NCERT Solutions - Application of Integral JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : NCERT Solutions - Application of Integral JEE Notes | EduRev

The document NCERT Solutions - Application of Integral JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Application of Integral

Question 1: Find the area of the region bounded by the curve y2x & the lines x = 1, x = 4 & the x-axis.

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

ANSWER : - The area of the region bounded by the curve, y2x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

Question 2: Find the area of the region bounded by y2 = 9xx = 2, x = 4 & the x-axis in the first quadrant.

ANSWER : - The area of the region bounded by the curve, y2 = 9xx = 2, and x = 4, and the x-axis is the area ABCD.

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

Question 3: Find the area of the region bounded by x2 = 4yy = 2, y = 4 and the y-axis in the first quadrant.

ANSWER : -  The area of the region bounded by the curve, x2 = 4yy = 2, and y = 4, and the y-axis is the area ABCD.

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

Question 4: Find the area of the region bounded by the ellipse NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : - The given equation of the ellipse, NCERT Solutions - Application of Integral JEE Notes | EduRev, can be represented as

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

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

∴ Area bounded by ellipse = 4 × Area of OAB

NCERT Solutions - Application of Integral JEE Notes | EduRev

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

Question 5: Find the area of the region bounded by the ellipse NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The given equation of the ellipse can be represented as

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

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

∴ Area bounded by ellipse = 4 × Area OAB

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, area bounded by the ellipse = NCERT Solutions - Application of Integral JEE Notes | EduRev

 

 Question 6: Find the area of the region in the first quadrant enclosed by x-axis, line NCERT Solutions - Application of Integral JEE Notes | EduRevand the circle NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The area of the region bounded by the circle, NCERT Solutions - Application of Integral JEE Notes | EduRev, and the x-axis is the area OAB.

The point of intersection of the line and the circle in the first quadrant is NCERT Solutions - Application of Integral JEE Notes | EduRev.

Area OAB = Area ΔOCA Area ACB

Area of OAC NCERT Solutions - Application of Integral JEE Notes | EduRev

Area of ABC NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, area enclosed by x-axis, the lineNCERT Solutions - Application of Integral JEE Notes | EduRev, and the circle NCERT Solutions - Application of Integral JEE Notes | EduRevin the first quadrant = NCERT Solutions - Application of Integral JEE Notes | EduRev

 

Question 7: Find the area of the smaller part of the circle x2y2a2 cut off by the line NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The area of the smaller part of the circle, x2y2a2, cut off by the line, NCERT Solutions - Application of Integral JEE Notes | EduRev, is the area ABCDA.

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

∴ Area ABCD = 2 × Area ABC

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, the area of smaller part of the circle, x2y2a2, cut off by the line, NCERT Solutions - Application of Integral JEE Notes | EduRev, is NCERT Solutions - Application of Integral JEE Notes | EduRev units.

 

Question 8: The area between xy2 and x = 4 is divided into two equal parts by the line xa, find the value of a.

ANSWER : -The line, xa, 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

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

 From (1) and (2), we obtain

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, the value of a is NCERT Solutions - Application of Integral JEE Notes | EduRev.

Question 9: Find the area of the region bounded by the parabola yx2 and NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The area bounded by the parabola, x2y,and the line,NCERT Solutions - Application of Integral JEE Notes | EduRev, can be represented as

The given area is symmetrical about y-axis.

∴ Area OACO = Area ODBO

The point of intersection of parabola, x2y, and line, yx, is A (1, 1).

NCERT Solutions - Application of Integral JEE Notes | EduRev

Area of OACO = Area ΔOAB – Area OBACO

NCERT Solutions - Application of Integral JEE Notes | EduRev

⇒ Area of OACO = Area of ΔOAB – Area of OBACO

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, required area = NCERT Solutions - Application of Integral JEE Notes | EduRevunits

Question 10: Find the area bounded by the curve x2 = 4y and the line x = 4y – 2

ANSWER : -The area bounded by the curve, x2 = 4y, and line, x = 4y – 2, is represented by the shaded area OBAO.

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

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

Coordinates of point NCERT Solutions - Application of Integral JEE Notes | EduRev.

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

NCERT Solutions - Application of Integral JEE Notes | EduRev

Similarly, Area OACO = Area OLAC – Area OLAO

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, required area = NCERT Solutions - Application of Integral JEE Notes | EduRev

Question 11: Find the area of the region bounded by the curve y2 = 4x and the line x = 3

ANSWER : -The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO.

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

The area OACO is symmetrical about x-axis.

∴ Area of OACO = 2 (Area of OAB)

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, the required area is NCERT Solutions - Application of Integral JEE Notes | EduRevunits.

 Question 12: Area lying in the first quadrant and bounded by the circle x2  +  y2 = 4 and the lines x = 0 and x = 2 is

A. π  

B. NCERT Solutions - Application of Integral JEE Notes | EduRev 

C. NCERT Solutions - Application of Integral JEE Notes | EduRev 

D. NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

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

NCERT Solutions - Application of Integral JEE Notes | EduRev

Thus, the correct answer is A.

 Question 13: Area of the region bounded by the curve y2 = 4xy-axis and the line y = 3 is

A.

B. NCERT Solutions - Application of Integral JEE Notes | EduRev

C. NCERT Solutions - Application of Integral JEE Notes | EduRev

D. NCERT Solutions - Application of Integral JEE Notes | EduRev

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : - The area bounded by the curve, y2 = 4xy-axis, and y = 3 is represented as

NCERT Solutions - Application of Integral JEE Notes | EduRev

Thus, the correct answer is B.

Question 14: Find the area of the circle 4x2 +  4y2 = 9 which is interior to the parabola x2 = 4y

ANSWER : -The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the point of

intersection asNCERT Solutions - Application of Integral JEE Notes | EduRev.

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

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 areNCERT Solutions - Application of Integral JEE Notes | EduRev.

Therefore, Area OBCO = Area OMBCO – Area OMBO

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, the required area OBCDO is NCERT Solutions - Application of Integral JEE Notes | EduRevunits.

Question 15: Find the area bounded by curves (x – 1)2 +  y2 = 1 and x2y 2 = 1

ANSWER : -The area bounded by the curves, (x – 1)2y2 = 1 and x2  + y 2 = 1, is represented by the shaded area as

On solving the equations, (x – 1)2 +  y2 = 1 and x2 +  y 2 = 1, we obtain the point of intersection as ANCERT Solutions - Application of Integral JEE Notes | EduRevand BNCERT Solutions - Application of Integral JEE Notes | EduRev.

NCERT Solutions - Application of Integral JEE Notes | EduRev

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 NCERT Solutions - Application of Integral JEE Notes | EduRev.

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, required area OBCAO = NCERT Solutions - Application of Integral JEE Notes | EduRevunits.

 

Question 16: Find the area of the region bounded by the curves yx2 + 2, yxx = 0 and x = 3

ANSWER : -The area bounded by the curves, yx2 +  2, yxx = 0, and x = 3, is represented by the shaded area OCBAO as

Then, Area OCBAO = Area ODBAO – Area ODCO

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

Question 17: Using integration finds the area of the region bounded by the triangle whose vertices are (–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 (ΔACB) = Area (ALBA) Area (BLMCB) – Area (AMCA) … (1)

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

Equation of line segment AB is

NCERT Solutions - Application of Integral JEE Notes | EduRev

Equation of line segment BC is

NCERT Solutions - Application of Integral JEE Notes | EduRev

Equation of line segment AC is

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, from equation (1), we obtain

Area (ΔABC) = (3 + 5 – 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 = 2x + 1, y = 3x + 1, and x = 4.

NCERT Solutions - Application of Integral JEE Notes | EduRev

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)

NCERT Solutions - Application of Integral JEE Notes | EduRev

Question 19: Smaller area enclosed by the circle x2 +  y2 = 4 and the line xy = 2 is

A. 2 (π – 2) 

B. π – 2 

C. 2π – 1 

 D. 2 (π 2)

 

ANSWER : -The smaller area enclosed by the circle, x2  + y2 = 4, and the line, x  + y = 2, is represented by the shaded area ACBA as

NCERT Solutions - Application of Integral JEE Notes | EduRev

It can be observed that,

Area ACBA = Area OACBO – Area (ΔOAB)

NCERT Solutions - Application of Integral JEE Notes | EduRev

Thus, the correct answer is B.

 Question 20: Area lying between the curve y2 = 4x and y = 2x is

A. NCERT Solutions - Application of Integral JEE Notes | EduRev

B. NCERT Solutions - Application of Integral JEE Notes | EduRev

C. NCERT Solutions - Application of Integral JEE Notes | EduRev     

D. NCERT Solutions - Application of Integral JEE Notes | EduRev

 

ANSWER : -The area lying between the curve, y2 = 4x and y = 2x, 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).

NCERT Solutions - Application of Integral JEE Notes | EduRev

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

NCERT Solutions - Application of Integral JEE Notes | EduRev

Thus, the correct answer is B.

 

Question 21: Find the area under the given curves and given lines:

(i) yx2x = 1, x = 2 and x-axis
 (ii) yx4x = 1, x = 5 and x –axis

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

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

NCERT Solutions - Application of Integral JEE Notes | EduRev

  1. The required area is represented by the shaded area ADCBA as

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

Question 22: Find the area between the curves yx and yx2

ANSWER : -The required area is represented by the shaded area OBAO as

NCERT Solutions - Application of Integral JEE Notes | EduRev

The points of intersection of the curves, yx and yx2, is A (1, 1).

We draw AC perpendicular to x-axis.

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

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

Question 23: Find the area of the region lying in the first quadrant and bounded by y = 4x2x = 0, y = 1 and y = 4

ANSWER : -The area in the first quadrant bounded by y = 4x2x = 0, y = 1, and y = 4 is represented by the shaded area ABCDA as

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

 Question 24: Sketch the graph of NCERT Solutions - Application of Integral JEE Notes | EduRevand evaluateNCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The given equation is NCERT Solutions - Application of Integral JEE Notes | EduRev

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

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

On plotting these points, we obtain the graph of NCERT Solutions - Application of Integral JEE Notes | EduRev as follows.

It is known that, NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev  

 

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

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

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

∴ Required area = Area OABO Area BCDB

NCERT Solutions - Application of Integral JEE Notes | EduRev

Question 26: Find the area enclosed between the parabola y2 = 4ax and the line ymx

ANSWER : -The area enclosed between the parabola, y2 = 4ax, and the line, ymx, is represented by the shaded area OABO as

The points of intersection of both the curves are (0, 0) and NCERT Solutions - Application of Integral JEE Notes | EduRev.

NCERT Solutions - Application of Integral JEE Notes | EduRev

We draw AC perpendicular to x-axis.

∴ Area OABO = Area OCABO – Area (ΔOCA)

NCERT Solutions - Application of Integral JEE Notes | EduRev

Question 27: Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 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)

NCERT Solutions - Application of Integral JEE Notes | EduRev

Question 28:Find the area of the smaller region bounded by the ellipse NCERT Solutions - Application of Integral JEE Notes | EduRevand the line NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The area of the smaller region bounded by the ellipse, NCERT Solutions - Application of Integral JEE Notes | EduRev, and the line, NCERT Solutions - Application of Integral JEE Notes | EduRev, is represented by the shaded region BCAB as

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

NCERT Solutions - Application of Integral JEE Notes | EduRev

Question 29:Find the area of the smaller region bounded by the ellipse NCERT Solutions - Application of Integral JEE Notes | EduRev & the line NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The area of the smaller region bounded by the ellipse, NCERT Solutions - Application of Integral JEE Notes | EduRev, and the line, NCERT Solutions - Application of Integral JEE Notes | EduRev, is represented by the shaded region BCAB as

NCERT Solutions - Application of Integral JEE Notes | EduRev

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

Question 30: Find the area of the region enclosed by the parabola x2y, the line yx + 2 and x-axis

ANSWER : -The area of the region enclosed by the parabola, x2y, the line, yx + 2, and x-axis is represented by the shaded region OABCO as

NCERT Solutions - Application of Integral JEE Notes | EduRev

The point of intersection of the parabola, x2y, and the line, yx + 2, is A (–1, 1).

∴ Area OABCO = Area (BCA) Area COAC

NCERT Solutions - Application of Integral JEE Notes | EduRev

Question 31: Using the method of integration find the area bounded by the curve NCERT Solutions - Application of Integral JEE Notes | EduRev

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

ANSWER : -The area bounded by the curve, NCERT Solutions - Application of Integral JEE Notes | EduRev, 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).

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

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

∴ Area ADCB = 4 × Area OBAO

NCERT Solutions - Application of Integral JEE Notes | EduRev

Question 32: Find the area bounded by curves NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The area bounded by the curves, NCERT Solutions - Application of Integral JEE Notes | EduRev, is represented by the shaded region as

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

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

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

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 ΔABC are A (2, 0), B (4, 5), and C (6, 3).

NCERT Solutions - Application of Integral JEE Notes | EduRev

Equation of line segment AB is

NCERT Solutions - Application of Integral JEE Notes | EduRev

Equation of line segment BC is

NCERT Solutions - Application of Integral JEE Notes | EduRev

Equation of line segment CA is

NCERT Solutions - Application of Integral JEE Notes | EduRev

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

NCERT Solutions - Application of Integral JEE Notes | EduRev

 

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

 

NCERT Solutions - Application of Integral JEE Notes | EduRev

2x +  y = 4, 3x – 2y = 6 and x – 3y +  5 = 0

 

ANSWER : -The given equations of lines are

2xy = 4 … (1)

3x – 2y = 6 … (2)

And, x – 3y + 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)

NCERT Solutions - Application of Integral JEE Notes | EduRev

Question 35: Find the area of the region NCERT Solutions - Application of Integral JEE Notes | EduRev

 

ANSWER : -The area bounded by the curves, NCERT Solutions - Application of Integral JEE Notes | EduRev, is represented as

The points of intersection of both the curves areNCERT Solutions - Application of Integral JEE Notes | EduRev.

The required area is given by OABCO.

NCERT Solutions - Application of Integral JEE Notes | EduRev

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

∴ Area OABCO = 2 × Area OBC

Area OBCO = Area OMC Area MBC

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

Therefore, the required area is NCERT Solutions - Application of Integral JEE Notes | EduRev units

 

Question 36: Area bounded by the curve yx3, the x-axis and the ordinates x = –2 and x = 1 is

A. – 9  
B. NCERT Solutions - Application of Integral JEE Notes | EduRev
C. NCERT Solutions - Application of Integral JEE Notes | EduRev  
D. NCERT Solutions - Application of Integral JEE Notes | EduRev

 

ANSWER : -

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

Thus, the correct answer is B.

 

Question 37:The area bounded by the curveNCERT Solutions - Application of Integral JEE Notes | EduRevx-axis and the ordinates x = –1 & x = 1 is given by [Hint: yx2 if x > 0 and y = –x2 if x < 0]

A. 0      
B. NCERT Solutions - Application of Integral JEE Notes | EduRev  
 C. NCERT Solutions - Application of Integral JEE Notes | EduRev 
 D. NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

Thus, the correct answer is C.

 

 Question 38: The area of the circle x2y2 = 16 exterior to the parabola y2 = 6x is

A. NCERT Solutions - Application of Integral JEE Notes | EduRev 
B. NCERT Solutions - Application of Integral JEE Notes | EduRev 
C. NCERT Solutions - Application of Integral JEE Notes | EduRev 
D. NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The given equations are

x2y2 = 16 … (1)

y2 = 6x … (2)

Area bounded by the circle and parabola

NCERT Solutions - Application of Integral JEE Notes | EduRev

Area of circle = π (r)2

= π (4)2

= 16π units

NCERT Solutions - Application of Integral JEE Notes | EduRev

Thus, the correct answer is C.

Question 39: The area bounded by the y-axis, y = cos x and y = sin x when NCERT Solutions - Application of Integral JEE Notes | EduRev

A. NCERT Solutions - Application of Integral JEE Notes | EduRev  
B. NCERT Solutions - Application of Integral JEE Notes | EduRev
 C. NCERT Solutions - Application of Integral JEE Notes | EduRev    
D. NCERT Solutions - Application of Integral JEE Notes | EduRev

NCERT Solutions - Application of Integral JEE Notes | EduRev

ANSWER : -The given equations are

y = cos x … (1)

And, y = sin x … (2)

Required area = Area (ABLA) area (OBLO)

NCERT Solutions - Application of Integral JEE Notes | EduRev

Integrating by parts, we obtain

NCERT Solutions - Application of Integral JEE Notes | EduRev

Thus, the correct answer is B.

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