CBSE Class 12 Mathematics Application of Integrals Notes Set 01

Important Notes and Formulas

Area Under the Curves

1. The area bounded by a cartesian curve y = f(x), x-axis and lines x = a and x = b is given as:
\( \text{Area} = \int_{a}^{b} y\,dx = \int_{a}^{b} f(x)\,dx \)

Here, we consider an arbitrary strip (vertical) of height y and width dx.

2. The area bounded by a cartesian curve x = f(y), y-axis and lines y = c and y = d is given as:
\( \text{Area} = \int_{c}^{d} x\,dy = \int_{c}^{d} f(y)\,dy \)

Here, we consider an arbitrary strip (horizontal) of length x and width dy.

Note:
1. If the position of the curve under consideration is below the x-axis, then since f(x) < 0 from x = a to x = b, as shown in the figure, the area bounded by the curve, x-axis and the ordinates x = a, x = b comes out to be negative. But it is only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute value, i.e. required area \( = \left| \int_{a}^{b} f(x)\,dx \right| \).

2. If some portion of the curve is above the x-axis and some below the x-axis, as shown in the figure, then \( A_1 \lt 0 \) and \( A_2 \gt 0 \). Therefore, the area A bounded by the curve y = f(x), x-axis and the ordinates x = a and x = b is given as \( A = |A_1| + A_2 \).

Solved Examples

Illustration 1. Using integration, sketch the graph of curve y = |x + 1| and evaluate area under the curve above x-axis and between x = −3 to x = 1. [6 Marks]
Answer:
\( y = |x + 1| = \begin{cases} x + 1; & \text{if } x \geq -1 \\ -(x + 1); & \text{if } x \lt -1 \end{cases} \) [1]

Required Area = Area of CDEC + Area of ABCA
\( = \int_{-3}^{-1} y\,dx + \int_{-1}^{1} y\,dx \) [1]
\( = \int_{-3}^{-1} -(x + 1)\,dx + \int_{-1}^{1} (x + 1)\,dx \) [1]
\( = -\left[\frac{x^2}{2} + x\right]_{-3}^{-1} + \left[\frac{x^2}{2} + x\right]_{-1}^{1} \) [1]
\( = -\left[\left(\frac{1}{2} - 1\right) - \left(\frac{9}{2} - 3\right)\right] + \left[\left(\frac{1}{2} + 1\right) - \left(\frac{1}{2} - 1\right)\right] \) [1]
\( = \frac{1}{2} + \frac{3}{2} + \frac{3}{2} + \frac{1}{2} = 4 \) square units [1]

Illustration 2. Using integration, find the area enclosed by the curve x = 3 cos t, y = 2 sin t. [Exemplar] [6 Marks]
Answer:
x = 3 cos t, y = 2 sin t
⇒ \( \frac{x}{3} = \cos t \) and \( \frac{y}{2} = \sin t \)
⇒ \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \) [2]
which is the equation of an ellipse.

From figure, we get [1]
the required area \( = 4\int_{0}^{3} \frac{2}{3}\sqrt{9 - x^2}\,dx \) [1]
\( = \frac{8}{3}\left[\frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\frac{x}{3}\right]_{0}^{3} = 6\pi \) sq. units [2]

Do Yourself

Do Yourself 1
(i) Draw a rough sketch of the given curve y = 1 + |x + 1|, x = −3, x = 3, y = 0 and find the area of the region bounded by them, using integration.
(ii) Make a rough sketch of the curve \( y = x^2 \), 0 ≤ x ≤ 3 and determine the area enclosed between the curve, the x-axis and the line x = 3.
(iii) Find the area of the region bounded by the curve \( x^2 = 4y \), the y-axis, the lines y = 2, y = 4 and which lies in the first quadrant.

Answers for Do Yourself

Do Yourself 1
(i) 16 square units
(ii) 9 square units
(iii) \( \frac{8}{3}\left(4 - \sqrt{2}\right) \) square units

Previous Year Questions

Long Answer Questions (6 Marks)

Question 1. Using integration, sketch the graph of y = |x + 3| and evaluate the area under the curve y = |x + 3| above x-axis and between x = −6 to x = 0. [CBSE 2011, 6M]
Answer:
[1½]
y = |x + 3|
Req. Area \( = \int_{-6}^{-3} (-x - 3)\,dx + \int_{-3}^{0} (x + 3)\,dx \) [1½]
\( = \left[-\frac{x^2}{2} - 3x\right]_{-6}^{-3} + \left[\frac{x^2}{2} + 3x\right]_{-3}^{0} \) [1½]
\( = -\frac{9}{2} + 9 + \frac{36}{2} - 18 - \frac{9}{2} + 9 = 9 \) square units [1½]