Answer to Question #232485 in Calculus for Ndilloh

Question

If f(0) = g(0) = 0 and f'' and g'' are continuous, show that

the integral of f(x)g''(x)dx (from 0 to a) = f(a)g'(a) - f'(a)g(a) + the integral of f''(x)g(x)dx (from 0 to a).

Answer 1

If $f(0)=g(0)=0$

Since, f'' and g'' are continuous

$\int^a_0 f(x) g''(x)dx = [f(x) \int g''(x)dx]^a_0 - [\int[\frac{d}{dx}f(x) \int g''(x)dx]dx]^a_0 \\ = f(a)g'(a) -f(0)g'(0) -[\int[[f'(x)g'(x)]]dx]^a_0 \\ = f(a) g'(a) -[\int[f'(x) g'(x)]]^a_0 \\ = f(a)g'(a) -f'(a)g(a) + \int^a_0 f''(x) g(x)dx$