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Question: \(\int_{}^{}{\lbrack f(x)g"(x) - f"(x)g(x)\rbrack}dx\) is equal to...

∫[f(x)g"(x)−f"(x)g(x)]dx\int_{}^{}{\lbrack f(x)g"(x) - f"(x)g(x)\rbrack}dx is equal to

A

f(x)g′(x)\frac{f(x)}{g'(x)}

B

f′(x)g(x)−f(x)g′(x)f'(x)g(x) - f(x)g'(x)

C

f(x)g′(x)−f′(x)g(x)f(x)g'(x) - f'(x)g(x)

D

f(x)g′(x)+f′(x)g(x)f(x)g'(x) + f'(x)g(x)

Answer

f(x)g′(x)−f′(x)g(x)f(x)g'(x) - f'(x)g(x)

Explanation

Solution

∫[f(x)g"(x)−f"(x)g(x)]dx=\int_{}^{}{\lbrack f(x)g"(x) - f"(x)g(x)\rbrack dx =} ∫f(x)g"(x)dx−∫f"(x)g(x)dx\int_{}^{}{f(x)g"(x)dx - \int_{}^{}{f"(x)g(x)dx}}

=[f(x)g′(x)−∫f′(x)g′(x)dx]−[g(x)f′(x)−∫g′(x)f′(x)dx]=f(x)g′(x)−f′(x)g(x)= \lbrack f(x)g'(x) - \int_{}^{}{f'(x)g'(x)}dx\rbrack - \lbrack g(x)f'(x) - \int_{}^{}{g'(x)f'(x)dx\rbrack} = f(x)g'(x) - f'(x)g(x)