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# 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

$\fn_cm arABCD=2\int_{1}^{4}y.dx$

$\fn_cm =2\int_{1}^{4}\sqrt{x}.dx$

$\fn_cm =2\left [ \frac{x^{3/2}}{3/2} \right ]_{1}^{4}$

$\fn_cm =\frac{4}{3}\left [ 4^{3/2} -1^{3/2}\right ]$

= 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]

$\fn_cm \int_{0}^{2p} \left ( \sqrt{2px} -\frac{x^{2}}{2p}\right )dx$

$\fn_cm =\sqrt{2p}\int_{0}^{2p}\sqrt{x}-\frac{1}{2p}\int_{0}^{2p}x^{2}$

$\fn_cm =\sqrt{2p}\int_{0}^{2p}x^{1/2}-\frac{1}{2p}\int_{0}^{2p}x^{2}$

$\fn_cm =\sqrt{2p}\left [ \frac{x^{3/2}}{3/2} \right ]_{0}^{2p}-\frac{1}{2p}\left [ \frac{x^{3}}{3} \right ]_{0}^{2p}$

$\fn_cm =\sqrt{2p}.\frac{2}{3}\left ( 2p \right )^{3/2}-\frac{1}{6p}\left ( 2p \right )^{3}$

$\fn_cm =2^{1/2}.p^{1/2}.2^{3/2}.p^{3/2}.\frac{2}{3}-\frac{1}{6p}.8p^{3}$

=(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

$\fn_cm \int_{0}^{\pi }y.dx$

$\fn_cm =\int_{0}^{\pi }sinx.dx$

$\fn_cm =\left [- cosx \right ]_{0}^{\pi }$

=-( 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

$\fn_cm \int_{0}^{1}\sqrt{x}dx-\int_{0}^{1}x^{2}dx$

$\fn_cm =\int_{0}^{1}x^{1/2}dx-\int_{0}^{1}x^{2}dx$

$\fn_cm =\left [ \frac{x^{3/2}}{3/2}-\frac{x^{3}}{3}\right ]_{o}^{1}$

=[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

$\fn_cm \int_{0}^{2}\sqrt{2^{2}-x^{2}}dx$

$\fn_cm =\left [ \frac{x}{2}\sqrt{2^{2}-x^{2}} +\frac{2^{2}}{2}sin^{-1}\frac{x}{2}\right ]_{0}^{2}$

$\fn_cm =\left [ \frac{x}{2}\sqrt{2^{2}-x^{2}} +2sin^{-1}\frac{x}{2}\right ]_{0}^{2}$

$\fn_cm =0 +2sin^{-1}1$

= 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

$\fn_cm \int_{0}^{2}\left [ x+6-x^{3} \right ]dx$

$\fn_cm =\left [ \frac{x^{2}}{2}+6x -\frac{x^{4}}{4}\right ]_{0}^{2}$

=(2²/2  +6×2 – 24/4]

4/2    +12 –   16/4

=2 +12 -4 = 10 square unit

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## NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

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