Prove the integration by parts formula: $\int^b_a{f'(x)g(x)dx} = (f(b)g(b)-f(a)g(a)) - \int^b_a{f(x)g'(x)dx}$

Question

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?

Answer 1

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.

Answer 2

The poster made the final comment, "...and showing why each term is Riemann integrable?" It does not look to me like the previous answers have specifically addressed this. I think the following breakdown will answer this particular part of the question.

Consider two functions $f$ and $g$ such that $f'$ and $g'$ are both continuous. Then we can write:

$$[f\cdot g]'= f'g+fg' \quad \text{Product Rule}$$

Next, integrate both sides:

$$\int [f\cdot g]'=\int f'g+fg'$$

Importantly, the only way we can apply the following theorem: $\int \phi+\omega=\int \phi + \int \omega$ is if we know that $\phi$ and $\omega$ are individually integrable. Well, there is a theorem that shows that if $\phi$ is a continuous function, then $\int \phi$ exists. In the context of our proof, $f$' is continuous. Further $g$ is differentiable, which means $g$ is continuous. This means that the function $f'g$ is continuous. A similar argument works for $fg'$.

As such, we can then write:

$$\int[f \cdot g]'=\int f'g+\int fg'$$

By the fundamental theorem of calculus, the left side is just $f \cdot g$, which let's us write:

$$fg=\int f' g+\int f g'$$

Rearranging gives us the desired form.

Answer 3

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.