Question

Calculate the area between the curve y=3x-x^2 and the axis between x=10 and x=6?

Answer 1

`int_6^10(3x -x^2)dx = -496/3`

Explanation:

In order to calculate the area bound by a curve, the `x`-axis and two lines with equations `x=a` and `x=b`, we integrate the curve between `a` and `b`.

`int_6^10 3x-x^2dx = 3int_6^10 x dx -int_6^10 x^2 dx = [3/2x^2-1/3x^3]_6^10 = 3/2(10)^2-1/3(10)^3-(3/2(6)^2-1/3(6)^3) = -496/3`