Solving Integration Question: f' Continuous on [a,b]

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I'm having major trouble with this question, can anyone assist me on this?

Let f be a function such that f' is continuous on [a,b]. Show that

\int_a^{b} f(t)f’(t) dt = 1/2 [f''(b) - f''(a)]

Hint: Calculate the derivative of F(x) = f''(x).
 
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Don't double post please
 
sorry, i thought a different forum would make a difference, since no one has an answer to my question yet.
 
trap said:
sorry, i thought a different forum would make a difference, since no one has an answer to my question yet.
Have you not read your other thread? People have been quite helpful.
 
Hi, are you sure those are second derivatives on the right hand side and not squares? Something like 1/2(f(b)^2-f(a)^2) instead?
 
i'm not sure if they are squares becoz the question reads f^2(b) - f^2(a)..so i thought they were second derivative..
 
trap said:
i'm not sure if they are squares becoz the question reads f^2(b) - f^2(a)..so i thought they were second derivative..
That is squares and I am sure about that because it is the answer. When your talking about the nth derivative you either use roman numerals or put the number in brackets.
 
Couldn't one simply integrate by parts to get the answer?
 
shmoe said:
Hi, are you sure those are second derivatives on the right hand side and not squares? Something like 1/2(f(b)^2-f(a)^2) instead?
Could you please explain how you come to that answer, I am having trouble seeing it.

Thanks.
 
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digink said:
Could you please explain how you come to that answer, I am having trouble seeing it.

Thanks.

Just use the hint applied to F(x)=(f(x))^2. Find the derivative of F(x) using the chain rule...


DeadWolfe-yes integration by parts will work fine.
 
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