# Help please? Integral question below. Don’t really understand the concept of integrals in general. Thanks!

Dec 8, 2017

This is a question involving the Fundamental Theorem of Calculus.

#### Explanation:

For $g \left(y\right) = {\int}_{3}^{y} f \left(x\right) \mathrm{dx}$, FTC part 1 gives us

$g ' \left(y\right) = f \left(y\right)$

Now we want $g ' ' \left(y\right) = f ' \left(y\right)$ where $f \left(y\right) = {\int}_{0}^{\sin} y \sqrt{1 + {t}^{2}} \mathrm{dt}$

Again, using FTC part 1, this time with the chain rule as well, we get:

$g ' ' \left(y\right) = f ' \left(y\right) = \sqrt{1 + {\sin}^{2} y} \cdot \frac{d}{\mathrm{dy}} \left(\sin y\right)$

$= \sqrt{1 + {\sin}^{2} y} \cos y$

At $y = \frac{\pi}{6}$, we get

$g ' ' \left(\frac{\pi}{6}\right) = \sqrt{1 + {\sin}^{2} \left(\frac{\pi}{6}\right)} \cos \left(\frac{\pi}{6}\right)$

$= \sqrt{1 + \frac{1}{4}} \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{15}}{4}$