- By hypothesis, $[a,b]$ is a closed real interval and $f,g:I\to\mathbb R$ are continouosly differentiable functions.
- Applying the product rule to the function $F=fg,$ we get $$F’(x)=f’(x)g(x)+f(x)g’(x).$$
- According to the fundamental theorem of calculus, this gives us

$$\int_a^bF’(x)dx=F(x)\;\Rule{1px}{4ex}{2ex}^b_a= f(x)g(x)\;\Rule{1px}{4ex}{2ex}^b_a=\int_a^bf’(x)g(x)dx+\int_a^bf(x)g’(x)dx.$$ - It follows

$$\int_a^bf(x)g’(x)dx=f(x)g(x)\;\Rule{1px}{4ex}{2ex}^b_a -\int_{a}^{b}g(x)f’(x)dx.$$

q.e.d

| | | | created: 2020-01-08 12:06:13 | modified: 2020-01-08 12:06:13 | by: | references: [581]

(none)

[581] **Forster Otto**: “Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen”, Vieweg Studium, 1983