text: if f(0) = g(0) = 0 and f'' and g'' are continuous, show that:? the integral (with upper limit a  and lower limit 0) of f(x)g''(x) dx = f(a)g'(a)  - f'(a)g(a) + the integral (with upper limit a  and lower limit 0) of f''(x)g(x) dx.  this is under the section called "integration by parts". according to integration by parts int(f(x)g(x)) = f(x)int(g(x))-int(f'(x)int(g(x)) now, consider  int (f(x)g''(x) )dx =f(x)g'(x)-int(f'(x)g'(x)) since int(f'(x)) = f(x) =f(x)g'(x)- { f'(x)g(x) - int(f''(x)g(x)) } limit from 0 to a ={f(x)g'(x) - f'(x)g(x) + int(f''(x)g(x)) } limit from 0 to a ={f(a)g'(a) - f'(a)g(a)} - {f(0)g'(0) - f'(0)g(0)} + {int(f''(x)g(x)) } limit from 0 to a ={f(a)g'(a) - f'(a)g(a)} + {int(f''(x)g(x)) } limit from 0 to a since f(0) = g(0) = 0  thats the proof.