$\displaystyle G(x) = \int_{1}^{x} xt \; dt$

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?

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- Nov 30th 2006, 05:40 PMvietintegral 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 PMThePerfectHacker
- Nov 30th 2006, 06:09 PMviet
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 PMThePerfectHacker
- Nov 30th 2006, 09:56 PMCaptainBlack
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