Mathematics Area between Two Curves CBSE-NCERT
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\color{green} ✍️ Area between Two Curves

### Area between Two Curves

● we are given two curves represented by y = f (x), y = g (x), where f (x) ≥ g(x) in [a, b] as shown in Fig .

● Here the points of intersection of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two curves.

● For setting up a formula for the integral, it is convenient to take elementary area in the form of vertical strips. As indicated in the Fig

f (x) – g(x) and width dx so that the elementary area

dA = [f (x) – g(x)] dx, and the total area A can be taken as

color{orange} {A = int_a^b [ f(x )- g(x )] dx}

Alternatively,

A = ["area bounded by" y = f (x), "x-axis and the lines"\ \ x = a, x = b]
– ["area bounded by" y = g (x), "x-axis and the lines" \ \ x = a, x = b]

color {red} {= int_a^b f(x) dx - int_a^b g(x) dx = int_a^b [ f(x) - g(x) ] dx} , where f(x) >= g(x)  in [ a,b]

● If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the Fig , then the area of the regions bounded by curves can be written as

Total Area = Area of the region ACBDA + Area of the region BPRQB

color{red} {= int_a^c [ f(x) - g(x) ] dx + int_c^b [ g(x) - f(x) ] dx}

Q 3106256178

Find the area of the region bounded by the two parabolas y = x^2 and y^2 = x.
Class 12 Chapter 8 Example 6
Solution:

The point of intersection of these two
parabolas are O (0, 0) and A (1, 1) as shown in
the Fig 8.15.
Here, we can set y^2 = x or y = sqrt x = f (x) and y = x^2
= g (x), where, f (x) ≥ g (x) in [0, 1].
Therefore, the required area of the shaded region

 = int_0^1 [ f (x) − g(x) ] dx

 = int_0^1 [ sqrt x - x^2 ] dx = [ 2/3 x^(3/2) - x^3/3 ]_0^1

 = 2/3 - 1/3 = 1/3
Q 3156356274

Find the area lying above x-axis and included between the circle
x^2 + y^2 = 8x and the parabola y^2 = 4x.
Class 12 Chapter 8 Example 7
Solution:

The given equation of the circle x^2 + y^2 = 8x can be expressed as
(x – 4)^2 + y^2 = 16. Thus, the centre of the
circle is (4, 0) and radius is 4. Its intersection
with the parabola y^2 = 4x gives

x^2 + 4x = 8x

or x^2 – 4x = 0

or x (x – 4) = 0

or x = 0, x = 4

Thus, the points of intersection of these
two curves are O(0, 0) and P(4,4) above the
x-axis.
From the Fig 8.16, the required area of
the region OPQCO included between these
two curves above x-axis is
= (area of the region OCPO) + (area of the region PCQP)

 = ∫_0^4 ydx + ∫_4^8 ydx

= 2 ∫_0^4 sqrt x dx + ∫_4^8 sqrt(4^2 −(x −4)^2) dx (Why?)

= 2 xx 2/3 [ x^(3/2]_0^4 + int_0^4 sqrt( 4^2 - t^2) dt , where,  x - 4 = t

 = (32)/3 + [ t/2 sqrt(4^2 - t^2) + 1/2 xx 4^2 xx sin^(-1) \ t/4 ]_0^4

 = (32)/3 + [ 4/2 xx 0 + 1/2 xx 4^2 xx sin^(-1) 1 ] = (32)/3 + [ 0 + 8 xx pi/2] = (32)/3+ 4 pi = 4/3 (8 + 3 pi)
Q 3106356278

AOBA is the part of the ellipse 9x^2 + y^2 = 36 in the first
quadrant such that OA = 2 and OB = 6. Find the area between the arc AB and the
chord AB.
Class 12 Chapter 8 Example 8
Solution:

Given equation of the ellipse 9x^2 + y^2 = 36 can be expressed as

x^2/2^2 + y^2/6^2 = 1 and hence, its shape is as given in Fig 8.17

Accordingly, the equation of the chord AB is

y – 0 = (6 - 0)/(0-2) (x - 2)

or y = – 3(x – 2)
or y = – 3x + 6
Area of the shaded region as shown in the Fig 8.17.

= 3∫_0^2 sqrt(4 − x^2) dx − ∫_0^2 (6 − 3x)dx (Why?)

 = 3 [ x/2 sqrt(4 − x^2) + 4/2 sin^(-1) \ x/2 ]_0^2 - [ 6x - (3x^2)/2 ]_0^2

 = 3 [ 2/2 xx 0 + 2 sin^(-1) (1)] - [ 12 - (12)/2 ] = 3 xx 2 xx pi/2 - 6 =3pi-6
Q 3136456372

Using integration find the area of region bounded by the triangle whose
vertices are (1, 0), (2, 2) and (3, 1).
Class 12 Chapter 8 Example 9
Solution:

Let A(1, 0), B(2, 2) and C(3, 1) be
the vertices of a triangle ABC (Fig 8.18).
Area of ΔABC
= Area of ΔABD + Area of trapezium
BDEC – Area of ΔA EC
Now equation of the sides AB, BC and
CA are given by

y = 2 (x – 1), y = 4 – x, y = 1/2 (x – 1), respectively.
Hence, area of Δ ABC = int _1^2 2 (x -2) dx int_2^3 (4-x) dx - int _1^3 (x-1)/2 dx

 = 2 [ x^2/2 - x]_1^2 + [ 4x - x^2/2]_2^3 - 1/2 [ x^2/2 - x]_1^3

 = 2 [ ( 2^2/2 - 2) - (1/2 -1) ] + [ ( 4 xx3 - 3^2/2 ) - ( 4 xx2 - 2^2/2 ) ] - 1/2 [ (3^2/2 - 3) - (1/2 -1) ]

 = 3/2
Q 3116556470

Find the area of the region enclosed between the two circles: x^2 + y^2 = 4
and (x – 2)^2 + y^2 = 4.
Class 12 Chapter 8 Example 10
Solution:

Equations of the given circles are
x^2 + y^2 = 4 ... (1)
and (x – 2)^2 + y^2 = 4 ... (2)
Equation (1) is a circle with centre O at the
origin and radius 2. Equation (2) is a circle with
centre C (2, 0)  and radius 2. Solving equations
(1) and (2), we have
(x –2)^2 + y^2 = x^2 + y^2

or x^2 – 4x + 4 + y^2 = x^2 + y^2

or x = 1 which gives y = ± sqrt3
Thus, the points of intersection of the given
circles are A(1, sqrt 3 ) and A′(1, – sqrt3 ) as shown in
the Fig 8.19.

Required area of the enclosed region OACA′O between circles
= 2 [area of the region ODCAO] (Why?)
= 2 [area of the region ODAO + area of the region DCAD]

= 2 [ int_0^1 ydx + int_1^2 ydx ]

 = 2 [ int_0^1 sqrt( 4 − (x − 2)^2) dx + int_1^2 sqrt( 4-x) dx]

 = 2 [1/2 (x-2) sqrt( 4 − (x − 2)^2) + 1/2 xx 4 sin^(-1) ((x-2)/2 ) ]_0^1

 + 2 [ 1/2 x sqrt( 4 - x^2 ) + 1/2 xx 4 sin^(-1) \ x/2 ] _1^2

 = [ ( - sqrt3 + 4 sin ^(-1) ( (-1)/2) - 4 sin ^(-1) (-1) ] + [ 4 sin ^(-1) 1 - sqrt 3 - 4 sin^(-1) \ 1/2 ]

 = [ ( - sqrt3 - 4 xx pi/4 ) + 4 xx pi/2 ] + [ 4 xx pi/2 - sqrt3 - 4 xx pi/6 ]

 = ( - sqrt3 - (2pi)/3 + 2 pi) + ( 9 2 pi - sqrt3 - (2 pi)/3 )

 = (8pi)/3 - 2 sqrt3