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`

---

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)`

---

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`

---

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`

---

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 `

---