The first method is fast but works because the area is that of a triangle, however the second method work for figures other than triangles.
Example 2: Find the area of the region enclosed between the curves y = x 2 - 2x + 2 and -x 2 + 6 .
Solution to Example 2:
We first graph the two equations and examine the area enclosed between the curves.
The region whose area is in question is limited above by the curve y = -x 2 + 6 and below by the curve y = x 2 - 2x + 2. The left endpoint and the right endpoint of the region are the point of intersection of the curves and can be found by equating y = x 2 - 2x + 2 and y = -x 2 + 6 and solving for x.
x2 - 2x + 2 = -x 2 + 6
2x2 - 2x - 4 = 0
(x + 1)(x - 2) = 0
which gives solutions
x = -1 and x = 2
Between the points of intersection, -x 2+6 is greater than or equal to x 2 - 2x + 2. Let f(x) = -x 2 + 6 and h(x) = x 2 - 2x + 2 and apply formula 1 above to find the area A of the region between the two curves. The limits of integration are the x coordinates of the points of intersection found above: -1 and 2
A = -12 [ f(x) - h(x) ] dx
[ (-x 2+6) - (x2 - 2x + 2) ] dx
[ -2x 2 + 2x + 4 ] dx
= [ -(2 / 3) x 3 + x 2 + 4x] -12 = 9
The area of the region enclosed between the curves y = x 2 - 2x + 2 and -x 2 + 6 is equal to 9.
1. Find the area of the region enclosed by y = (x-1) 2 + 3 and y = 7
2. Find the area of the region bounded by x = 0 on the left, x = 2 on the right, y = x 3 above and y = -1 below.
Answers to Above Exercises
More references on
integrals and their applications in calculus.