Integration by parts for Riemann integrable fucntion Suppose $f$ is Riemann integrable on $[a,b]$ and $g$ be continuously differentiable on $[a,b]$. 
Define $$F(x):= \int_a ^x f(t) dt,\ \ \forall x\in[a,b]$$
We know $F$ is continuous hence Riemann integrable.
Is the following true? 
$$\int_c^dg(t)f(t)dt = [g(d)F(d)-g(c)F(c)]- \int_c^d g'(t)F(t)dt\\for\ \ d,c\in[a,b]$$
i.e. can I use some "sort" of integration by parts?
Thanks in advance.
p.s.
I have no exposure to measure theory.
 A: Strong sufficient conditions for integration-by-parts found in calculus books are $F$ and $g$ have continuous derivatives.  If $f$ is continuous we can be sure that $F' = f$ and it follows that for Riemann integrals
$$\tag{*}\int_c^dg(t) f(t) \, dt = g(d)F(d) - g(c)F(c) - \int_c^d F(t) g'(t) \, dt.$$
Weaker conditions for Lebesgue integrals are that $F$ and $g$  are absolutely continuous in which case (*) is true for Lebesgue integrals.
Assuming you want to work with Riemann integrals, the result is also true with your conditions.  We have to be careful because the assumption is only that $f$ is Riemann integrable.
Here is a way to prove this using Riemann-Stieltjes integration and the more general integration-by-parts result for such integrals.  Since $F$ has bounded variation and $g$ is continuous we have that $g$ is R-S integrable with respect to $F$ and 
$$\tag{**}\int_c^d g(t) f(t) \, dt = \int_c^dg dF = g(d)F(d) - g(c) F(c) - \int_c^d F \, dg.$$
Since $g$ is continuously differentiable your result follows. Using the mean value theorem of differential calculus it can be shown that
$$\int_c^dF\, dg = \int_c^d F(t) g'(t) \, dt.$$
A bit more work is needed to justify all of the steps.  The first equality in (**) can be proved using Riemann-Stieltjes sums.  The second equality is integration-by-parts for R-S integrals.
A: This is not really an answer as it uses Lebesgue integration.
The result is true, but takes work to prove using Riemann apparatus.
However, it is straightforward to see using the Lebesgue integral:
Since $f$ is Riemann integrable it is bounded and hence $F$ is absolutely continuous (ac.) and $F'(x) = f(x) $ for ae. $x$. Since $g$ is
$C^1$ it is ac. and hence the function $\phi(x) = g(x) F(x)$ is ac. We see that $\phi'(x) = g'(x) F(x) + g(x) f(x)$ for ae. $x$.
Hence we have the desired result (except that we are using the Lebesgue integral): $\phi(d)-\phi(c) = \int_c^d \phi'(x) dx = \int_c^d g'(x) F(x)dx + \int_c^d g(x) f(x)  dx$.
To finish, note that the functions $x \mapsto g'(x) F(x)$, $x \mapsto g(x) f(x)$ are bounded and continuous ae. hence the integrals are the same as the
Riemann integral.
A: Try integration by parts:
$$\int_c^d g(t)f(t)dt $$ 
$$=\int_a^d g(t)f(t)dt - \int_a^c g(t)f(t)dt$$
$$=\left.\left[g(t)\int f(t)dt - \int g'(t)\left\{\int f(t) dt\right\} \right]\right|_a^d - \left.\left[g(t)\int f(t)dt - \int g'(t)\left\{\int f(t) dt\right\} \right]\right|_a^c$$
$$\ldots \ldots$$
$$\ldots \ldots$$
$$= [g(d)F(d)-g(c)F(c)]- \int_c^d g'(t)F(t)dt$$
Try to fill up the gaps. Its easy.
Hope this helps you.
