Thread: Calculating the second derivative of an integral

1. Calculating the second derivative of an integral

Question:

$\displaystyle F(x) = \int_0^x[{(2t+3)\int_2^t{f(u)du}]dt$

Calculate F''(2).

I believe:

F'(x) = $\displaystyle [{(2x+3)\int_2^x{f(u)du}]$

but I'm stuck now.

2. Re: Calculating the second derivative of an integral

Now use the product rule! You know both the derivate of (2x + 3), AND the derivative of the integral (by the FTC, which you already used)...

You're almost there!

3. Re: Calculating the second derivative of an integral

This is what I got next.

(2x+3)(f(x)dx)+2$\displaystyle \int_2^x {f(u)du}$

and the integral term becomes 0 because it is over the interval 2 to 2, but what happens in the remaining term?

4. Re: Calculating the second derivative of an integral

You're close (see P's reply)...
Just plug in 2 for x (and lose the "dx"...)

5. Re: Calculating the second derivative of an integral

I'm down to $\displaystyle {7*f(2)}$. Is this as far as I can go?

6. Re: Calculating the second derivative of an integral

No, I don't know what $\displaystyle {f(x)}$. Thanks for your help.

7. Re: Calculating the second derivative of an integral

Originally Posted by CountingPenguins
Question:

$\displaystyle F(x) = \int_0^x[{(2t+3)\int_2^t{f(u)du}]dt$
Calculate F''(2).
I believe:
F'(x) = $\displaystyle [{(2x+3)\int_2^x{f(u)du}]$
Correct now:

$\displaystyle F''=2\int_0^x {f(u)du} + 2xf(x) + 3f(x)$