Class 12 Maths Important Questions-Application of Integrals - Future Study Point

Class 12 Maths Important Questions-Application of Integrals

application of integral imp.questions

Class 12 Maths Important Questions-Application of Integrals

application of integral imp.questions

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

Q1 important questions application of integral

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

Q2. application of integral imp.question

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

Q1 imp.questions class 12 application of integral

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)

Q4 imp.questions-application of itegral

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)

Q5 imp.questions application of integral

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

Q6.imp.question application of integral

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 1- Number SystemChapter 9-Areas of parallelogram and triangles
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Chapter 4- Carbon and its CompoundsChapter 12- Electricity
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Chapter 1-Relations and FunctionsChapter 9-Differential Equations
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