How do you find the area bounded by $y=6x-x^2$ and $y=x^2-2x$ ?

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Answer 1 · 2017-04-06T19:20:29

Use $A = int_a^b(y_1(x)-y_2(x))dx$ where $y_1(x) >= y_2(x)$

Find the x coordinates of endpoints of the area.

$6x - x^2 = x^2- 2x$

$0 = 2x^2-8x$

$x = 0 and x = 4$

This means that:

$a = 0 and b = 4$

Evaluate both at 2 and observe which is greater:

$y = 6(2)-(2)^2 = 8$

$y = 2^2 - 2(2) = 0$

The first one is greater so we subtract the second from the first in the integral:

$int_0^4(6x-x^2) - (x^2 - 2x)dx =$

$int_0^4(8x-2x^2)dx =$

${:4x^2-2/3x^3|_0^4 =$

$4(4)^2-2/3(4)^3- 4(0)^2 + 2/3(0)^2 = 64/3$

How do you find the area bounded by y=6x-x^2y=6x-x^2 and y=x^2-2xy=x^2-2x?