# FTC2 - Not sure how to deal with this one

• Mar 31st 2011, 07:13 PM
joatmon
FTC2 - Not sure how to deal with this one
Find g'(x) by using Part 2 of the Fundamental Theorem of Calculus.

$g(x) = \int^x_0 (1+\sqrt{t})dt$

Here is what I have done so far:

$g(x) = t + \frac{1}{2\sqrt{t}}]^x_0$
$g(x) = x + \frac{1}{2\sqrt{x}} - (0 + \frac{1}{2\sqrt{0}})$

Assuming that these two steps are correct (are they?), how do handle the fact that $\frac{1}{2\sqrt{0}}$ is undefined?

Thanks!
• Mar 31st 2011, 07:23 PM
pickslides
Those steps aren't quite right, consider

$\displaystyle \int t^{\frac{1}{2}} ~dt= \frac{2}{3}\times t^{\frac{3}{2}}+C$
• Mar 31st 2011, 07:28 PM
TheChaz
Obviously, you are able to compute the integral and then take the derivative. This will not always be the case, and eventually (!) you will become familiar with the following:
$g(x) = \int^x_a f(t)dt = f(x)$

Oh... it's on wikipedia!
Fundamental theorem of calculus - Wikipedia, the free encyclopedia

Just replace the "t" inside the integral with x.
If your upper limit has an "interesting" derivative, you'll need to use the chain rule.
• Mar 31st 2011, 08:18 PM
joatmon
I took the derivative instead of the antiderivative. Duh. Thanks, pickslides, for pointing that out. Chaz, the question specifically requires that I use part 2 of the FTC. Your post uses part 1 (at least that's what our book calls them). The question actually makes us do it both ways. I got the answer easily using part 1, but when I did it with part 2, I got hung up. Now I know why.

Thanks to both of you.