Area Bounded Region : Type II

Type II: The area of the shaded portion is given by

$\displaystyle \int_{a}^{c} f_1 (x) dx + \int_{c}^{b} f_2 (x) dx$

where x = c is a solution of f1(x) = f2(x) , a < c < b

Illustration : Find the area included between the curves y = sin–1 x, y = cos–1x and the x–axis.

Solution: Clearly we have to find the area of the shaded region OPBO.

The point P of intersection of

y = sin–1x and y = cos–1x is obtained by solving

sin–1x = cos–1x

=> π/2 – cos–1x = cos–1x

=> 2 cos–1x = π/2

=> cos–1x = π/4

$\displaystyle x = \frac{1}{\sqrt{2}}$

The required area $\displaystyle = \int_{0}^{1/\sqrt{2}} sin^{-1}x dx + \int_{1/\sqrt{2}}^{1} cos^{-1}x dx$

On integrating ,

Required Area   $\displaystyle = (\sqrt{2}-1) sq. units$

Illustration :   Find the area of the region defined by the set S = S3 – S1 – S2, where

S1= {(x, y): x2 + 2y≤ 2} , S2 ={(x, y): 2x2 + y2  ≤ 2} and S3 = {(x, y): x2 + y2 ≤  2}

Solution: $\displaystyle S_1 \equiv \frac{x^2}{2} + \frac{y^2}{1} \le 1$

$\displaystyle S_2 \equiv x^2 + \frac{y^2}{2} \le 1$

$\displaystyle S_3 \equiv x^2 + y^2 \le 2$

One of the points of intersection of $\large S_1 \equiv \frac{x^2}{2} + \frac{y^2}{1} = 1$ and $\displaystyle S_2 \equiv x^2 + \frac{y^2}{2} = 1$ is $A (\sqrt{\frac{2}{3}} , \sqrt{\frac{2}{3}} )$

Area in the first quadrant, bounded by $\displaystyle y = x \; , x^2 + \frac{y^2}{2} = 1$ is

$\displaystyle \int_{0}^{\sqrt{\frac{2}{3}}} (\sqrt{2}\sqrt{1-x^2} – x )dx$

$\displaystyle = \frac{1}{\sqrt{2}} sin^{-1} \sqrt{\frac{2}{3}} + \frac{1}{3} – \frac{1}{3}$

$\displaystyle = \frac{1}{\sqrt{2}} sin^{-1} \sqrt{\frac{2}{3}}$

Using symmetry the shaded area i.e. S = S3 – S1 – S2 is

$\displaystyle = 8 (\frac{2 \pi}{8} – \frac{1}{\sqrt{2}} sin^{-1} \sqrt{\frac{2}{3}})$

$\displaystyle = 2\pi – 4 \sqrt{2} sin^{-1} \sqrt{\frac{2}{3}}$

Illustration : Find the area bounded by the curves x = y2 and x = 3 – 2 y2

Solution: The vertices of the parabolas x = y2 and x = 3 – 2 y2 are respectively O (0, 0) and M (3, 0). Thus given curves intersect at (1, 1) and (1, –1)

So the required area = area(OPMQO) = 2 area (OPMO)

$\displaystyle = 2 (\int_{0}^{1} \sqrt{x} dx + \int_{1}^{3} \sqrt{\frac{3-x}{2} } dx )$

$\displaystyle = 2 [\frac{2}{3}x^{3/2}]_{0}^{1} – 2 \frac{1}{\sqrt{2}} \frac{2}{3}[(3-x)^{3/2}]_{1}^{3}$

$\displaystyle = 2 (\frac{2}{3} – 0) – \frac{4}{3\sqrt{2}}(0-2^{3/2})$

$\displaystyle = \frac{4}{3} + \frac{4}{3\sqrt{2}}2 \sqrt{2} = \frac{12}{3}= 4 sq.units$

Illustration : Find the area bounded by the curve |x| + y = 1 and the x–axis.

Solution: The given curve is |x| + y = 1

i.e. x + y = 1 , when x ≥ 0

and , –x + y = 1 , when x <  0

The required area  = area (CAOC) + area (OABO)

$\displaystyle = \int_{-1}^{0} y dx + \int_{0}^{1} y dx$

$\displaystyle = \int_{-1}^{0} (x+1) dx + \int_{0}^{1} (1-x) dx$

$\displaystyle = [\frac{x^2}{2}+ x]_{-1}^{0} + [x-\frac{x^2}{2}]_{0}^{1}$

= 1/2 + 1/2 = 1 sq.unit

Note: Obviously, y = 1 – |x| is an even function. Hence graph of y = 1 – |x| is symmetrical about the y–axis.

Thus the required area $\displaystyle = 2 \int_{0}^{1}(1-x) dx$

Illustration : Find the area bounded by y = (x–1) (x–2) (x–3), the x–axis, 1 ≤ x ≤ 3

Solution: y(1) = 0, y(2) = 0, y(3) = 0

From wavy curve y is positive from x = 1 to x = 2 and negative from x = 2 to x = 3

So area is $\displaystyle = \int_{1}^{3} |y| dx$

$\displaystyle = \int_{1}^{2} (x–1) (x–2) (x–3) dx + \int_{2}^{3} – (x–1) (x–2) (x–3) dx$

$\displaystyle = \int_{1}^{2} (x^3 – 6 x^2 + 11 x – 6 ) dx – \int_{2}^{3} (x^3 – 6 x^2 + 11 x – 6 ) dx$

= 1/4 + 1/4 = 1/2

Exercise :

(i) Find the area of the region bounded by [x] and {x}, where [.] and {.} are greatest integer function and fractional part of x respectively.

(ii) Find the area bounded by the curves $\displaystyle y = x + \sqrt{x^2}$ and xy = 1, the x–axis and the line x = 2.

(iii) Find the area bounded by the curves y = ln |x|, y–axis and y = 1 – |x|.

(iv) Find the area bounded by $\displaystyle y = sin x , 0 \le x \le \frac{\pi}{4}$ , $\large y = cosx x , \frac{\pi}{4} \le x \le \frac{\pi}{2}$ and the x–axis.