### Exercise 8.1

### Solve The Following Questions.

**Question 1. Find the area of the region bounded by the curve y^{2} = x and the lines x = 1, x = 4 and the x-axis.**

**Solution :**

The area of the region bounded by the curve, *y*^{2} = *x*, the lines,* x* = 1 and* x* = 4, and the *x*-axis is the area ABCD.**Question 2. Find the area of the region bounded by y^{2} = 9x, x = 2, x = 4 and the x-axis in the first quadrant.**

**Solution :**

The area of the region bounded by the curve, *y*^{2} = 9*x*, *x* = 2, and* x* = 4, and the *x*-axis is the area ABCD.

**Question 3. ****Find the area of the region bounded by x^{2} = 4y, y = 2, y = 4 and the y-axis in the first quadrant.**

**Solution :**

The area of the region bounded by the curve, *x*^{2} = 4*y*, *y* = 2, and* y* = 4, and the *y*-axis is the area ABCD.

**Question 4. Find the area of the region bounded by the ellipse ****Solution :**

The given equation of the ellipse, , can be represented as

It can be observed that the ellipse is symmetrical about *x*-axis and *y*-axis.

∴ Area bounded by ellipse = 4 × Area of OAB

Therefore, area bounded by the ellipse = 4 × 3π = 12π units

**Question 5. Find the area of the region bounded by the ellipse ****Solution :**

The given equation of the ellipse can be represented as

It can be observed that the ellipse is symmetrical about *x*-axis and *y*-axis.

∴ Area bounded by ellipse = 4 × Area OAB

Therefore, area bounded by the ellipse = 4 x3π/2 = 6π units.**Question 6. **Find the area of the region in the first quadrant enclosed by *x*-axis, line x = √3y and the circle x^{2} + y^{2} = 4

**Solution :**

The area of the region bounded by the circle, x^{2} + y^{2} = 4,x = √3y and the *x*-axis is the area OAB.

The point of intersection of the line and the circle in the first quadrant is (√3,1).

Area OAB = Area ΔOCA + Area ACB

**Question 7. Find the area of the smaller part of the circle x^{2} + y^{2} = a^{2} cut off by the line x = a/√2**

**Solution :**

The area of the smaller part of the circle, *x*^{2} +* y*^{2} = *a*^{2}, cut off by the line, x = a/√2, is the area ABCDA.

It can be observed that the area ABCD is symmetrical about *x*-axis.

∴ Area ABCD = 2 × Area ABC

Therefore, the area of smaller part of the circle, *x*^{2} +* y*^{2} = *a*^{2}, cut off by the line,**Question****8. The area between x = y^{2} and x = 4 is divided into two equal parts by the line x = a, find the value of a.**

**Solution :**

The line, *x* = *a*, divides the area bounded by the parabola and *x* = 4 into two equal parts.

∴ Area OAD = Area ABCD

It can be observed that the given area is symmetrical about *x*-axis.

⇒ Area OED = Area EFCD

**Question****9. Find the area of the region bounded by the parabola y = x^{2} and y = |x|**

**Solution :**

The area bounded by the parabola,

*x*

^{2}=

*y*,and the line,y = |x|, can be represented as

The given area is symmetrical about *y*-axis.

∴ Area OACO = Area ODBO

The point of intersection of parabola, *x*^{2} = *y*, and line, *y *= *x*, is A (1, 1).

Area of OACO = Area ΔOAM – Area OMACO

Therefore, required area = 2[1/6] = 1/3 units

**Question****10. Find the area bounded by the curve x^{2} = 4y and the line x = 4y – 2**

**Solution :**The area bounded by the curve,

*x*

^{2}= 4

*y*, and line,

*x*= 4

*y*– 2, is represented by the shaded area OBAO.

Let A and B be the points of intersection of the line and parabola.

Coordinates of point A are (-1, 1/4).

Coordinates of point B are (2, 1).

We draw AL and BM perpendicular to *x*-axis.

It can be observed that,

Area OBAO = Area OBCO + Area OACO … (1)

Then, Area OBCO = Area OMBC – Area OMBO

Similarly, Area OACO = Area OLAC – Area OLAO

Therefore, required area =units**Question****11. ****Find the area of the region bounded by the curve y^{2} = 4x and the line x = 3**

**Solution :**

The region bounded by the parabola, *y*^{2} = 4*x*, and the line, *x* = 3, is the area OACO.

The area OACO is symmetrical about *x*-axis.

∴ Area of OACO = 2 (Area of OAB)

Therefore, the required area is 8√3 units.**Question****12. Choose the correct answer:**

**Area lying in the first quadrant and bounded by the circle x^{2} + y^{2} = 4 and the lines x = 0 and x = 2 is**

**A. π**

**B. π/2**

**C. π/3**

**D. π/4**

**Solution :**

The area bounded by the circle and the lines, *x* = 0 and *x* = 2, in the first quadrant is represented as

Therefore, option (A) is correct.**Question****13. Choose the correct answer:**

**Area of the region bounded by the curve y^{2} = 4x, y-axis and the line y = 3 is**

**A. 2**

**B. 9/4**

**C. 9/3**

**D. 9/2**

**Solution :**

The area bounded by the curve, *y*^{2} = 4*x*, *y*-axis, and *y* = 3 is represented as

Therefore, option (B) is correct.

### Exercise 8.2

### Solve The Following Questions.

**Question 1. Find the area of the circle 4 x^{2} + 4y^{2} = 9 which is interior to the parabola x^{2} = 4y**

**Solution :**

The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, 4*x*^{2} + 4*y*^{2} = 9, and parabola, *x*^{2} = 4*y*, we obtain the point of intersection as B(√2,1/2) and D (-√2,1/2).

It can be observed that the required area is symmetrical about *y*-axis.

∴ Area OBCDO = 2 × Area OBCO

We draw BM perpendicular to OA.

Therefore, the coordinates of M are (√2,0).

Therefore, Area OBCO = Area OMBCO – Area OMBO

Therefore, the required area OBCDO is

**Question 2. Find the area bounded by curves ( x – 1)^{2} + y^{2} = 1 and x^{2} + y^{ 2} = 1**

**Solution :**

The area bounded by the curves, (*x* – 1)^{2} + *y*^{2} = 1 and *x*^{2} + *y*^{ 2} = 1, is represented by the shaded area as

On solving the equations, (*x* – 1)^{2} + *y*^{2} = 1 and *x*^{2} + *y*^{ 2} = 1, we obtain the point of intersection as A (1/2,√3/2) and B (1/2,√3/2) .

It can be observed that the required area is symmetrical about *x*-axis.

∴ Area OBCAO = 2 × Area OCAO

We join AB, which intersects OC at M, such that AM is perpendicular to OC.

The coordinates of M are (1/2,0).

Therefore, required area OBCAO = units.

**Question 3. Find the area of the region bounded by the curves y = x^{2 }+ 2, y = x, x = 0 and x = 3**

**Solution :**

The area bounded by the curves, *y* = *x*^{2 }+ 2, *y *= *x*, *x* = 0, and *x* = 3, is represented by the shaded area OCBAO as

Then, Area OCBAO = Area ODBAO – Area ODCO

**Question 4. Using integration, find the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).**

**Solution :**

BL and CM are drawn perpendicular to *x*-axis.

It can be observed in the following figure that,

Area (ΔACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)

Therefore, from equation (1), we obtain

Area (ΔABC) = (3 + 5 – 4) = 4 units.

**5. Using integration, find the area of the triangular region whose sides have the equations y = 2x +1, y = 3x + 1 and x = 4.**

**Solution :**

The equations of sides of the triangle are *y* = 2*x* +1, *y* = 3*x* + 1, and *x *= 4.

On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C (4, 9).

It can be observed that,

Area (ΔACB) = Area (OLBAO) –Area (OLCAO)

**Question 6. Choose the correct answer:**

**Smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2 is**

**(A)2 (π – 2)**

**(B). π – 2**

**(C). 2π – 1**

**(D). 2 (π + 2)**

**Solution :**

The smaller area enclosed by the circle, *x*^{2} + *y*^{2} = 4, and the line, *x* + *y* = 2, is represented by the shaded area ACBA as

It can be observed that,

Area ACBA = Area OACBO – Area (ΔOAB)

Therefore, option (B) is correct.

**Question 7. Choose the correct answer:**

**Area lying between the curves y^{2} = 4x and y = 2x is**

**(A) 2/3**

**(B) 1/3**

**(C) 1/4**

**(D) 3/4**

**Solution :**

The area lying between the curve, *y*^{2} = 4*x* and *y* = 2*x*, is represented by the shaded area OBAO as

The points of intersection of these curves are O (0, 0) and A (1, 2).

We draw AC perpendicular to *x*-axis such that the coordinates of C are (1, 0).

∴ Area OBAO = Area (OCABO) – Area (ΔOCA)

Therefore, option (B) is correct.

### Miscellaneous Exercise

### Solve The Following Questions.

**Question 1. Find the area under the given curves and given lines:**

**(i) y = x^{2}, x = 1, x = 2 and x-axis**

**(ii) y = x^{4}, x = 1, x = 5 and x –axis**

**Solution :**

(i)The required area is represented by the shaded area ADCBA as

(ii)The required area is represented by the shaded area ADCBA as

**Question 2. Find the area between the curves y = x and y = x^{2}**

**Solution :**

The required area is represented by the shaded area OBAO as

The points of intersection of the curves, *y* = *x* and *y* = *x*^{2}, is A (1, 1).

We draw AC perpendicular to *x*-axis.

∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)

**Question 3. Find the area of the region lying in the first quadrant and bounded by y = 4x^{2}, x = 0, y = 1 and y = 4**

**Solution :** The area in the first quadrant bounded by *y* = 4*x*^{2}, *x* = 0, *y* = 1, and *y *= 4 is represented by the shaded area ABCDA as

**Question 4. Sketch the graph of y = |x + 3| and evaluate **

**Solution :**

The given equation is y = |x + 3|

The corresponding values of *x *and *y* are given in the following table.

On plotting these points, we obtain the graph of y = |x + 3| as follows.

**Question 5. Find the area bounded by the curve y = sin x between x = 0 and x = 2π**

**Solution :**

The graph of *y* = sin *x* can be drawn as

∴ Required area = Area OABO + Area BCDB

**Question 6. Find the area enclosed between the parabola y^{2} = 4ax and the line y = mx**

**Solution :**

The area enclosed between the parabola, *y*^{2} = 4*ax*, and the line,* y *= *mx*, is represented by the shaded area OABO as

The points of intersection of both the curves are (0, 0) and .

We draw AC perpendicular to *x*-axis.

∴ Area OABO = Area OCABO – Area (ΔOCA)

**Question 7. Find the area enclosed by the parabola 4 y = 3x^{2} and the line 2y = 3x + 12**

**Solution :**

The area enclosed between the parabola, 4*y* = 3*x*^{2}, and the line, 2*y* = 3*x* + 12, is represented by the shaded area OBAO as

The points of intersection of the given curves are A (–2, 3) and (4, 12).

We draw AC and BD perpendicular to *x-*axis.

∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)

**Question 8. Find the area of the smaller region bounded by the ellipse and the line**

**Solution :**

The area of the smaller region bounded by the ellipse, ** = 1**, and the line, , is represented by the shaded region BCAB as

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

**Question 9. Find the area of the smaller region bounded by the ellipse and the line **

**Solution :**

The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

**Question 10. Find the area of the region enclosed by the parabola x^{2} = y, the line y = x + 2 and x-axis**

**Solution :**

The area of the region enclosed by the parabola, *x*^{2} = *y*, the line, *y* = *x* + 2, and *x*-axis is represented by the shaded region OACO as

The point of intersection of the parabola, *x*^{2} = *y*, and the line, *y* = *x* + 2, is A (–1, 1) and C(2, 4).

**Question 11.Using the method of integration, find the area enclosed by the curve |x| + |y| = 1**

**[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 11]**

**Solution :**

The area bounded by the curve, |x| + |y| = 1, is represented by the shaded region ADCB as

The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0).

It can be observed that the given curve is symmetrical about *x*-axis and *y*-axis.

∴ Area ADCB = 4 × Area OBAO

**Question 12. Find the area bounded by curves {(x, y) : y ≥ x ^{2} and y = |x|}.**

**Solution :**

The area bounded by the curves, {(x, y) : y ≥ x^{2} and y = |x|}**.**, is represented by the shaded region as

It can be observed that the required area is symmetrical about *y*-axis.

**Question 13.Using the method of integration, find the area of the triangle whose vertices are A (2, 0), B (4, 5) and C (6, 3).**

**Solution :**

Vertices of the given triangle are A (2, 0), B (4, 5) and C (6, 3).

Equation of side AB is

Equation of side BC

Equation of side CA is

Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)

**Question 14.Using the method of integration, find the area of the region bounded by the lines: 2 x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0**

**Solution :**

The given equations of lines are

2*x* + *y* = 4 … (1)

3*x* – 2*y* = 6 … (2)

And, *x* – 3*y *+ 5 = 0 … (3)

The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on *x*-axis.

Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)

**Question 15. Find the area of the region {(x, y) : y2 ≤ 4x, 4×2 + 4y2 ≤ 9}.**

**Solution :**

The area bounded by the curves, {(x, y) : y2 ≤ 4x, 4×2 + 4y2 ≤ 9}, is represented as

The points of intersection of both the curves are (1/2,√2) and (1/2, -√2).

The required area is given by OABCO.

It can be observed that area OABCO is symmetrical about *x*-axis.

∴ Area OABCO = 2 × Area OBC

Area OBCO = Area OMC + Area MBC

Therefore, the required area is units.

**Question 16.Choose the correct answer:**

**Area bounded by the curve y = x^{3}, the x-axis and the ordinates x = –2 and x = 1 is**

**(A) -9**

**(B) -15/4**

**(C) 15/4**

**(D) 17/4**

**Solution :**

Therefore, option (D) is correct.

**Question 17.Choose the correct answer:**

**The area bounded by the curve y = x|x|, axis and the ordinates x = –1 and x = 1 is given by:**

**[Hint: y = x^{2} if x > 0 and y = –x^{2} if x < 0]**

**(A) 0**

**(B) 1/3**

**(C) 2/3**

**(D) 4/3**

**Solution :**

Therefore, option (C) is correct.

**Question 18.Choose the correct answer:**

**The area of the circle x^{2} + y^{2} = 16 exterior to the parabola y^{2} = 6x is**

**Solution :**

The given equations are

*x*^{2} + *y*^{2} = 16 … (1)

*y*^{2} = 6*x*… (2)

Area bounded by the circle and parabola

Thus, the correct answer is C.

**Question 19.Choose the correct answer:**

**The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ π/2.**

**(A) 2(√2 – 1)**

**(B) √2 – 1**

**(C) √2 + 1**

**(D) √2**

**Solution :**

The given equations are

*y* = cos *x* … (1)

And,* y *= sin *x* … (2)

Required area = Area (ABLA) + area (OBLO)

Required area = Area (AB

Therefore, option (B) is correct.