Last integral question for the day...?

Question

This one is a bit different from the other problems...so here it goes:

Given f" is continuous
f(1) =2
f'(1) = 5
f(4) = 7
f'(4) = 3

Find integral (upper: 4, lower:1) xf"(x)dx

I take it that I'm supposed to value of f"(4) and f"(1) and find the difference of them, but how do you find a function for f and f'?

All right then...

u = x
du = dx
dv = f"(x) dx
v= f'(x) ?

xf'(x) - integral f'(x)dx
xf'(x) - f(x) ?
Then what do I do? Evaluate?

Answer 1

I think you are putting the cart before the horse. I suspect that you need to do an integration by parts first, and then to evaluate each term at 4 and 1, and THEN subtract.

Answer 2

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