# What is the area under the curve in the interval [0, 3] for the function y=x^2+1?

Jun 26, 2018

$= 12$

#### Explanation:

For this we must use integration

Here we are trying to find:

${\int}_{0}^{3} \left({x}^{2} + 1\right) \mathrm{dx}$

Use the reverse power rule:

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + \text{constant }$

$= {\left[{x}^{2 + 1} / \left(2 + 1\right) + {x}^{1} / \left(0 + 1\right)\right]}_{0}^{3}$

$= {\left[\frac{1}{3} {x}^{3} + x\right]}_{0}^{3}$

If we are trying to find ${\left[F \left(x\right)\right]}_{a}^{b}$

We just do $F \left(b\right) - F \left(a\right)$

$\implies \left\{\left(\frac{1}{3} \cdot {3}^{3}\right) + 3\right\} - \left\{\left(\frac{1}{3} \cdot {0}^{3}\right) + 0\right\}$

$= 12$