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# Math Examples of Definite Integral

• ## Find the area bounded by the following equations: L (x) = 4 x – x2 and n (x) = 5 - 2x?

To get the Intersection points, equating the above equations, we get:
4 x – x 2 = 5- 2 x,
( x – 1 ) ( x – 5 ) = 0,
It implies, x = 1 , 5,
Hence the intersection points are (1,3) and (5,-5).
Integrating by applying limits, we get,
Area ( A ) = ∫ 1 5 [ ( 4 x – x 2) - ( 5 – 2 x) ] dx,
=∫ 1 5 ( 6 x –x 2 – 5 )dx,
=[ 3 x 2 – 1 / 3 x 3 – 5 x ] 1 5,
=32 / 3.
Hence,t he required area=32/3.

## Find the area between the curves n = 1 and n = 2 sin Ńł?

The graph can be divided into two regions.

The first region ( b 1 ) is the area of circle n = 2 sin Ńł.
The second region ( b 2 ) is the area between n = 2 sin Ńł and n = 1.
required area=area of b1- area of b2,
the Area of a Circle= π r 2,

where 'r' is the radius of the circle.
Hence area of 'b1' is π . b 1 = π
To find the required area, we have to find the points of Intersection by equating the two equations as follows:
2 sin Ńł = 1 / 2 → Ńł = π / 6 and Ńł = 5 π / 6,
The equation is symmetric with respect to y-axis, and the limits are from π/2 to π/6.
Integrating and applying limits, we get,
Area = 2 ∫ π / 6 π / 2 1 / 2 ( 4 sin 2 Ńł – 1 ) dŃł,
= ∫ π / 6 π /4 (( 1 / 2 ) ( 1 – cos 2 Ńł ) -1 )dŃł,
=2 ∫ π / 6 π / 2 ( 1 – 2 cos 2 Ńł ) dŃł = [ Ńł – sin 2 Ńł ] π / 6 π / 2,
=π / 3 + √ 3 / 2.

## Find the area of the curve x = 2 - 2 sin Ńł?

It is a trigonometric function, its graph can be made as follows:

The given function is symmetric to the y-axis,so the value of Ńłis in between – π / 2 and π / 2,multiply the area by 2,we get
Area = 2 ∫ - π / 2 π / 2 1 / 2 ( 2 -2 sin Ńł ) 2 dŃł,
= ∫ - π / 2 π / 2 ( 4 - 8 sin Ńł + 4 sin 2 Ńł ) 2 dŃł,
= ∫ - π / 2 π / 2 ( 6 – 8 sin Ńł – 2 cos 2 Ńł ) dŃł,
= [ 6 Ńł + 8 cos Ńł – sin 2 Ńł ] – π / 2 π / 2,
=6π.

## Find the area between the given curves, x 2 + y 2 = 4 and ( x – 2 ) 2+ y 2 = 4?

Equations are given as:
x+ y = 4 -------------(equation 1)
(x-2)2+y2=4 -------------(equation 2)
The equation 1 represents a Circle of radius 2 and Centre at (0,0).
The equation 2 represents a circle with centre at (0,0) and radius is equal to 2.
To find Intersection Point, we have to equate these two equations,

( x – 2 ) + y = x + y 2,
It implies, x = 1,
Putting x = 1, we will get,
Y = +- √ 3.
Hence, the points of intersection are H ( 1 , √ 3 ) and H ’ ( 1 , - √ 3 ).
The required area=2[area of the region oabco]
As area of oabco = area of odcdo.
=2 [ ∫ y d x + ∫ y . dx],
=2 [ ∫ √ 4 - ( x - 2 ) . d x + ∫ √ 4- x . d x ],
= 2 [ 1 / 2 ( x – 2 ) √ 4 - ( x – 2 ) + 1 / 2 * 4 sin – 1 ( x – 2 / 2 ) ] + 2 [1 / 2 x √ 4 – x + 1 / 2 * 4 sin – 1 x / 2 ] 2,
= [ ( - √ 3 + 4 sin - 1 ( - 1 / 2 ) ) - 4 sin – 1 ( - 1 ) ] + [ 4 sin – 1 1 - √ 3 - 4 sin – 1 1 / 2 ],
= 8 π / 3 – 2 √ 3.

## Find the area of the region bounded by the curves y=x and x2 +y 2 = 32, common with first quadrant?

Equations given,
Y = x - - - - - - - - - - - -- - -( equation 1 )
+ x = 32 - - - - - - - - - - - - - - - ( equation 2 )
To find the Intersection Point of two curves, solving both,
We get the points as: (4,4).
First, area of the region =∫ y dx = ∫ x . dx,
=1 / 2 [ x 4,
=8 units ----------------(equation 4)
Now, area of second region = ∫ 4 √ 2 √ 32 – x . dx
=[1/2x√32-x2+1/2*32*sin-1x/4√2]44√2,
=(1/24√2*0+1/2*32*sin-1)-(4/2√32-16+1/2*32*sin-11/√2),
=8Đź-(8+4Đź)=4Đź-8 ------------------------(equation 5),
Hence, total area=8+4Đź-8,
=4Đź.

## Find the area enclosed by the ellipse: X / m 2+ y / n 2 = 1?

From the equation, we can easily get the Intersection points of ellipse with ‘x’ and ‘y’ axis as (m,0) and (0,n) respectively.
From the graph, it can be seen that ellipse is divided into four equal area.

As it is symmetrical to both ‘x’ and ‘y’ axis,
Hence, total area=4*(area of one part)
= 4∫0ay.dx ------(equation 2)
From equation 1, we can calculate ‘y’ as,
Y = + - n / m √ m - n 2,
We are taking first quadrant, hence ‘y’ will be positive,
Taking positive value of ‘y’ in equation 2, we get-
4 ∫ n / m√ m – x .dx,
=4 n / m [ x / 2 √ m – x + m / 2 sin – 1 x / m ] (applying inverse trigonometric Integration formulas)
= 4 n / m [ ( m / 2 * 0 + m / 2 sin – 1 1 ) = 0 ],
= 4 n m π / 4 m,
=π m n.
Hence, the total area of elliptical surface= πmn.

## Find the area of the region enclosed between the following curves, a = b  2  – 2 b + 2 and a = - b  2 + 6 ?

First, we have to find the Intersection Point of the two equations,
We can find it by equating the given equations,

2 – 2 b + 2 = - b  2 + 6,
it implies, 2 b  – 2 b - 4 = 0,

hence,
( b + 1 ) ( b – 2 ) = 0,
the intersection points are,
b = -1 and b = 2 .
Now, the area which we have to find is above a=-b2+6 and below the equation,
a = b – 2 b + 2.
Now, in between the points of intersection,
+ 6 ≥ b - 2 b + 2,
as calculated above, the intersection points are b=-1,b=2.
These are the limits on Integration when we are calculating the area,
Applying limits we get,
∫ -1 2[ ( - b + 6 )- ( b -2 b + 2) ] db,
= ∫ -1 2[ - 2 b + 2 b + 4 ] d b,
= [ - ( 2 / 3 ) b + b + 4 b ] – 1 2 ,
=9.
Hence the total area between the two given curves =9 units.

## Find the area of the Region bounded by the following set of equations: m = n 3 – 6 n 2 – 1 6 n and m = 8 n + 2 n 2 – n 3?

We can find the Intersection points of the curves by solving the given equations, equating both curves, we get-
n- 6 n – 1 6 n = 8 n + 2 n –n 3,
It implies, n ( n – 4 n – 1 2 ) = 0,
→n ( n – 6 ) ( n + 2 ) = 0.
Hence, the intersection points are n = - 2, n = 0 and n=6.
These points decides the ranges of Integration.
The graph can be taken in two pieces. The piece to the left of y-axis and above x-axis is denoted by equation:
M = n – 6 n – 16 n.
The piece to the left of y-axis and below x-axis is:
m = 8 n + 2 n - n 3.
The piece to the right of y-axis and above x-axis is:
M = 8 n + 2 n – n 3.
The piece to the right of y-axis below x-axis is:
M = n - 6 n - 16 n.
Now, to calculate total area, we have to integrate within limits as follows:
-20((n3-6n2-16n)-(8n+2n2-n3)).dn + ∫06((8n+2n2-n3)-(n3-6n2-16n)).dn = ∫-20((2n3-8n2-24n).dn + ∫06(-2n3+8n2+24n).dn
=[1/2n4-8/3n3-12n2]-2+ [-1/2n4+8/3n3+12n2]06
=1136/3 = 378.66667.

## Calculate the area of the region given by the following lines: a + b = 2, a – b = - 1, a + 2 y = 2 ?

First of all, we have to plot the graph to watch the region,
Now, we have to find the Intersection Point of the three lines.

F ( a ) = 2 - a , G ( a ) = a + 1 , h ( a ) = - 1 / 2 a + 1.
To get the intersection point, we have to equate all the three equations,
We get the points as (0,1),(2,0)and(1/2,3/2).
The whole region ∑ can be further subdivided into two sub regions ∑1 and ∑2.
The region ∑is represented by the area above g(a) and below h(a) satisfying the equation: 0≤a≤2.
To calculate the area of the region, we have to integrate in The Range from1/2 to 2.
Area ( ∑ ) = ∫ 1 / 2 ( g ( a ) – h ( a ) ) d a = ( ∫ 1 / 2 ( ( a + 1 ) - ( -1 / 2 a + 1 ) ) d a
= ∫ 1 / 2 3 / 2 a .d a= 45/ 16.
Similarly, area of region ∑is:
Area (∑ ) = ∫ 1 / 2 ( f ( a ) – h ( a ) ) . d a = ∫ 1 / 2 ( ( 2- a) - ( - 1 / 2 a + 1 ) ). D a
= ∫ 1 / 2 ( 1 - a / 2 ) . d a= 9 / 16.
Now total area is sum of its sub regions-
Hence, Area ( ∑ ) = Area ( ∑ ) + A rea(∑2)= 45 / 1 6 + 9 /1 6 = 54/16 = 3.375.

## Find the area of the region which is represented by the equations, F ( a ) = a 2 H ( a ) = 8 √ a ?

The region is bounded by F(a) and H(a).
Now, we have to find the boundary points to plot the graph.
We will get the vertical side lines by solving the given equations as:
F(a)=H(a).

It implies, a = 8 Ń´ a ,
Solving the above, we get,
a=0 and a=4.
Hence we have get area as:
Area ( ∑ ) = ∫ 0√ a – a ) da,
= [16/3.a3 /2–1/3.a3]40 ,
=64/3.
Area (∑)=64/3.
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