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?

• Take a look at chapter 6 of Baby Rudin. Nov 18, 2016 at 1:02

$${\displaystyle {\frac {d}{dx}}{\Big (}u(x)v(x){\Big )}=v(x){\frac {d}{dx}}\left(u(x)\right)+u(x){\frac {d}{dx}}\left(v(x)\right).\!}$$ 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.