Area Bounded Region : Type I

Basic Concepts :

Let y = f1 (x) and y = f2 (x) be two given curves, which are continuous in [a, b]. Suppose we have to find the area of the plain region bounded by the two curves y = f1(x) and y = f2(x) between the ordinates x = a and x = b (i.e. the area of the region A1B1C1D1).

For this purpose we take an elemental strip ABCD of width dx and height AD (or BC).

Clearly AD ≅ BC = f1(x) ~ f2(x).

Now the area of this strip,

dA = (f1(x)~f2(x)) dx

Thus the required area,

$ \displaystyle A = \int_{a}^{b} (f_1(x) \sim f_2(x)) dx$

Remarks:

♦If f1(x) ≥ f2(x) ∀ x ∈ [ a, b] then , $ \displaystyle A = \int_{a}^{b} (f_1(x) – f_2(x)) dx$

♦If f1(x) ≤ f2(x) ∀ x ∈ [ a, b] then ., $ \displaystyle A = \int_{a}^{b} (f_2(x) – f_1(x)) dx$

♦If f1(x) ≥ f2(x) ∀ x ∈ [ a , c] and f1(x) ≤ f2(x) ∀ x ∈ [ c , b] , then

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

♦We can finally conclude that the area bounded by the two curves between x = a and x = b is

$ \displaystyle \int_{a}^{b} (curve \; lying \; above – curve \; lying \; below) dx $

♦If f2(x) = 0 ∀ x ∈ R,

$ \displaystyle \int_{a}^{b}f_1(x) dx $ , would give us area bounded by y = f1(x) and the x–axis between the ordinates x = a and x = b (here we are assuming that f1(x) ≥ 0 ∀ x ∈ [ a, b] ).

Similarly , $ \displaystyle A = \int_{a}^{b} (f_1(y) \sim f_2(y)) dy $

would give us the area bounded by the curves x = f1(y) and x = f2(y)
between the lines y = c and y = d.

Types of Problems :

We know the method(s) of finding areas of triangle, quadrilateral, polygon, circle etc. In this chapter we shall learn how to find the area of the regions bounded by curves of the following types.

Type I: y = f(x), the x-axis and the lines x = a and x = b where f is a continuous function in [a, b].

Type II: The curves y = f1(x), x ∈ (a, c), y = f2(x) , x ∈ (c, b) , the x-axis and the lines x = a , x = b ; a < c< b

Type III: The curves y = f1(x), y = f2(x) and the lines x = a , x = b.

Type IV: Bounding curves are represented by function defined through given conditions.

Type I:

(i) The area bounded by y = f(x) above the x– axis and the coordinates x = a and x = b (represented by shaded portion ) is given by

$ \displaystyle A = \int_{a}^{b} f(x) dx = \int_{a}^{b} y dx $

Illustration : Calculate the area bounded by the curve y = x(3 – x)2 , the x-axis and the ordinates of the maximum and minimum points of the curve.

Solution:

Since y = x(3 – x)2

Now for points of maxima or minima, we have dy/dx = 0

=> (3 – x)2 – 2x(3 – x) = 0

=> (3 – x)(3 – x – 2x) = 0

∴ x = 1, 3

Now , $\frac{d^2 y}{dx^2} = 3(x – 3) + 3(x – 1)$

= 6(x – 2)

At x = 1 :

$\frac{d^2 y}{dx^2} < 0 $

=> x = 1 is point of maxima

At x = 3 :

$\frac{d^2 y}{dx^2} > 0 $

=> x = 1 is point of minima.

∴ Required area

$ \displaystyle A = \int_{1}^{3} x(3-x)^2 dx $

$ \displaystyle A = \int_{1}^{3} (9x – 6x^2 + x^3 ) dx $

$ \displaystyle A = [\frac{9x^2}{2} – \frac{6x^3}{3} + \frac{x^4}{4}]_{1}^{3} $

= 4

Illustration : Find the area of one of the loops of |y| = |sin x|.

Solution:

We have the given curve

|y| = |sin x|

Required area

$ \displaystyle A = 2\int_{0}^{\pi}sinx dx $

$ \displaystyle = 2(-cosx)_{0}^{\pi} $

= 4

(ii) The area bounded by y = f(x) below the x-axis and the ordinates x = a and x = b (represented by the shaded portion) is given by

$ \displaystyle |\int_{a}^{b}f(x)dx | $

or, $ \displaystyle \int_{a}^{b}|f(x)|dx $

or , $ \displaystyle -\int_{a}^{b}f(x)dx $

Illustration : Find the area bounded by the curve y = 2x/(1 + x2 ) and the lines x = − 1 , x = 2 and the x–axis .

Solution: Required area =

$ \displaystyle \int_{-}^{2}|\frac{2x}{1+x^2}|dx $

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

$ \displaystyle = -ln(1+x^2)|_{-1}^{0} + ln(1+x^2)|_{0}^{2} $

= ln 2 + ln 5

= ln 10

(iii) If the curve y = f(x) crosses the x–axis at the points a1 , a2 , ….. , an
where a < a1 < a2 < ………. < an < b


then the area bounded by y= f(x), x = a, x = b and x–axis is given by

$ \displaystyle |\int_{a}^{a_1}f(x)dx|+|\int_{a_1}^{a_2}f(x)dx|+ …+|\int_{a_n}^{b}f(x)dx| $

Illustration . Find the area bounded by the parabola y2 = 4x and the straight line x + y = 3

Solution: The line x + y = 3 cuts the x–axis at (3, 0) and the y-axis at (0, 3). Also, the given curve is y2= 4x.


Solving for y, we get y2 = 4 (3 − y)

=> y = − 6 , 2

The required area $ \displaystyle = \int_{-6}^{2} (x_1 -x_2) dy $

$ \displaystyle = \int_{-6}^{2} (3-y – \frac{y^2}{4}) dy $

$ \displaystyle = [3y – \frac{y^2}{2} – \frac{y^3}{12}]_{-6}^{2} $

= (10/3) + 18 = 64/3 sq. units.

(iv) The area bounded by x = f(y), on the right of y-axis, y-axis and the abscissa y = a and y = b (represented by the shaded portion) is given by  $ \displaystyle \int_{a}^{b} x dy $

If the area completely lies on the left of y-axis then the required area

$ \displaystyle = |\int_{a}^{b} x dy| $

$ \displaystyle = \int_{a}^{b} |x| dy $

$ \displaystyle = -\int_{a}^{b} x dy $

Illustration : Calculate the area bounded by y = x2 on the left of y-axis, the y-axis and the lines y = 1 , y = 4

Solution: The required area

$ \displaystyle |\int_{1}^{4} -\sqrt{y} dy | $

$ \displaystyle \frac{2}{3}y^{3/2}|_{1}^{4} $

$ \displaystyle \frac{2}{3}(8-1) $

= 14/3 sq.units

Note: y = x2

=> x = ±√y

Since we are considering portion of y = x2 on the left of y-axis , we take x = − √y

Exercise :

(i) Find the area enclosed by |x| + |y| = 1

(ii) Find the area of the region bounded by C: y = tan x, tangent drawn to C at x = π/4 and the x – axis.

(iii) Find the area bounded by y2 = 4ax and the tangents at the ends of its latus rectum.

(iv) Calculate the area bounded by the curve y(y − 1) = x and the y-axis.

(v) Find the point P on the parabola y2 = 4ax such that area bounded by the parabola, the x-axis and the tangent at P is equal to that of bounded by the parabola, the x- axis and the normal at P.

Also Read :

→ Area Bounded Region : Type I
→ Area Bounded Region : Type II
→ Area Bounded Region : Type III
→ Area Bounded Region : Type IV

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