Utilizing Integration for Area Calculations

Applications of integration Applications of integration - - Area Area

Area under a Curve as a Definite Integral If f(x) is nonnegative and integrable over a closed interval [a, b], then the area under the curve ? = ?(?) over [a, b] is the integral of from a to b, ? ? = ? ? ?? ? Finding Area Using Antiderivatives To find the area between the graph of ? = ?(?) and the x-axis over the interval [a, b], do the following: 1. Subdivide [a, b] at the zeros of . 2. Integrate over each subinterval. 3. Add the absolute values of the integrals.

Example Calculate the area bounded by the x-axis and the parabola ? = 6 ? ?2 Sol. We find where the curve crosses the x-axis by setting ? = 0 = 6 ? ?2= (3 + ?)(2 ?) Which gives x=-3 and x=2 as sketched below:

The curve is nonnegative on [-3,2]. The area is 2 2 (6 ? ?2).?? = 6? ?2 2 ?3 = 12 2 8 18 9 2+ 9 3 3 3 3 2 (6 ? ?2).?? = 20.8333 3 Example Figure below shows the graph of the function ? ? = sin? between ? = 0 ??? ? = 2? Compute (a) The definite integral of (x) over [0,2?] (b) The area between the graph of (x) and the x-axis over [0,2?]

Sol. The definite integral for ? ? = sin? is given by: 2? 2?= 1 1 = 0 sin??? = cos?0 0 The definite integral is zero because the portions of the graph above and below the x-axis make cancelling contributions. The area between the graph of (x) and the x-axis over [0,2?] is calculated by breaking up the domain of sin x into two pieces: 1. the interval [0,?] over which it is nonnegative. 2. the interval [?,2?] over which it is nonpositive. ? ?= 1 1 = 2 sin? ?? = cos?0 0

2? 2?= 1 1 sin? ?? = cos?? = 2 ? The second integral gives a negative value. The area between the graph and the axis is obtained by adding the absolute values ???? = 2 + 2 = 4 Example Find the area of the region between the x-axis and the graph of f(x) ? ? = ?3 ?2 2? , 1 ? 2

First find the zeros of . Since ? ? = ?3 ?2 2? = ? ?2 ? 2 = ? ? + 1 ? 2 The values where f will be zero are x = 0,-1 and 2 The values subdivide the [-1,2] into two intervals [-1,0] on which ? 0 and [0,2] on which ? 0 . We integrate f over each subinterval and add the absolute values of the calculated integrals. 0 0 ?4 4 ?3 1 4+1 5 (?3 ?2 2?)?? = 3 ?2 = 0 3 1 = 12 1 1 2 2 ?4 4 ?3 = 4 8 3 4 0 = 8 (?3 ?2 2?)?? = 3 ?2 3 0 0 The total enclosed area is obtained by adding the absolute values of the calculated integrals, 5 8 3 =37 12 ???? = 12+

Example Find the area of the shaded region The area of the rectangle bounded by the lines ? = 2, ? = 0 , ? = ? ??? ? = 0 is ?rea of a rectangle = L W = 2? The area under the curve ? ?= ? + 0 0 = ? (1 + cos?)?? = ? + sin?0 0 Therefore the area of the shaded region is 2? ? = ?

Example Find the area of the shaded region Sol. The area of the rectangle bounded by the lines ? =? 6 , ? =5? 6, ? = sin? 6= sin5? 6=1 2 ??? ? = 0 ?rea of a rectangle = L W =1 5? 6 ? =? 2 6 3 The area under the curve

5? 6 5? 6= cos5? 6 cos? sin? ?? = cos? = ? 6 6 ? 6 3 2 3 2 = = 3 Therefore the area of the shaded region is 3 ? 3