Class 12 Maths Important Questions-Application of Integrals

Q1.Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4 and x – axis.

Ans. The given curve  y² = x and the lines x =1,x =4

Also given x -axis (i.e y =0)

Drawng the graph of the curve (i.e parabola) y² = x and the lines x =1,x =4 and y=0

The given curve is

y² = x

y = ±√x

Area of the region covered by the parabola and the lines x =1 ,x=4, and X-axis is symmetrical about the X-axis which is shown by the shaded region

∴ Area of ABCD = 2 area ABFE

Area ABFE = Integration of y with respect to x within the limits x =1 and x =2

= 4/3(2³ -1) = 4/3(8 -1) =28/3

Therefore the area of the shaded region is (28/3) square unit

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Q2.Find the area of the region bounded by the curves y²=2px and x²=2py.

Ans. Solving equations y²=2px and x²=2py,we get

Putting the value y = √(2px) in the equation x²=2py

x²=2p√(2px)

Squaring both sides

x⁴ =4p²× 2px

x⁴ -8p³x =0

x(x³ -8p³) =0

x =0

x³ -8p³ =0

x³ = 8p³

x = 2p

When x = 0 then y=0 and when x =2p then y²=2p(2p)⇒ y =2p

Therefore area bounded by both curves is shown as shaded region in the following graph

The area of the shaded region = The integration of y with respect to x within the limits x=0 and x=2p

From the curve y²=2px ⇒y =√2px and from the curve x²=2py⇒y=x²/2p

The function representing the shaded region = [√(2px) – x²/2p]

 

=(8p²/3) – (4p²/3)

=4p²/3

The area covered by both curves is (4p²/3) square unit

Q3.Find the area of the curve y = sinx, between 0 and π.

Ans. Drawing the curve y = sinx

Area of the curve y = sinx, between 0 and π = Area of the shaded region

Area of the shaded region=Integraion of the function y with respect to x within the limit 0 and π

=-( cos π – cos 0)

= -(-1 -1)

=-(-2) = 2

Hence area of the shaded region is 2 square unit

Q4.Find the area of the region bounded by the two parabolas y = x² and y² = x

Ans. The given curves are y = x² and y² = x

Solving both curves

Putting the value y = x² in the equation y² = x

x⁴=x

x⁴-x=0

x(x³ – 1) =0

x =0, x =1

Putting x =0,y =0 and putting x =1,y=1 indicates that both curves meet at (0,0) and (1,1)

y = x²  and y² = x⇒y = √x

The function representing the shaded region = √x -x² since √x ≥ x²

Area of the shaded region = integration of the function (√x -x²) within the limit 0 and 1

=[2/3 – 1/3] = 1/3

Hence area bounded by the given curves is 1/3 square unit

Q5. Find the area enclosed by the circle x² + y² = 4 and the straight line x + y =2 .

Ans. Solving the equation of the circle and the line

x² + y² = 4⇒ y =√( 4-x²) and y = 2- x

√( 4-x²) = 2- x

Squaring both sides

4 -x² = 4 + x² – 4x

2x² – 4x = 0

x² – 2x = 0

x( x -2) = 0

x =0 and x =2

Putting x = 0 and x =2 in the equation x + y =2

When x = 0,y =2 and when x =2 ,y = 0

The straight line and the circle intersects at (0,2) and (2,0)

Area of the shaded region = Area of the quadrant – Area of the triangle

Area of the quadrant = Integration of the y = √(2² – x²) w.r.t x within the limits x =0 and x =2

= 2×π/2 = π

Area of the quadrant = π square unit

Area of the triangle = (1/2)×2×2 = 2 square unit

Area of the shaded region = (π -2)square unit

Q6.Find the area of the region bounded by the curve y = x³, y = x +6 and y-axis.

Ans. Solving the equation y = x³, y = x +6

Substituting y =x³ in second equation

x³ = x +6

x³ -x -6 =0

x =2 satisfies the above equation

Factorizing the above equation by the remiender theorem

x²(x -2) +2x² -x -6 =0

x²(x -2) +2x(x -2) +4x -x -6 =0

x²(x -2) +2x(x -2) +3x-6 =0

x²(x -2) +2x(x -2) +3(x-2) =0

(x -2)( x² +2x +3) =0

Since the roots of the equation x² +2x +3 are not real,therefore putting x =2 in y = x +6,we get y =8

Hence the curve and the given line intersect at only one point (2,8)

Drawing the graph of y = x³, y = x +6 and y-axis

The shaded region represents the function [(x +6)-x³]

The area of the shaded region =Integrating the function [(x +6)-x³] w r t x within the limits x =0 and x =2

=(2²/2  +6×2 – 2⁴/4]

4/2    +12 –   16/4

=2 +12 -4 = 10 square unit 

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Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
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Chapter 3-Linear equations	Chapter 11- Construction
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Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
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Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

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Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
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Chapter 7- Integrals
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