Answer:
[tex]A = [\frac{32}{3}][/tex]
Step-by-step explanation:
Given
[tex]y_1 = 2x - x^2[/tex]
[tex]y_2 = 2x - 4[/tex]
Required
Determine the area bounded by the curves
First, we need to determine their points of intersection
[tex]2x - x^2 = 2x - 4[/tex]
Subtract 2x from both sides
[tex]-x^2 = -4[/tex]
Multiply through by -1
[tex]x^2 = 4[/tex]
Take square root of both sides
[tex]x = 2[/tex] or [tex]x = -2[/tex]
This Area is then calculated as thus
[tex]A = \int\limits^a_b {[y_1 - y_2]} \, dx[/tex]
Where a = 2 and b = -2
Substitute values for [tex]y_1[/tex] and [tex]y_2[/tex]
[tex]A = \int\limits^a_b {(2x - x^2) - (2x - 4)} \, dx[/tex]
Open Brackets
[tex]A = \int\limits^a_b {2x - x^2 - 2x + 4} \, dx[/tex]
Collect Like Terms
[tex]A = \int\limits^a_b {2x - 2x- x^2 + 4} \, dx[/tex]
[tex]A = \int\limits^a_b {- x^2 + 4} \, dx[/tex]
Integrate
[tex]A = [-\frac{x^{3}}{3} +4x](2,-2)[/tex]
[tex]A = [-\frac{2^{3}}{3} +4(2)] - [-\frac{-2^{3}}{3} +4(-2)][/tex]
[tex]A = [-\frac{8}{3} +8] - [-\frac{-8}{3} -8][/tex]
[tex]A = [\frac{-8+ 24}{3}] - [\frac{8}{3} -8][/tex]
[tex]A = [\frac{-8+ 24}{3}] - [\frac{8-24}{3}][/tex]
[tex]A = [\frac{16}{3}] - [\frac{-16}{3}][/tex]
[tex]A = [\frac{16}{3}] + [\frac{16}{3}][/tex]
[tex]A = [\frac{16 + 16}{3}][/tex]
[tex]A = [\frac{32}{3}][/tex]
Hence, the Area is:
[tex]A = [\frac{32}{3}][/tex]
