Given a function g, continue to any x in R : g(1)=5 and the Integral of g(t)dt=2 in {0,1}

If f(x)=1/2(Integral of(x-t)^2)g(t)dt) in {0,x}

Prube that

f'(x)=x(integral(g(t)dt- t·g(t)dt) in {0,x}

then find f''(1) and f'''(1)

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- Nov 6th 2012, 01:09 PMnigromantefirst and second theorem
Given a function g, continue to any x in R : g(1)=5 and the Integral of g(t)dt=2 in {0,1}

If f(x)=1/2(Integral of(x-t)^2)g(t)dt) in {0,x}

Prube that

f'(x)=x(integral(g(t)dt- t·g(t)dt) in {0,x}

then find f''(1) and f'''(1) - Nov 6th 2012, 09:41 PMchiroRe: first and second theorem
Hey nigromante.

Can you show us what you have tried (partial attempts are OK) as well as any ideas you have to solve the problem?

Hint: Use the fundamental theorem of calculus to link f(x) with f'(x). - Nov 17th 2012, 02:59 PMDark SunRe: first and second theorem
Nailed it, I know this threads been dead, but I got stuck on it at first glance:

$\displaystyle f(x)=\frac{1}{2}\int _0^x(x^2-2xt+t^2)g(t)dt$ $\displaystyle =\frac{1}{2}[x^2\int _0^xg(t)dt-2x\int _0^xtg(t)dt+\int _0^xt^2g(t)dt]$

$\displaystyle f'(x)=\frac{1}{2}[2x\int _0^xg(t)dt+x^2g(x)-2\int _0^xtg(t)dt-2x^2g(x)+x^2g(x)]=x\int _0^xg(t)dt-\int _0^xtg(t)dt$ - Nov 18th 2012, 06:50 PMnigromanteRe: first and second theorem
THanks, I get the same, but now I have a problem with the third derivate, on the second I get: f''(1)=Integrate of g(t)dt, {0,1}=2, but f'''(1) I think is zero, but it supose to be "5".