Let f be everywhere continuous and set
F(x) = $\displaystyle \int _0\,^x\![t \int _1\,^t\!f(u)du] dt$
Find F'(x) and F''(x)
How do I deal with the double interal? I have $\displaystyle \int _0\,^x\![t f(t)] dt$, is it the right first step?
Let f be everywhere continuous and set
F(x) = $\displaystyle \int _0\,^x\![t \int _1\,^t\!f(u)du] dt$
Find F'(x) and F''(x)
How do I deal with the double interal? I have $\displaystyle \int _0\,^x\![t f(t)] dt$, is it the right first step?
$\displaystyle F(x) = \int _0^x\left[t \int _1^t f(u)~du \right] ~dt$
let $\displaystyle G(t) = t \int _1^t f(u)~du$, then we have
$\displaystyle F(x) = \int _0^x G(t) ~dt$.
By FTC,
$\displaystyle F'(x) = G(x)$
and $\displaystyle F''(x) = \frac{d}{dx} \left(x \int _1^x f(u)~du \right)= \int _1^x f(u)~du + xf(x)$