So I need to find $\displaystyle\int_0^3f(x)g'(x)dx $, given that $f(x)=x^2$, and $g(x)$ satisfies the values in the table below $\left(\displaystyle\frac {d}{dx}g(x) = g'(x)\right)$.
I have tried every reasonable combination of integration by subsitution, and have tried or thought about doing integration by parts the second time, but still can't seem to figure out an answer.
EDIT: This is what I have tried:
$u=f(x), u'=f'(x)$
$v'=g'(x), v=g(x)$
$\displaystyle\int_0^3f(x)g'(x)dx=f(x)g(x)\big|_{0}^{3}-\displaystyle\int _0^3 g(x)f(x)'dx$
I can't integrate the second term on the right hand side, so by trying integration by parts again on this integral
$u=f(x), u'=f''(x)$ and $v'=g(x), v=\displaystyle\int g(x)dx$ would require me to integrate $g(x)$. Doing it the other way round (i.e. $u=g(x), u'=g'(x)...)$ would simply take me back to square one.
Also, I am not aware of the Stieltjes integral.