What is the area under the curve f(x)=x^2+2f(x)=x2+2 from 1 to 3?
1 Answer
Aug 12, 2015
Explanation:
To determine the area under the curve for this function you need to integrate it on the interval
"area" = F(x) = int_1^3f(x)dxarea=F(x)=∫31f(x)dx
In your case, you would get
F(x) = int_1^3(x^2 + 2)dx = int_1^3x^2dx + 2int_1^3dxF(x)=∫31(x2+2)dx=∫31x2dx+2∫31dx
This is equivalent to
F(x) = (1/3x^3 + 2x + C)|_1^3F(x)=(13x3+2x+C)∣∣∣31
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)=(13⋅33+2⋅3+C)−(13⋅13−2⋅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)
