Question

If F(x)=integral of sqrt(t^3+1) dt from 0 to x, then F'(2)=?

Answer 1

\[F(x)=\int_0^x\sqrt{t^3+1}dt\\
f'(x)=\sqrt{x^3+1}{d\over dx}(x)=?\]

Answer 2

this is differentiation under the integral scheme

Answer 3

This is one of the rare instances that we can apply the first fundamental theorem of Calculus- near-directly.
Allow me to state it for you...
If \[\huge F(x) = \int_a^xf(t)dt\]
Then
\[\huge F'(x) = f(x)\]

Answer 4

Of course, this assumes the continuity of f(t) over some interval containing [a,x] but that's a formality. Care to proceed, @Idealist ?

Answer 5

The derivative I got is 3t^2/2sqrt(t^3+1) but do I do with that? Do I plug 2 into the derivative?

Answer 6

No, see,
\[\huge F(x)= \int_a^xf(t)dt =\int_0^x\sqrt{t^3 +1} \ dt \]
So in this set up, what is f(t) ?

Answer 7

I don't understand what you wrote in the formula, sorry.

Answer 8

Okay, I hope you can see this correspondence...
\[\huge F(x)= \int_a^x\color{red}{f(t)}dt =\int_0^x\color{red}{\sqrt{t^3 +1}} \ dt\]

Answer 9

Wait a minute, so don't you just plug 2 into sqrt(t^3+1)?

Answer 10

Yes, quite so :)

Answer 11

Thank you. I'm sorry that I took a bit to understand.

Answer 12

No worries :)