Question

$\quad\int_{0}^{1} \left[f \left(x\right)g'' \left(x\right)-f''\left(x\right)g \left(x\right)\right]$ dx is equal to : [Given f(0) = g(0) = 0]

• A. f (1) g(1) -f(1) g??l)
• B. f(l) g??l) + (l) f??(l)
• C. f (l) g??l) - f??l) g(l)
• D. none of these

Answer 1

The Correct Option is C

Integrating by parts. $\quad\int f \left(x\right) g'' \left(x\right)dx-\int f''\left(x\right)g\left(x\right)dx$ $=f \left(x\right)g'\left(x\right)-\int f '\left(x\right)g'\left(x\right)dx$ $-f '\left(x\right)g\left(x\right)+\int f '\left(x\right)g'\left(x\right) dx$ $=f\left(x\right)g'\left(x\right)-f'\left(x\right)g\left(x\right)$ $Hence,\, \int_{0}^{1}f \left(x\right)g''\left(x\right)dx-\int_{0}^{1}f''\left(x\right)g\left(x\right) dx$ $=f\left(1\right)g'\left(1\right)-f'\left(1\right)g\left(1\right)-f \left(0\right)g'\left(0\right)+f'\left(0\right)g\left(0\right)$ $f\left(1\right)g'\left(1\right)-f'\left(1\right)g\left(1\right)$