Question
$$\int_0^1 {\left[ {f\left( x \right)g''\left( x \right) - f''\left( x \right)g\left( x \right)} \right]} dx$$ is equal to :
[Given $$f\left( 0 \right) = g\left( 0 \right) = 0$$ ]
A.
$$f\left( 1 \right)g\left( 1 \right) - f\left( 1 \right)g'\left( 1 \right)$$
B.
$$f\left( 1 \right)g'\left( 1 \right) + f'\left( 1 \right)g\left( 1 \right)$$
C.
$$f\left( 1 \right)g'\left( 1 \right) - f'\left( 1 \right)g\left( 1 \right)$$
D.
none of these
Answer :
$$f\left( 1 \right)g'\left( 1 \right) - f'\left( 1 \right)g\left( 1 \right)$$
Solution :
Integrating by parts.
$$\eqalign{
& \int {f\left( x \right)g''\left( x \right)dx - \int {f''\left( x \right)g\left( x \right)} } dx \cr
& = f\left( x \right)g'\left( x \right) - \int {f'\left( x \right)g'\left( x \right)dx - f'\left( x \right)g\left( x \right) + } \int {f'\left( x \right)g'\left( x \right)dx} \cr
& = f\left( x \right)g'\left( x \right) - f'\left( x \right)g\left( x \right) \cr} $$
Hence, $$\int_0^1 {\left[ {f\left( x \right)g''\left( x \right) - f''\left( x \right)g\left( x \right)} \right]} dx$$
$$\eqalign{
& = f\left( 1 \right)g'\left( 1 \right) - f'\left( 1 \right)g\left( 1 \right) - f\left( 0 \right)g'\left( 0 \right) + f'\left( 0 \right)g\left( 0 \right) \cr
& = f\left( 1 \right)g'\left( 1 \right) - f'\left( 1 \right)g\left( 1 \right) \cr} $$