Class 12 Maths Application of Integrals Area under Simple Curves

Area under Simple Curves

Let we want to find the area bounded be the curve y = f(x), x-axis and the ordinates x = a and x = b.

Class_12_Maths_Applications_Of_Integrals_Area_Under_Simple_Curves

From the figure, we assume that area under the curve is composed of large number of very thin vertical strips. Let us take an arbitrary strip of height y and width dx, then dA (area of the elementary strip) = y dx, where, y = f(x). This area is called the elementary area which is located at an arbitrary position within the region which is specified by some value of x between a and b.

Total area A of the region between x-axis, ordinates x = a, x = b and the curve y = f(x) is calculated by adding up the elementary areas of thin strips across the region PQRSP.

The area A under the curve f(x) bounded by x = a and x = b is given by

A = ab dA = ab y dx = ab f(x) dx

Class_12_Maths_Applications_Of_Integrals_Area_Under_Simple_Curves_1

The area A of the region bounded by the curve x = g (y), y-axis 

and the lines y = c, y = d as shown in the figure, is given by

A = cd x dy = cd g(y) dy

Example: Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.

Solution:

The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.

Class_12_Maths_Applications_Of_Integrals_Area_Under_Simple_Curves_2

So, area of ABCD = ʃ14 y dx

                               = ʃ14 √x dx

                               = [x3/2/(3/2)]14 

                               = (2/3)[43/2 - 13/2]

                               = (2/3)[8 – 1]

                               = (2/3) * 7

                               = 14/3 units

Example: Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Solution:

The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.

Class_12_Maths_Applications_Of_Integrals_Area_Under_Simple_Curves_3

So, area of ABCD = ʃ24 x dy

                               = ʃ24 2√y dy

                               = 2[y3/2/(3/2)]24 

                               = (4/3)[43/2 - 23/2]

                               = (4/3)[8 – 2√2]

                               = (32 – 8√2)/3 units

 

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