Find Area Under Curve
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.
Area = x1 x2 [ f(x) - h(x) ] dx
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
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.
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
[ 2x - 0 ] dx = 2 0 2 x dx
= [ 2 (x2 / 2) ] 0 2 = 4
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 = -1 2 [ f(x) - h(x) ] dx
= -1 2
[ (-x 2+6) - (x2 - 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.
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.
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Updated: 2 April 2013
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