Find Area Under Curve

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 = ò_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

ò₀ ² [ 2x - 0 ] dx = 2 ò₀ 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

Which gives. 2x² - 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 = ò_-1 ² [ f(x) - h(x) ] dx

= ò_-1 ² [ (-x ²+6) - (x² - 2x + 2) ] dx

= ò_-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.