(Finding area using integration) Applications of Integration (Finding area using integration)

Objectives: Express the area under a curve/between curves as a definite integral. Compute the area under a curve/between curve using integration.

The Fundamental Theorem of Calculus Find the area between the graph of f and the x-axis on the interval [0, 3].

The Fundamental Theorem of Calculus Find the area between the graph of f and the x-axis on the interval [0, 3].

Area Under the Curve Find the area of the shaded region.

Area Under the Curve Area  sum of the areas of the rectangles = um (height)  (base) = b = a

Area Under the Curve = The area really given by the antiderivative? The definite integral of f from a to b is the limit of the Riemann sum as the lengths of the rectangle approach zero.

The general ideas The area under the curve is split into strips. The area of each strip is then approximated by 2 rectangles, one above and one below the curve as shown. The exact area of the strip under the curve lies between the area of the 2 rectangles.

Using 10 rectangles below and 10 above to estimate an area below a curve, we have . . . Greater accuracy would be given with 20 rectangles below and above . . . For an exact answer we let the number of rectangles approach infinity.

Note:

Definite integration results in a value. Areas Definite integration results in a value. It can be used to find an area bounded, in part, by a curve e.g. gives the area shaded on the graph 1 The limits of integration . . .

x = 0 is the lower limit x = 1 is the upper limit Areas Definite integration results in a value. It can be used to find an area bounded, in part, by a curve 1 e.g. gives the area shaded on the graph 1 The limits of integration . . . . . . give the boundaries of the area. x = 0 is the lower limit ( the left hand boundary ) x = 1 is the upper limit (the right hand boundary )

The units are usually unknown in this type of question Finding an area Since 1 2 3 + = x y the shaded area equals 3 The units are usually unknown in this type of question

Finding an area A B For parts of the curve below the x-axis, the definite integral is negative, so

Finding an area A Area A

Finding an area B Area B

SUMMARY An area is always positive. To find an area, we need to know whether the curve crosses the x-axis. For areas above the axis, the definite integral gives the area. For areas below the axis, we need to change the sign of the definite integral to find the area.

Exercise Find the areas described in each question. 1. The area between the curve the x-axis and the lines x = 1 and x = 3. 2. The area between the curve , the x-axis and the x = 2 and x = 3.

Solutions: 1. A B 2.

Area between the curve and the y axis To find this we rearrange the integral as d x δy c

Finding Integrals between the Y axes and f(x) we rearrange the Integral as follows Rearrange as x=y2+2

Area between the curve and the y axis Rearrange as x=y2+2 Area = 12 2/3 units square

Area between curves/functions Find the coordinates of the points of intersection of the curve and line shown. Find the area enclosed by the curve and line. Solution: The points of intersection are given by

Substitute in The area required is the area under the curve between 0 and 1 . . . . . . minus the area under the line (a triangle ) Method 1 Area under the curve Area of the triangle Required area

Method 2 Instead of finding the 2 areas and then subtracting, we can subtract the functions before doing the integration. We get Area

; or Area

Ex. Find the area of the region bounded by the graphs of f(x) = x2 + 2, g(x) = -x, x = 0, and x = 1 . Area = Top curve – bottom curve f(x) = x2 + 2 g(x) = -x

Find the area of the region bounded by the graphs of f(x) = 2 – x2 and g(x) = x First, set f(x) = g(x) to find their points of intersection. y= 2 – x2 y = x 2 – x2 = x 0 = x2 + x - 2 0 = (x + 2)(x – 1) x = -2 and x = 1

Find the area of the region between the graphs of f(x) = 3x3 – x2 – 10x and g(x) = -x2 + 2x Again, set f(x) = g(x) to find their points of intersection. y = 3x3 – x2 – 10x 3x3 – x2 – 10x = -x2 + 2x 3x3 – 12x = 0 3x(x2 – 4) = 0 x = 0 , -2 , 2 y = -x2 + 2x

Now, set up the two integrals and solve.

Find the area of the region bounded by the graphs of x = 3 – y2 and y = x - 1 Area = Right - Left

Objectives: Express the area under a curve/between curves as a definite integral. Compute the area under a curve/between curve using integration.