# Assume that f is a function with a continuous second derivative f''(x) and suppose that f(1) = 2; f(4) = 6; f'(1) = 3 and f'(4) = 5. Evaluate integrate from 1 to 4 of xf''(x)dx

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You should use integration by parts such that:

`int udv = uv - int vdu`

Considering `u = x` and `dv = f''(x)` yields:

`u = x => du = dx`

`dv = f''(x)dx => v = f'(x)`

`int_1^4 xf''(x)dx = xf'(x)|_1^4- int_1^4 f'(x)dx`

`int_1^4 xf''(x)dx = xf'(x)|_1^4 - f(x)|_1^4`

`int_1^4 xf''(x)dx = 4*f'(4) - 1*f'(1) - f(4) + f(1)`

Notice that the problem provides the following values `f(1) = 2; f(4) = 6; f'(1) = 3` and `f'(4) = 5` such that:

`int_1^4 xf''(x)dx = 4*5 - 3 - 6 + 2 `

`int_1^4 xf''(x)dx = 13`

**Hence, evaluating the definite integral `int_1^4 xf''(x)dx,` under the given conditions, yields `int_1^4 xf''(x)dx = 13` .**

Using integration by parts we have

Int [xf''(x)] = [x f'(x) ] - Int(f'(x)) = 4*f'(4) - 1*f'(1) - f(4) + f(1)

= 4*5 - 1*3 - 6 + 2 =13.