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