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**Q.1. Find the area of the region bounded by the curves y ^{2} = 9x, y = 3x.**

We have, y

Solving the two equations, we have

(3x)

â‡’ 9x

âˆ´ x = 0, 1 Area of the shaded region

= ar (region OAB) â€“ ar (DOAB)

Hence, the required area =sq. units.

We are given that: x

and y

From eqn. (i) we get y =

Putting the value of y in eqn. (ii) we have

â‡’ x

â‡’ x (x

Required area = Area of the region (OCBA â€“ ODBA)

Hence, the required area =sq. units.

We are given that: y = x

Solving y = x

x + 6 = x

â‡’ x

â‡’ x

â‡’ (x â€“ 2) (x

x

âˆ´ Required area of the shaded region

10 sq. units.

We have y

â‡’

â‡’

â‡’ x

â‡’ x(x

âˆ´ x = 0, x = 4

Required area

Hence, the required area =

Given that: y

and y = x...(ii)

Solving eqns. (i) and (ii) we have

x

x (x â€“ 9) = 0 âˆ´ x = 0, 9

Required area

Hence, the required area =

Here, x

âˆ´ x

â‡’ x

â‡’ x

â‡’ x(x â€“2) + 1 (x â€“ 2) = 0

â‡’ (x â€“ 2) (x + 1) = 0

âˆ´ x = â€“1, 2

Graph of y = x + 2

Area of the required region

Hence, the required area =

Here, y

y

âˆ´ y = Â±4

Required area

Hence, the area of the region =

Given that {(x, 0) : y =

â‡’ y

â‡’ x

Required area

[Since circle is symmetrical about y-axis]

Hence, the required area = 2Ï€ sq. units.

Given the curves y = 2âˆšx , x = 0 and x = 1.

y = 2âˆšx â‡’ y

Required area

Hence, required area =

Given that: 2y = 5x + 7, x-axis, x = 2 and x = 8.

Let us draw the graph of 2y = 5x + 7 â‡’ y =

Area of the required shaded region

Hence, the required area = 96 sq. units.**Q.11. Draw a rough sketch of the curve y =in the interval [1, 5]. Find the area under the curve and between the lines x = 1 and x = 5.****Ans.**

Here, we have y =

â‡’ y^{2} = x â€“ 1 (Parabola)

Area of the required region

Hence, the required area =**Q.12. Determine the area under the curve y = included between the lines x = 0 and x = a.****Ans.**

Here, we are given y =

â‡’ y^{2} = a^{2} â€“ x^{2}

â‡’ x^{2} + y^{2} = a^{2}

Area of the shaded region

Hence, the required area =**Q.13. Find the area of the region bounded by y = âˆšx and y = x.****Ans.**

We are given the equations of curve y = âˆšx and line y = x.

Solving y = âˆšx â‡’ y^{2} = x and y = x, we get

x^{2} = x â‡’ x^{2} â€“ x = 0

â‡’ x (x â€“ 1) = 0 âˆ´ x = 0, 1

Required area of the shaded region

Hence, the required area =**Q.14. Find the area enclosed by the curve y = â€“x ^{2} and the straight lilne x + y + 2 = 0.**

We are given that y = â€“x

and the line x + y + 2 = 0

Solving the two equations, we get

x â€“ x

â‡’ x

â‡’ x

â‡’ x (x â€“ 2) + 1 (x â€“ 2) = 0

â‡’ (x â€“ 2) (x + 1) = 0

âˆ´ x = â€“1, 2

Area of the required shaded region

â‡’

â‡’

â‡’

â‡’

Given that: y = âˆšx , x = 2y + 3, first quadrant and x-axis.

Solving y = âˆšx and x = 2y + 3, we get

y = â‡’ y

â‡’ y^{2} â€“ 2y â€“ 3 = 0 â‡’ y^{2} â€“ 3y + y â€“ 3 = 0

â‡’ y(y â€“ 3) + 1 (y â€“ 3) = 0

â‡’(y + 1) (y â€“ 3) = 0

âˆ´ y = â€“1, 3

Area of shaded region

= 18 - 9 = 9 sq. units

Hence, the required area = 9 sq. units.**Long Answer (L.A.)****Q.16. Find the area of the region bounded by the curve y ^{2} = 2x and x^{2} + y^{2} = 4x.**

Equations of the curves are given by

x

and y

â‡’ x

â‡’ x

â‡’ (x â€“ 2)

Clearly it is the equation of a circle having its centre (2, 0) and radius 2.

Solving x

x

â‡’ x

â‡’ x

â‡’ x (x â€“ 2) = 0

âˆ´ x = 0, 2

Area of the required region

[âˆ´ Parabola and circle both are symmetrical about x-axis.]

sq. units

Hence, the required area =

Required area =

The coordinates of the vertices of Î”ABC are given by A(â€“1, 1), B (0, 5) and C (3, 2).

Equation of AB is y â€“ 1 =

â‡’ y â€“ 1 = 4x + 4

âˆ´ y = 4x + 4 + 1 â‡’ y = 4x + 5 ...(i)

Equation of BC is y â€“ 5 =

â‡’ y â€“ 5 = â€“x

âˆ´ y = 5 â€“ x ...(ii)

Equation of CA is

â‡’

âˆ´

Area of Î”ABC

Given that:

{(x, y) : y

Equation of Parabola is

y

and equation of circle is

x

Solving eqns. (i) and (ii) we get

x

â‡’ x

â‡’ x

â‡’ x(x + 8a) â€“ 2a (x + 8a) = 0

â‡’(x + 8a) (x â€“ 2a) = 0

âˆ´ x = 2a and x = â€“ 8a. (Rejected as it is out of region)

Area of the required shaded region

Hence, required area =

Given that: x + 2y = 2 ...(i)

y â€“ x = 1 ...(ii)

and 2x + y = 7 ...(iii)

Solving eqns. (ii) and (iii) we get

y = 1 + x

âˆ´ 2x + 1 + x = 7

3x = 6

â‡’ x = 2

âˆ´ y = 1 + 2 = 3

Coordinates of B = (2, 3)

Solving eqns. (i) and (iii)

we get

x + 2y = 2

âˆ´ x = 2 â€“ 2y

2x + y = 7

2(2 â€“ 2y) + y = 7

â‡’ 4 â€“ 4y + y = 7 â‡’ - 3y = 3

âˆ´ y = â€“1 and x = 4

âˆ´ Coordinates of C = (4, â€“ 1) and coordinates of A = (0, 1).

Taking the limits on y-axis, we get

Hence, the required area = 6 sq. units.

Given that y = 4x + 5 ...(i)

y = 5 â€“ x ...(ii)

and 4y = x + 5 ...(iii)

Solving eq. (i) and (ii) we get

4x + 5 = 5 â€“ x

â‡’ x = 0 and y = 5

âˆ´ Coordinates of A = (0, 5)

Solving eq. (ii) and (iii)

y = 5 â€“ x

4y = x + 5

5y = 10

âˆ´ y = 2 and x = 3

âˆ´ Coordinates of B = (3, 2)

Solving eq. (i) and (iii)

y = 4x + 5

4y = x + 5

â‡’ 4 (4x + 5) = x + 5

â‡’ 16x + 20 = x + 5 â‡’ 15x = - 15

âˆ´ x = â€“1 and y = 1

âˆ´ Coordinates of C = (â€“1, 1).

âˆ´ Area of required regions

Hence, the required area =

Given equation of the curve is y = 2 cos x

âˆ´ Area of the shaded region

Given equations are y = 1 + |x + 1|, x = â€“3

and x = 3, y = 0

Taking y = 1 + |x + 1|

â‡’ y = 1 + x + 1

â‡’ y = x + 2 and y = 1 â€“ x â€“ 1

â‡’ y = â€“x

On solving we get x = â€“1

Area of the required regions

Hence, the required area = 16 sq. units.

Given that y-axis, y = cos x, y = sin x,

Required area

Hence, the correct option is (c).

Given that: The equation of parabola is x

and equation of straight line is x = 4y â€“ 2 ...(ii)

Solving eqn. (i) and (ii) we get

â‡’ x = x

â‡’ x

â‡’ x (x â€“ 2) + 1 (x â€“ 2) = 0 â‡’ (x â€“ 2) (x + 1) = 0 âˆ´ x = â€“1, x = 2

Required area =

Hence, the correct option is (d).

Here, equation of curve is y =

Required area

Hence, the correct option is (a).

Solution .

Given equation of circle is x

and the line is y = x and the x-axis.

Solving the two equations we have

x

â‡’ 2x

â‡’ x

âˆ´ x = Â± 4

Required area

Hence, the correct option is (b).

Solution .

Given that: y = cos x, x = 0, x = Ï€

Required area

Hence, the correct option is (a).

Solution .

Given equation of parabola is y

and equation of straight line is 2y = x ...(ii)

Solving eqns. (i) and (ii) we get

â‡’ x(x â€“ 4) = 0 âˆ´ x = 0, 4

Required area

Hence, the correct answer is (a).

Solution .

Given equation of curve is y = sin x between x = 0 and x =

Area of required region

Hence, the correct answer is (d).

Solution .

Given equation of ellipse is

âˆ´ Since the ellipse is symmetrical about the axes.

âˆ´ Required area

Hence, the correct answer is (a).

Solution .

Given equation of circle is

x

Since the circle is symmetrical about the axes.

Hence, the correct answer is (b).

Solution .

Given equation of lines are

y = x + 1, x = 2 and x = 3

Required area

Hence, the correct option is (a).

**(a) 4 sq. units (b) 3/2 sq. units **

**(c) 6 sq. units (d) 8 sq. units**

Solution .

Given equations of lines are x = 2y + 3, y = 1 and y = â€“1

Required area

Hence, the correct answer is (c).

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