Question

Find the area enclosed by the line y=4x and the curve y=x^2?

Answer 1

The area is `32/3` square units.

Explanation:

Start by finding the coordinates of intersection. These will be our bounds of integration, aka `a` and `b` in `int_a^b`.

`4x = x^2`

`0 = x^2 - 4x`

`0 = x(x -4)`

`x = 0 and 4`

We now must determine which graph lies above which. At `x = 2`, `y = 4x = 4(2) = 8` and at `x = 2, y= x^2 = 2^2 = 4`. This means that in `0 < x < 4`, we will have the line above the parabola.

Our expression to integrate, therefore, will be

`A = int_0^4 4x - x^2dx`

Where `A` represents the area we are wanting to find.

`A = [2x^2 - 1/3x^3]_0^4`

`A = 2(4)^2 - 1/3(4)^3 - (2(0)^2 - 1/3(0)^3)`

`A = 32 - 64/3`

`A = 32/3`

Hopefully this helps!