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Suppose that (1)=−2 , f(4)=5, f'(1)=4 , f'(4)=-7 and f'' is continuous.

Find the value of ∫41″().

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Answer:

integrates by parts:

U = x dV = f''

dU = dx V = f'

The integral becomes:

x f' - integral f' dx

= x f' - f

limit as x->4 : 4 * f'(4) - f(4) = 4 * -3 - 7

= -12-7 = -19

limit as x->1 : 1 * f'(1) - f(1) = 1 * 9 - 8 = 1

subtracting: -19-1 = -20

Explanation:

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