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If f(0)=f(1) = 0 and f'' exists
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If f(0)=f(1) = 0 and f'' exists, show that \[\int\limits_{0}^{1}f''(x)f(x)dx =-\int\limits_{0}^{1}[f'(x)]^2dx\]
Do Integration by parts \[ \int_0^1 u dv =\left [ uv \right]_0^1- \int_0^1 v du \]
Put \[ dv = f''(x) dx;\quad u=f(x) \\ v= f' ;\quad du =f'(x) dx\\ \int\limits_{0}^{1}f''(x)f(x)dx = \left[ f(x) f'(x) \right]_0^1-\int\limits_{0}^{1} f'(x) f'(x)dx\\ \int\limits_{0}^{1}f''(x)f(x)dx =0-\int\limits_{0}^{1} f'(x) f'(x)dx\\ \int\limits_{0}^{1}f''(x)f(x)dx = -\int\limits_{0}^{1} [f'(x)]^2 dx \]
Nice one ^.^
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