# How do you use the second fundamental theorem of Calculus to find the derivative of given int (t^2 +3t+2)dt from [-3, x]?

Feb 17, 2016

The second Fundamental Theorem of Calculus enables us to find the function defined by $g \left(x\right) = {\int}_{-} {3}^{x} \left({t}^{2} + 3 t + 2\right) \mathrm{dt}$. We will then differentiate.

#### Explanation:

Find an antiderivative of ${t}^{2} + 3 t + 2$ and evaluate from $- 3$ to $x$

g(x) = int_-3^x (t^2 +3t+2)dt = ]_-3^x

$= \left({\left(x\right)}^{3} / 3 + \frac{3 {\left(x\right)}^{2}}{2} + 2 \left(x\right)\right) - \left({\left(- 3\right)}^{3} / 3 + \frac{3 {\left(- 3\right)}^{2}}{2} + 2 \left(- 3\right)\right)$

$= {x}^{3} / 3 + \frac{3 {x}^{2}}{2} + 2 x + \frac{3}{2}$

So, $g ' \left(x\right) = {x}^{2} + 3 x + 2$

Note
We get the same answer much more quickly by using part 1 of the fundamental theorem.