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)
Nailed it, I know this threads been dead, but I got stuck on it at first glance:
$\displaystyle f(x)=\frac{1}{2}\int _0^x(x^2-2xt+t^2)g(t)dt$ $\displaystyle =\frac{1}{2}[x^2\int _0^xg(t)dt-2x\int _0^xtg(t)dt+\int _0^xt^2g(t)dt]$
$\displaystyle f'(x)=\frac{1}{2}[2x\int _0^xg(t)dt+x^2g(x)-2\int _0^xtg(t)dt-2x^2g(x)+x^2g(x)]=x\int _0^xg(t)dt-\int _0^xtg(t)dt$