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Question: \[\int_{0}^{b - c}{f^{''}(x + a)dx =}\]...

∫0b−cf′′(x+a)dx=\int_{0}^{b - c}{f^{''}(x + a)dx =}

A

f′(a)−f′(b)f^{'}(a) - f^{'}(b)

B

f′(b−c+a)−f′(a)f^{'}(b - c + a) - f^{'}(a)

C

f′(b+c−a)+f′(a)f^{'}(b + c - a) + f^{'}(a)

D

None of these

Answer

f′(b−c+a)−f′(a)f^{'}(b - c + a) - f^{'}(a)

Explanation

Solution

∫0b−cf"(x+a)dx=[f′(x+a)]0b−c=f′(b−c+a)−f′(a)\int_{0}^{b - c}{f"(x + a)dx} = \lbrack f'(x + a)\rbrack_{0}^{b - c} = f'(b - c + a) - f'(a).