Prove the integration by parts formula: $\int^b_a{f'(x)g(x)dx} = (f(b)g(b)-f(a)g(a)) - \int^b_a{f(x)g'(x)dx}$ Let $f,g : [a,b] \rightarrow \mathbb{R}$ be differentiable such that $f'.g' : [a,b] \rightarrow \mathbb{R}$ are continuous. Prove the integration by parts formula:  
$\int^b_a{f'(x)g(x)dx} = (f(b)g(b)-f(a)g(a)) - \int^b_a{f(x)g'(x)dx}$
Can someone show me how to do this with the product rule $(fg)'$, and showing why each term is Riemann integrable?
 A: Suppose u(x) and v(x) are two continuously differentiable functions. The product rule gives,
$${\displaystyle {\frac {d}{dx}}{\Big (}u(x)v(x){\Big )}=v(x){\frac {d}{dx}}\left(u(x)\right)+u(x){\frac {d}{dx}}\left(v(x)\right).\!}$$
Integrating both sides with respect to x,
$$\int {\frac {d}{dx}}\left(u(x)v(x)\right)\,dx=\int u'(x)v(x)\,dx+\int u(x)v'(x)\,dx$$
Now using the definition of indefinite integral,
$$u(x)v(x)=\int u'(x)v(x)\,dx+\int u(x)v'(x)\,dx$$
Rearranging,
$$\int u(x)v'(x)\,dx=u(x)v(x)-\int u'(x)v(x)\,dx$$
Replace $u$ by $g$ and $v$ by $f$ and then apply the limits to get your required expression.
A: This is  a proof I have written based off a similar answer on Quora, which you can view here: https://www.quora.com/What-is-a-good-intuitive-proof-of-integration-by-parts/answer/David-Rutter-2
Proof:
Note that G’(x) = g(x), and F’(x) = f(x).
We remember that by the product rule of derivatives, $$ \frac{d}{dx} F(x)G(x)=f(x)G(x)+F(x)g(x). $$
If we take the Integral from a to b of both sides, we get $$F(b)G(b)-F(a)G(a)=\int^b_a f(x)G(x) dx+ \int^b_a F(x)g(x) dx.$$
This can be rewritten as $$ \int^b_a F(x)g(x) dx=F(b)G(b)-F(a)G(a)- \int ^b_a f(x)G(x) dx, $$ as required.
