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.
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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!
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?
You're close (see P's reply)... Just plug in 2 for x (and lose the "dx"...)
I'm down to $\displaystyle {7*f(2)}$. Is this as far as I can go?
No, I don't know what $\displaystyle {f(x)}$. Thanks for your help.
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: $\displaystyle F''=2\int_0^x {f(u)du} + 2xf(x) + 3f(x)$
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