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**Application of Integral**

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

**ANSWER : - **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 & the x-axis in the first quadrant.**

**ANSWER : - **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.**

**ANSWER : - **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 **

**ANSWER : - **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.

âˆ´ Area bounded by ellipse = 4 Ã— Area of OAB

Therefore, area bounded by the ellipse = 4 Ã— 3Ï€ = 12Ï€ units

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

**ANSWER : -**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.

âˆ´ Area bounded by ellipse = 4 Ã— Area OAB

Therefore, area bounded by the ellipse =

**Question 6: Find the area of the region in the first quadrant enclosed by x-axis, line and the circle **

**ANSWER : -**The area of the region bounded by the circle, , and the *x*-axis is the area OAB.

The point of intersection of the line and the circle in the first quadrant is .

Area OAB = Area Î”OCA Area ACB

Area of OAC

Area of ABC

Therefore, area enclosed by *x*-axis, the line, and the circle in the first quadrant =

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

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

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

âˆ´ Area ABCD = 2 Ã— Area ABC

Therefore, the area of smaller part of the circle, *x*^{2} + *y*^{2} = *a*^{2}, cut off by the line, , is units.

**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.**

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

âˆ´ Area OAD = Area ABCD

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

â‡’ Area OED Area EFCD

From (1) and (2), we obtain

Therefore, the value of *a* is .

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

**ANSWER : -**The area bounded by the parabola, *x*^{2} = *y*,and the line,, can be represented as

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

âˆ´ 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 Î”OAB â€“ Area OBACO

â‡’ Area of OACO = Area of Î”OAB â€“ Area of OBACO

Therefore, required area = units

**Question 10: Find the area bounded by the curve x^{2} = 4y and the line x = 4y â€“ 2**

**ANSWER : -**The area bounded by the curve, *x*^{2} = 4*y*, and line, *x* = 4*y* â€“ 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 .

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 â€¦ (1)

Then, Area OBCO = Area OMBC â€“ Area OMBO

Similarly, Area OACO = Area OLAC â€“ Area OLAO

Therefore, required area =

**Question 11: Find the area of the region bounded by the curve y^{2} = 4x and the line x = 3**

**ANSWER : -**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.

âˆ´ Area of OACO = 2 (Area of OAB)

Therefore, the required area is units.

**Question 12: 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.** Ï€

**B.**

**C.**

**D. **

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

Thus, the correct answer is A.

**Question 13: Area of the region bounded by the curve y^{2} = 4x, y-axis and the line y = 3 is**

**A.** 2

**B.**

**C.**

**D. **

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

Thus, the correct answer is B.

**Question 14: Find the area of the circle 4 x^{2} + 4y^{2} = 9 which is interior to the parabola x^{2} = 4y**

**ANSWER : -**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.

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

âˆ´ Area OBCDO = 2 Ã— Area OBCO

We draw BM perpendicular to OA.

Therefore, the coordinates of M are.

Therefore, Area OBCO = Area OMBCO â€“ Area OMBO

Therefore, the required area OBCDO is units.

**Question 15: Find the area bounded by curves ( x â€“ 1)^{2} + y^{2} = 1 and x^{2} + y ^{2} = 1**

**ANSWER : -**The area bounded by the curves, (*x* â€“ 1)^{2} + *y*^{2} = 1 and *x*^{2} + *y* ^{2} = 1, is represented by the shaded area as

On solving the equations, (*x* â€“ 1)^{2} + *y*^{2} = 1 and *x*^{2} + *y* ^{2} = 1, we obtain the point of intersection as Aand B.

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

âˆ´ Area OBCAO = 2 Ã— Area OCAO

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

The coordinates of M are .

Therefore, required area OBCAO = units.

**Question 16: Find the area of the region bounded by the curves y = x^{2} + 2, y = x, x = 0 and x = 3**

**ANSWER : -**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 â€“ Area ODCO

**Question 17: Using integration finds the area of the region bounded by the triangle whose vertices are (â€“1, 0), (1, 3) and (3, 2).**

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

It can be observed in the following figure that,

Area (Î”ACB) = Area (ALBA) Area (BLMCB) â€“ Area (AMCA) â€¦ (1)

Equation of line segment AB is

Equation of line segment BC is

Equation of line segment AC is

Therefore, from equation (1), we obtain

Area (Î”ABC) = (3 + 5 â€“ 4) = 4 units

**Question 18: Using integration find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.**

**ANSWER : -**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 (Î”ACB) = Area (OLBAO) â€“Area (OLCAO)

**Question 19: Smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2 is**

**A. 2 (Ï€ â€“ 2) **

**B. Ï€ â€“ 2 **

**C. 2Ï€ â€“ 1 **

** D. 2 (Ï€ 2)**

**ANSWER : -**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 â€“ Area (Î”OAB)

Thus, the correct answer is B.

**Question 20: Area lying between the curve y^{2} = 4x and y = 2x is**

**A. **

**B. **

**C. **

**D. **

**ANSWER : -**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).

âˆ´ Area OBAO = Area (Î”OCA) â€“ Area (OCABO)

Thus, the correct answer is B.

**Question 21: 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 â€“axis

**ANSWER : - **(i) The required area is represented by the shaded area ADCBA as

- The required area is represented by the shaded area ADCBA as

**Question 22: Find the area between the curves y = x and y = x^{2}**

**ANSWER : -**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.

âˆ´ Area (OBAO) = Area (Î”OCA) â€“ Area (OCABO) â€¦ (1)

**Question 23: Find the area of the region lying in the first quadrant and bounded by y = 4x^{2}, x = 0, y = 1 and y = 4**

**ANSWER : -**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 24: Sketch the graph of and evaluate**

**ANSWER : -**The given equation is

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

On plotting these points, we obtain the graph of as follows.

It is known that,

**Question 25: Find the area bounded by the curve y = sin x between x = 0 and x = 2Ï€**

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

âˆ´ Required area = Area OABO Area BCDB

**Question 26: Find the area enclosed between the parabola y^{2} = 4ax and the line y = mx**

**ANSWER : -**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.

âˆ´ Area OABO = Area OCABO â€“ Area (Î”OCA)

**Question 27: Find the area enclosed by the parabola 4 y = 3x^{2} and the line 2y = 3x + 12**

**ANSWER : -**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 (â€“2, 3) and (4, 12).

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

âˆ´ Area OBAO = Area CDBA â€“ (Area ODBO Area OACO)

**Question 28:Find the area of the smaller region bounded by the ellipse and the line **

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

âˆ´ Area BCAB = Area (OBCAO) â€“ Area (OBAO)

**Question 29:Find the area of the smaller region bounded by the ellipse & the line **

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

âˆ´ Area BCAB = Area (OBCAO) â€“ Area (OBAO)

**Question 30: **Find the area of the region enclosed by the parabola *x*^{2} = *y*, the line *y* = *x* + 2 and *x*-axis

**ANSWER : -**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 OABCO as

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

âˆ´ Area OABCO = Area (BCA) Area COAC

**Question 31: Using the method of integration find the area bounded by the curve **

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

**ANSWER : -**The area bounded by the curve, , is represented by the shaded region ADCB as

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

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

âˆ´ Area ADCB = 4 Ã— Area OBAO

**Question 32: Find the area bounded by curves **

**ANSWER : -**The area bounded by the curves, , is represented by the shaded region as

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

**Question 33: Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A (2, 0), B (4, 5) and C (6, 3)**

**ANSWER : -**The vertices of Î”ABC are A (2, 0), B (4, 5), and C (6, 3).

Equation of line segment AB is

Equation of line segment BC is

Equation of line segment CA is

Area (Î”ABC) = Area (ABLA) Area (BLMCB) â€“ Area (ACMA)

**Question 34: Using the method of integration find the area of the region bounded by lines:**

2*x* + *y* = 4, 3*x* â€“ 2*y* = 6 and *x* â€“ 3*y* + 5 = 0

**ANSWER : -**The given equations of lines are

2*x* + *y* = 4 â€¦ (1)

3*x* â€“ 2*y* = 6 â€¦ (2)

And, *x* â€“ 3*y* + 5 = 0 â€¦ (3)

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

Area (Î”ABC) = Area (ALMCA) â€“ Area (ALB) â€“ Area (CMB)

**Question 35: Find the area of the region **

**ANSWER : -**The area bounded by the curves, , is represented as

The points of intersection of both the curves are.

The required area is given by OABCO.

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

âˆ´ Area OABCO = 2 Ã— Area OBC

Area OBCO = Area OMC Area MBC

Therefore, the required area is units

**Question 36: Area bounded by the curve y = x^{3}, the x-axis and the ordinates x = â€“2 and x = 1 is**

**A.** â€“ 9 **B.** **C.** **D. **

**ANSWER : -**

Thus, the correct answer is B.

**Question 37:The area bounded by the curve, x-axis and the ordinates x = â€“1 & x = 1 is given by [Hint: y = x^{2} if x > 0 and y = â€“x^{2} if x < 0]**

**A.** 0 **B.**

**C.**

**D. **

**ANSWER : -**

Thus, the correct answer is C.

**Question 38: The area of the circle x^{2} + y^{2} = 16 exterior to the parabola y^{2} = 6x is**

**A.** **B.** **C.** **D. **

**ANSWER : -**The given equations are

*x*^{2} + *y*^{2} = 16 â€¦ (1)

*y*^{2} = 6*x* â€¦ (2)

Area bounded by the circle and parabola

Area of circle = Ï€ (*r*)^{2}

= Ï€ (4)^{2}

= 16Ï€ units

Thus, the correct answer is C.

**Question 39: The area bounded by the y-axis, y = cos x and y = sin x when **

**A.** **B.**

**C.** **D. **

**ANSWER : -**The given equations are

*y* = cos *x* â€¦ (1)

And, *y* = sin *x* â€¦ (2)

Required area = Area (ABLA) area (OBLO)

Integrating by parts, we obtain

Thus, the correct answer is B.

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