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

Question

Ezekiel Alexander · Accepted Answer

#"area" = 38/3#

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

#"area" = F(x) = int_1^3f(x)dx#

In your case, you would get

#F(x) = int_1^3(x^2 + 2)dx = int_1^3x^2dx + 2int_1^3dx#

This is equivalent to

#F(x) = (1/3x^3 + 2x + C)|_1^3#

Evaluate this for the extremes of integration to get

#F(x) = (1/3 * 3^3 + 2 * 3 + C) - (1/3 * 1^3 - 2 * 1 + C)#

#F(x) = 9 + 6 + color(red)(cancel(color(black)(C))) - 1/3 - 2 - color(red)(cancel(color(black)(C)))#

#F(x) = (27 + 18 -1 - 6)/3 = color(green)(38/3)#

Adam Adkins · Answer

undefined To find the area under the curve of $ f(x) = x^2 + 2 $ from $ x = 1 $ to $ x = 3 $, you need to integrate the function with respect to $ x $ over the given interval:

$$
\int_{1}^{3} (x^2 + 2) \, dx
$$

Integrating the function:

$$
\int_{1}^{3} (x^2 + 2) \, dx = \left[\frac{x^3}{3} + 2x\right]_{1}^{3}
$$

Evaluating the definite integral:

$$
\left(\frac{3^3}{3} + 2(3)\right) - \left(\frac{1^3}{3} + 2(1)\right) = \left(9 + 6\right) - \left(\frac{1}{3} + 2\right) = 15 - \frac{7}{3} = \frac{38}{3}
$$

So, the area under the curve $ f(x) = x^2 + 2 $ from $ x = 1 $ to $ x = 3 $ is $ \frac{38}{3} $ square units.