# How do you use the second fundamental theorem of Calculus to find the derivative of given int cos(sqrt( t))dt from [6, x]?

Mar 15, 2016

See the explanation below.

#### Explanation:

The second Fundamental Theorem of Calculus (a.k.a The Fundamental Theorem of Calculus, Part 2) tells us that

${\int}_{6}^{x} f \left(t\right) \mathrm{dt} = F \left(x\right) - F \left(6\right)$

where $F$ is an antiderivative of $f$.

That is, where $F ' \left(x\right) = f \left(x\right)$

Now, $F \left(6\right)$ is a constant, so its derivative is $0$.

Therefore,

$\frac{d}{\mathrm{dx}} \left(F \left(x\right) - F \left(6\right)\right) = \frac{d}{\mathrm{dx}} \left(F \left(x\right)\right) - \frac{d}{\mathrm{dx}} \left(F \left(6\right)\right)$

$= \frac{d}{\mathrm{dx}} \left(F \left(x\right)\right) - 0$

$= F ' \left(x\right) = f \left(x\right) = \cos \left(\sqrt{x}\right)$

Note that I did not find an expression for $F \left(x\right)$ because I was not asked to. If you are expected to do that, use integration by parts for

$2 \int {\underbrace{\sqrt{x}}}_{u} {\underbrace{\cos \left(\sqrt{x}\right) \frac{1}{2 \sqrt{x}} \mathrm{dx}}}_{\mathrm{dv}}$