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#