# Thread: Find F'(x) and F''(x)

1. ## Find F'(x) and F''(x)

Let f be everywhere continuous and set
F(x) = $\int _0\,^x\![t \int _1\,^t\!f(u)du] dt$
Find F'(x) and F''(x)

How do I deal with the double interal? I have $\int _0\,^x\![t f(t)] dt$, is it the right first step?

2. Yes, that's the first step.

Now is just a matter of simple application of the FTC.

3. Originally Posted by Krizalid
Yes, that's the first step.

Now is just a matter of simple application of the FTC.

So, is the answer F'(x) = xf(x), then what is F''(x)?

4. Originally Posted by 450081592
Let f be everywhere continuous and set
F(x) = $\int _0\,^x\![t \int _1\,^t\!f(u)du] dt$
Find F'(x) and F''(x)

How do I deal with the double interal? I have $\int _0\,^x\![t f(t)] dt$, is it the right first step?
$F(x) = \int _0^x\left[t \int _1^t f(u)~du \right] ~dt$

let $G(t) = t \int _1^t f(u)~du$, then we have

$F(x) = \int _0^x G(t) ~dt$.

By FTC,

$F'(x) = G(x)$

and $F''(x) = \frac{d}{dx} \left(x \int _1^x f(u)~du \right)= \int _1^x f(u)~du + xf(x)$