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QUESTION: 1

The area between the curves y = x^{2} and y = x^{3} is:

Solution:

y = x^{2} and y = x^{3}

To find point of intersections :

x^{2} = x^{3}

x^{2}(x-1) = 0

So, x = 0,1

POI are (0,0) & (1,1)

= [x^{3}/3 – x^{4}/4]_{x=1} - [x^{3}/3 – x^{4}/4]_{x=0}

= (1/3 – 1/4) – (0)

= 1/12 sq units

QUESTION: 2

The area bounded by the curve y^{2} = 16x , x = 1, y = 3 and X-axis is:

Solution:

QUESTION: 3

The shaded area is equal to:

Solution:

QUESTION: 4

The curves x^{2} + y^{2}=16 and y^{2} = 6x intersects at

Solution:

x^{2} + y^{2}=16 and y^{2} = 6x

So, to figure out the P.O.I

x^{2} + 6x = 16

x^{2} + 6x – 16 = 0

(x+8) (x-2) = 0

x = -8, 2

For x = -8

y = (6*(-8))^{1/2 }

= (-48)^{1/2}

No real value for y

For x = 2

y = (6*2)^{1/2 }

= (12)^{1/2}

= 2*(3)^{1/2 }

So, P.O.I is (2 , 2*(3)^{1/2} )

QUESTION: 5

Find the area of region bounded by curve

Solution:

QUESTION: 6

The area of the region bounded by y = x^{2} – 2x and y = 4 – x^{2} is.

Solution:

To find area between y = x^{2} – 2x and y = 4 – x^{2},

We need to find POI.

x^{2} – 2x = 4 – x^{2}

2x^{2} - 2x – 4 = 0

x^{2} – x – 2 = 0

(x-2)(x+1) = 0

x = -1, 2

= [(-2x^{3})/3 + x^{2} + 4x]_{x=2} - [(-2x^{3})/3 + x^{2} + 4x]_{x= -1}

= -16/3 + 4 + 8 – (2/3 + 1 - 4)

= 9 sq units

QUESTION: 7

The shaded area enclosed between the parabolas with equations y = 1 + 10x – 2x^{2}and y = 1 + 5x – x^{2} is equal to:

Solution:

QUESTION: 8

The area of the smaller region lying above the x-axis and included between the circle x^{2} + y^{2} = 2x and the parabola y^{2} = x.

Solution:

= π/4 – 2/3

QUESTION: 9

The area bounded by the curves f(x) = x^{2} + 1 and g(x) = x – 1 on the interval [1,3] is:

Solution:

QUESTION: 10

The area shaded in the given figure can be calculated by which of the following definite integral?

Solution:

### Area between curves

Video | 04:09 min

### Example Area between two curves - Application of Integrals

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### Example Area between two curves - Application of Integrals

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### Examples : Area Between 2 Curves

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