What is the area of the region bounded by the two curves: y = x^2 and y = |x|

Answer 1

We know that to find the (signed) area between a function curve and the x-axis, over the interval [a,b] we are supposed to evaluate ∫_[a,b]f(x)dx. There is no difference in finding the area between two curves, except that the lower function is no longer the curve y=0.

However, we do need to find the intersections of the graphs of y=x² and y=|x| in order to set the limits of integration, The graphs intersect in 3 places, (-1,-1), (0,0), and (1,1) The area between the two curves on [-1,0] is the same as the area between the curves on [0,1], and on [0,1], |x|=x ≥ x². So the requested area is:

2* ∫_[0,1](x-x²)dx = 2*(0.5x²-(1/3)x³)₀¹= 1/3

Answer 2

To find the area between two curves we need two main pieces of information: which function is the 'upper function' and where they intersect. We can set the two equations equal to each other to determine that they intersect at (-1, 1), (0,0), and (1, 1). This can be verified through a simple graph as well. We know that y = x^2 and y = |x| are both symmetric about the y axis, so when answering this question we will only deal with the functions to the right of the y axis and multiply the result by 2. in the interval [0, 1], |x| > x^2. this can be verified by plugging any point in [0, 1] ((.5)^2 = .25 < |.5| = .5). We then use the formula for area between two curves using the integral from 0 to 1 of the upper function - lower function. ( we can use x in place of |x| since |x| = x on [0, 1]

∫ (x - x^2) dx

=x^2/2 - x^3/3 evaluated on [0, 1]

(1/2 - 1/3) - (0/2 -0/3) = 3/6 - 2/6 = 1/6

This is the area between the two curves on the right of the y axis, so we multiply by two since the functions are symmetric to get 1/3