Area under Curves – Tricks for Solving Area under Curves | Class 12 Math Notes Study Material Download Free PDF

 

Important Concepts on Area under the Curves

If the area under one curve and another curve can be found, then the area between the curves can be calculated.

Consider two curves to be y = f (x) and y = g (x) between the interval x = a to x = b is

$\begin{array}{l}Area={\int }_a^b|f(x)-g(x)dx\end{array}$

• The area of the ellipse = πab.
• The shift in the coordinate axis leads to the area enclosed by a curve remaining invariant.
• To check the symmetry of the curve following steps are followed:

1. The curve is said to be symmetrical about the x-axis when y is changed to -y, and there is no change in the curve.
2. The curve is said to be symmetrical about the y-axis when x is changed to -x, and there is no change in the curve.
3. If the curve remains unchanged when x is changed to -x and y is changed to -y, then it is symmetrical to both axes.

• If x is interchanged to y, resulting in no change of the curve, then it is symmetrical about the line y = x.

The standard steps that are followed in drawing the curve are as follows:

1) A check for symmetry is to be done.

• Symmetry about the x-axis: The curve is symmetrical about the x-axis if the curve remains unchanged when y = -y.
• Symmetry about the y-axis: The curve is symmetrical about the y-axis if the curve remains unchanged when x = -x.
• Symmetrical about both axes: The curve is symmetrical about both axes if the curve remains unchanged when x = -x and y = -y.
• Symmetrical about the line y = x: The curve is said to be symmetrical about the line y = x if the curve remains unchanged on interchanging x and y. 

2) The point of intersection of the axes is to be noted.
3) Find dy / dx to find stationary points. Find the intervals to determine whether the function is increasing or decreasing.

4) Find y when x → ∞ or -∞.

How to Find the Area under Curves

A few examples are illustrated below on finding areas enclosed by different curves.

Example 1: What is the area bounded by the curve y = log x, x−axis and the ordinates x = 1, x = 2?

Solution:

Given curve y = log x and x = 1, x = 2

Hence, the required area

$\begin{array}{l}={[{\int }_1^{2}logxdx=(xlogx-x)]}_1^2\end{array}$

= 2 log 2 − 1 = (log 4 − 1) sq. unit

Example 2: If the ordinate x = a divides the area bounded by the curve y = (1 + [8 / x2]), the x−axis and the ordinates x = 2, x = 4 into two equal parts, then what is the value of a?

Solution:

Let the ordinate at x = a divide the area into two equal parts.

$\begin{array}{l}AMNB={\int }_4^2=[1+\frac{8}{x^2}]dx=[x-\frac{8}{x}]{{}_2}^4=4\end{array}$

and

$\begin{array}{l}ACDM={\int }_2^a=[1+\frac{8}{x^2}]dx=2\end{array}$

On solving, we get a = ± 2√2

Since a > 0, a = 2√2

Example 3: If the area bounded by the curves y2 = 4ax and y = mx is a2 / 3, then find the value of m.

Solution:

The two curves y2 = 4ax and y = mx intersect at (4a / m2, 4a / m), and the area enclosed by the two curves is given by

$\begin{array}{l}{\int }_0^{4a/m^2}[\sqrt{4ax-mx}]dx=\frac{a^2}{3}\end{array}$

[8 / 3] * [a2 / m3] = a2/ 3

⇒ m3 = 8

⇒ m = 2

Example 4: Find the ratio of the areas bounded by the curves y = cos x and y = cos 2x between x = 0, x = π/3 and x−axis.

Solution:

$\begin{array}{l}{A}_1={\int }_0^{\pi /3}cosxdx=\frac{\sqrt{3}}{2}\end{array}$

$\begin{array}{l}{A}_2={\int }_0^{\pi /3}cos2xdx=\frac{\sqrt{3}}{4}\end{array}$

Hence, the ratio is A1 : A2 = 2 : 1

Example 5: What is the area of the smaller part between the circle x2 + y2 = 4 and the line x = 1?

Solution:

Area of smaller part

$\begin{array}{l}=2{\int }_1^2\sqrt{4-x^2}dx=2[\frac{x}{2}\sqrt{4-x^2}+2si{n}^{-1}\frac{x}{2}]\end{array}$

in the interval [1, 2]

$\begin{array}{l}=2[2\ast \frac{\pi }{2}-[\frac{\sqrt{3}}{2}-2\ast \frac{\pi }{6}]\end{array}$

$\begin{array}{l}=2[\pi -[\frac{\sqrt{3}}{2}-\frac{\pi }{3}]\end{array}$

$\begin{array}{l}=\frac{8\pi }{3}-\sqrt{3}\end{array}$

Example 6: What is the area in the first quadrant between x2 + y2 = π2 and y = sinx?

Solution:

The area of the circle in the first quadrant is π* (π2) / 4, i.e., π3 / 4.

Also, the area bounded by curve y = sinx and x-axis is 2 sq. unit.

Hence, the required area is [π3 / 4] − 2 = [π3 − 8] / 4

Example 7: The area enclosed by the curves y = sin x + cos x and y = |cos x – sin x| over the interval [0, π/2] is

Solution:

$\begin{array}{l}{y}_1=sinx+cosx=\sqrt{2}sin(x+\frac{\pi }{4})\\ y_2=|sinx-cosx|=\sqrt{2}sin|sin\frac{\pi }{4}-x|\\ \text{Area}={\int }_0^{\frac{\pi }{4}}[(sinx+cosx)-(cosx-sinx)]dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}[(sinx+cosx)-(sinx-cosx)]dx\\ =4-2\sqrt{2}\\ =2\sqrt{2}(\sqrt{2}-1)\end{array}$

Example 8: Let g(x) = cos x2, f(x) = √x and α, β (α < β) be the roots of the quadratic equation 18x2 – 9πx + π2 = 0. Then, the area bounded by the curve y = (gof) (x) and the lines x = α, x = β and y = 0 is

Solution:

$\begin{array}{l}18x^2-9\pi x+{\pi }^2=0;gof(x)=cosx\\ (3x-\pi )(6x-\pi )=0\\ \alpha =\frac{\pi }{6},\beta =\frac{\pi }{3}\\ A={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}cosxdx=\frac{\sqrt{3}-1}{2}\end{array}$