle G'(x)=x(x)+G(x)=x^2+G(x)$
• Nov 30th 2006, 06:09 PM
viet
so answer is $\displaystyle x^2+G(x)$
im not sure how to get $\displaystyle G(x)$, isnt it just $\displaystyle x$?
• Nov 30th 2006, 06:14 PM
ThePerfectHacker
Quote:

Originally Posted by viet
so answer is $\displaystyle x^2+G(x)$
im not sure how to get $\displaystyle G(x)$, isnt it just $\displaystyle x$?

No, because you end with after using the product rule,
$\displaystyle G'(x)=x^2+\int_1^x t dt$
But,
$\displaystyle G(x)=\int_1^x tdt$
Thus,
$\displaystyle G'(x)=x^2+G(x)$
• Nov 30th 2006, 09:56 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
No
$\displaystyle G'(x)=x(x)+G(x)=x^2+G(x)$

Which is a first order ODE for $\displaystyle G$, which has a general solution of the form:

$\displaystyle G(x)=Ae^{x}+P(x)$

where $\displaystyle P(x)$ is the second degree polynomial for which:

$\displaystyle P'(x)-P(x)=x^2$

which gives:

$\displaystyle P(x)=-x^2-2x-2$,

and so:

$\displaystyle G(x)=Ae^{x}-x^2-2x-2$

But we also have the initial condition $\displaystyle G(1)=0$ so:

$\displaystyle A=5/e$

So finaly we have:

$\displaystyle G(x)=(5/e) e^{x}-x^2-2x-2$

or:

$\displaystyle \boxed{G(x)=5 e^{x-1}-x^2-2x-2}$

and so:

$\displaystyle \boxed{G'(x)=5 e^{x-1}-2x-2}$

Now there is probably an error in there somewhere, but you should get the idea.

RonL