Class 12 Maths Important Questions-Application of Integrals
Q1.Find the area of the region bounded by the curve y^{2} = x and the lines x = 1, x = 4 and x – axis.
Ans. The given curve y^{2} = 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) y^{2} = x and the lines x =1,x =4 and y=0
The given curve is
y^{2} = 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
x^{4} =4p²× 2px
x^{4} -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 = x^{2} and y^{2} = x
Ans. The given curves are y = x^{2} and y^{2} = x
Solving both curves
Putting the value y = x² in the equation y^{2} = x
x^{4}=x
x^{4}-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 y^{2} = 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 – 2^{4}/4]
4/2 +12 – 16/4
=2 +12 -4 = 10 square unit
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