# What is the area under the curve f(x)=x^2+2 from 1 to 3?

Aug 12, 2015

$\text{area} = \frac{38}{3}$

#### Explanation:

To determine the area under the curve for this function you need to integrate it on the interval $\left[- 1 , 3\right]$

$\text{area} = F \left(x\right) = {\int}_{1}^{3} f \left(x\right) \mathrm{dx}$

In your case, you would get

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

This is equivalent to

$F \left(x\right) = \left(\frac{1}{3} {x}^{3} + 2 x + C\right) {|}_{1}^{3}$

Evaluate this for the extremes of integration to get

$F \left(x\right) = \left(\frac{1}{3} \cdot {3}^{3} + 2 \cdot 3 + C\right) - \left(\frac{1}{3} \cdot {1}^{3} - 2 \cdot 1 + C\right)$

$F \left(x\right) = 9 + 6 + \textcolor{red}{\cancel{\textcolor{b l a c k}{C}}} - \frac{1}{3} - 2 - \textcolor{red}{\cancel{\textcolor{b l a c k}{C}}}$

$F \left(x\right) = \frac{27 + 18 - 1 - 6}{3} = \textcolor{g r e e n}{\frac{38}{3}}$