|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.
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 = aꭍb dA = aꭍb y dx = aꭍb f(x) dx
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 = cꭍd x dy = cꭍd 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.
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.
So, area of ABCD = ʃ14 y dx
= ʃ14 √x dx
= (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.
The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.
So, area of ABCD = ʃ24 x dy
= ʃ24 2√y dy
= (4/3)[43/2 - 23/2]
= (4/3)[8 – 2√2]
= (32 – 8√2)/3 units