What is the area under the curve of #x - x^2# in the positive #(x,y)# region?
1 Answer
Oct 13, 2015
When you define an integral like this, you simply subtract the equation that is higher from the equation that is lower. We just have to find the intersections with the x-axis.
Solving for the intersections,
#x = x^2#
#x = 1, 0#
Thus,
#int_0^1 (x-x^2) - 0dx#
#= {:[x^2/2 - x^3/3]|:}_(0)^(1)#
#= [1^2/2 - 1^3/3] - [0]#
#= [3/6 - 2/6]#
#= color(blue)(1/6)#
graph{(y - x + x^2)(y)sqrt(0.25 - (x - 0.5)^2) <= 0 [-2, 2, -2, 2]}
