How to find the area under (and bewteen) curves using definite integrals; tutorials, with examples and detailed solutions are presented . A set of exercises with answers is presented at the end.
Review: formulas to calculate the area between 2 curves using definite integrals.
1 - If f and h are functions such that f(x) >= h(x) for all x in the interval [x1 , x2], the area shown below is given by
Area = _{x1}^{x2} [ f(x) - h(x) ] dx
2 - If z and x are functions such that z(y) >= x(y) for all y in the interval [ y1 , y2 ], the area shown below is given by
Area = _{y1}^{y2} [ z(y) - x(y)] dy
A - Find Area Under a curve
Example 1: Find the area of the region bounded by y = 2x, y = 0, x = 0 and x = 2.(see figure below).
Solution to Example 1:
Solution 1: This problem may be solved using the formula for the area of a triangle. area = (1/2)* base * height = (1/2)*2*4 = 4.
Solution 2:
We shall now use definite integrals to find the area defined above. If we let f(x) = 2x and g(x) = 0, according to 1 above, the area is given by the definite integral
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.
x^{2} - 2x + 2 = -x^{ 2} + 6
Which gives.
2x^{2} - 2x - 4 = 0
or
(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 = _{-1}^{2} [ f(x) - h(x) ] dx
= _{-1}^{2}
[ (-x^{ 2}+6) - (x^{2} - 2x + 2) ] dx
= _{-1}^{2}
[ -2x^{ 2} + 2x + 4 ] dx
= [ -(2 / 3) x^{ 3} + x^{ 2} + 4x] _{-1}^{2} = 9
The area of the region enclosed between the curves y = x^{ 2} - 2x + 2 and -x^{ 2} + 6 is equal to 9.
Exercises:
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
1. 22/3
2. 6
More references on
integrals and their applications in calculus.