Showing posts with label integration. Show all posts
Showing posts with label integration. Show all posts

## HOW TO FIND AREA BOUNDED BY THREE LINES AND CIRCLES , AREA UNDER CURVES BY INTEGRATION METHOD

How to find common area of three lines and one circles which are  intersecting at different  points with the help of an example.

## Given Lines and Curves

Consider one circle and three lines whose equations are  given below
x - 1 )2 y2 12            ......................(1)

y = x                            .....................(2)

y = -√3 (x-2)                      .....................(3)

y = 0                                   .....................(4)

Let us draw these lines and circle in coordinate planes, We can compare the equation of circle  with standard form of circle to find  the coordinate of  centre of the circle is  (1,0)  and radius of both the circles is 1.

## How To Draw Figure 1st of all check whether these lines  intersect with circle or not . And if these lines intersect with each other or with circle then what are their coordinates of points of intersections.

## Solve (1) and (2)

Putting the value of  'y' from (2) in equation (1), we get
x - 1 )2 x2 12

x2  + 12 -2×1×x +x2 12

-2x +x2 = 0
⇒   2x(-1+x) = 0
Either  2x= 0 or (-1+x) = 0
x = 0 or x = 1
Now putting the values of x in (2) we get
x = 0 when x = 0  and y = 1 when x = 1
Therefore points of intersection of (1) and (2) are
O( 0 , 0 ) and A( 1, 1 )

## Solve (1) and (3)

Putting the value of  'y' from (3) in equation (1), we get
x - 1 )2 [-√3 (x-2)]2 12
⇒  x 2 1 2  - 2x +  3 (x-2)2 = 1

x 2 1- 2x +  3 [ x 2 +4 - 4x] =1

x 2 1- 2x +  3x 2 + 12 - 12x -1 = 0
4x 2 - 14x +12 = 0

2x 2 - 7x + 6 = 0 ,    By Factorisation Method

2x 2 - 4x - 3x  + 6 = 0

2x(x-2) -3( x - 2) = 0

(x-2)( 2x - 3) = 0

Either (x-2) = 0   or ( 2x - 3) = 0
x = 2 and x = 3/2
To find the values of y , put both the values of  'x' in (3) .i.e. in  y = -√3 (x-2)

when x = 2 ,    then  y = 0
and    x = 3/2 , then  y = √3 /2

Therefore points of intersection of (1) and (3) are
C( 2 , 0 ) and B( 3/2√3 /2 )

## How to Find Required Area

Required Area = Shaded Area = Area of ΔOAL + Area of Curve ABMLA+ Area of Δ BCM
After simplification , we get

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## Final words

Thanks for visiting this website and spending your valuable time to read this post regarding how to find area bounded by three lines and circle  .If you liked this post , don't forget to   share it with your friends to benefit them also ,we shall meet in next post , till then bye and take care......

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## How to find common area of two parabolas  , Area under two parabolas , Area of region bounded by two parabolas .

Let us consider two parabolas whose equations are given by
y2 =  4ax  --------------  (1)
x2 =  4ay ----------------  (2)

To check whether these parabolas intersect with each others or not And if they intersect then what is/are their point/s of intersection.

## How to find Points of Intersection

To find coordinate of points of intersection ,we have to solve equation (1) and (2)

Consider   eq (2)
x2 =  4ay
⇒ y  x2 /4a   ---------------(3)

Putting the value of "y" in equation  (1) ,we get Area under Curves
(x2 /4a)2 =  4ax

x4/16a2 =  4ax

⇒    x4=  64xa3
x464xa3= 0

Taking 'x' common

x(x364a3) = 0
Either  x = 0    or x364a3 = 0
⇒ x = 0    or    (x)3(4a)3 = 0
⇒ x = 0    or    (x)3(4a)3 = 0

⇒ x = 0    or    (x-4a)[ (x)2(4a)2 + (x)(4a)]   = 0

⇒ x = 0    or    (x-4a) =  0  or  [ (x)2(4a)2 + (x)(4a)]   = 0
⇒ x = 0    or    x =  4a  or   (x)2(4a)2 + (x)(4a)   = 0

Since   x2+ 4a.x + 16a2   = 0   have no real  roots ,because its discriminant is negative, therefore this quadratic equation have complex roots. And these roots are rejected .

### To find values of y

Now putting both  values of  "x"  in eq (3) i. e .  x2 /4a   ,we get
1st  put x = 0
y = 0 / 4a = 0         when x = 0 then y = 0
and put x = 4
y =  (4a)2 /(4a)
y =  4a                    ⇒ when x = 4a then y = 4a

Hence two points of intersection of (1) and (2)   O(0,0) and A (4a , 4a) .
Now draw two parabolas using their points of intersections as drawn in given picture.

## How to Find Required Area

Now to find the area enclosed between two Parabolas.
Required Area = shaded Area =Area OLAMO - Area ONAMO

Watch this video to remove your doubts if any

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## Final words

Thanks for visiting this website and spending your valuable time to read this post regarding how to find area bounded by two parabolas  .If you liked this post , do share it with your friends to benefit them also ,we shall meet in next post , till then bye and take care......

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## USING METHOD OF INTEGRATION ,HOW TO FIND AREA OF TRIANGLE BOUNDED BY THREE LINES

Using method of integration find the area of triangle,using the method of integration find the area of the region bounded by the lines,area of triangle by integration method

## Using Method of Integration , How to find the area of triangle bounded by three lines

2x + y = 0 , 3x - 2y = 6  and x - 3y  + 5 = 0

## Solution

Given lines are
2x + y  = 4   --------  (1)
3x - 2y = 6  ---------  (2) and
x - 3y  = -5  ---------  (3)

If these lines are intersecting then we have to find their coordinates of points of intersection .

### To Find Coordinate of Point A

Multiply (1) by 2 and adding to (2) , we get
4x + 2y + 3x - 2y = 8 + 6
7x = 14 ⇒ x =2
Putting x = 2 in (1) , we get Area under Curve
2(2) + y = 4
4 + y = 4 ⇒ y = 0
∴ (1) and (2) meets at point A(2,0).

### To Find Coordinate of Point B

To find point of intersection (2) and (3);
Multiply (3) by -3 and adding to (2) , we get
3x - 2y -3x +9y  = 6 +15
7y = 21  ⇒ y = 3

Putting y = 3 in (3) , we get
x-3(3)  = -5
⇒ x = -5 +9  ⇒ x = 4
∴ (2) and (3) meets at point B(4,3).

### To Find Coordinate of Point C

To find point of intersection (1) and (3);
Multiply (1) by 3 and adding to (3) , we get
6x + 3y + x - 3y = 12 - 5
7x = 7  ⇒ x =1

Putting x = 1 in (1) , we get
2(1) + y  = 4
⇒ y =4 - 2  ⇒ y = 2
∴ (1) and (3) meets at point C(1,2).

we get points of intersection of (1) and (2)  A(2,0),  points of intersection of (2) and (3) B(4,3) and points of intersection of (1) and (3) C(1,2).

## Required Area = Shaded Area   =  Area DCBED - Area DCAD -Area ABEA

Also read my previous post  How to find area bounded by two circles

For better understanding watch this video

## Conclusion

Thanks  for visiting this website and spending your precious time to read how to find area of triangle Using method of integration find the area of triangle,using the method of integration find the area of the region bounded by the lines,area of triangle by integration method.

If you are a mathematician Don't forget to visit my Mathematics You tube channel ,Mathematics Website and Mathematics Facebook Page , whose links are given below

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## HOW TO FIND AREA OF TWO CIRCLES INTERSECTING EACH OTHERS,

Here we are going to discuss how to find common area of two circles which are overlapping or intersecting at two points with the help of an example
Let us consider two circles whose equations are  given below
x2 + y2 =  12                   .......................(1)
x - 1 )2 y2 12          .....................(2) Let us draw these circles in coordinate planes, We can compare these equations with standard form of circle to find  the coordinate of  centre of both the circles are (0,0) and (1,0) respectively and radius of both the circles are 1.

If these two circles intersect with each other then we have to find their point/s of intersection.
To find points of intersection subtracting equation (1) from eq (2) , we get

X - 1 )2 Y2 - X2 Y=  12 - 1
⇒( X - 1 )2  - X2 =  0
X 2  +1 2  - 2×(1)×(X) X2 =  0

⇒1 - 2X = 0
⇒ X = 1/2 ,
Now to find the values of y put the value of x in equation # 1
(1/2)2 Y=  12
Y1- (1/4) = 3/4

Y = 土⇃(3/4)
Therefore two points of intersection are B(1/2 , ⇃(3/4)) , C ( (1/2 , -⇃(3/4))

To understand better the solution of  this problem watch this  video

Required area = shaded  Area ,
we can divide shaded area into four equal parts , As each parts is symmetrical , Therefore to find shaded area  it is sufficient to find the area of any one of four part and then then multiply it with 4.

Hence  Required area = 4 area OBLO = 4 Area BALB
To avoid tedious calculations choose 2nd part to integrate
I= Required area = 4 Area BALB    ------------- (3)
After simplification , we have

My previous post HOW TO FIND AREA OF THE CIRCLE  WHICH IS INTERIOR TO THE PARABOLA

## Final words

Thanks for visiting this website and spending your valuable time to read this post regarding how to find area bounded by two circles .If you liked this post , do share it with your friends to benefit them also we shall meet in next post , till then bye and take care....

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## HOW TO PROVE TRIGONOMETRIC IDENTITIES || TRIGONOMETRY

Proof of trigonometric identities , trigonometric identities problems, proving trigonometric identities formulas,these trigonometric identities of class 10, fundamental trigonometric identities,trigonometric identities class 11 and its formation with the help of some examples.

## How to prove Identity

cos 6x = 32cos6 x - 48.cos4 x   + 18.cos2 x  - 6.cos2 x  - 1

## Proof

1st of all  rewrite 3x as 3.2x

L.H.S. = cos 6x =  cos (3.2x)

Now using the result cos 3θ = 4cos3 θ - 3 cos θ  -----(1)

Replacing θ as 2x in (1), we get

L.H.S. = 4cos3 2x - 3 cos 2x  -----------(2)

Now using the result  1+ cos 2θ = 2 cos2 θ

⇒ cos 2θ = 2 cos2 θ -1

Replacing cos 2x = 2 cos2 x -1 in (2), we get

L.H.S.= 4 {2cos2 x -1}3 - 3 {cos2 x  -1}

Now using the result {a - b }3 = {a}3 - b }3  -  3{a }2 .b   + 3(a). b2

cos 6x    = 4[ {2cos2 x  }3 - { 1 }3  -  3{2cos2 x  }2 .1   +3.(2cos2 x) .12 ] - 3 . {cos2 x  -1}

Taking the product of powers to simplify it

cos 6x  =   4[ 8cos6 x  - 1 - 12cos4 x  + 6cos2 x]  - 3{2cos2 x-1}

Multiply by 4 in 1st term and multiply by -3 in 2nd term

cos 6x  = 32cos6 x  - 4 - 48cos4 x  + 24cos2 x  - 6cos2 x + 3

Adding the like powers terms and arranging in descending order

cos 6x   = 32cos6 x - 48cos4 x  + 18cos2 x  - 6cos2 x  - 1

Hence the Proof

## tan (2x) =  2tan x  1 - tan2 x

Proof

We know that

tan (A+B) =  tan A +  tan B1 - tan A tan B

Put A = B  = x in above formula . then it becomes

tan (x+x) =  tan x +  tan x1 - tan x tan x

tan (2x) =  2tan x  1 - tan2 x
Hence the Proof

## Proof

As we know that sin (A + B) = sin A cos B + cos A sin B..  ...(1)

Put A = B  = x in ...   (1)

sin (x + x) = sin x cos x + cos x sin x

sin (2x) = sin x cos x +  sin x cos x

sin (2x) = 2 sin x cos x

Hence the Proof

## Proof

As we know that cos (A + B) = cos A cos B - sin A sin B..  ...(1)
Put X = A = B in (1) , we get

cos (x + x) = cos x cos x - sin x sin x

cos 2x = cos2 x - sin2 x

cos 2x = cos2 x - sin2 x

Hence the Proof

Using the result
1+cos 2θ = 2cos2 θ
cos 2θ = 2cos2 θ -1 -------------(1)
Replacing θ with 2x in eq (1)
1+ cos 4x = 2cos2 2x
cos 4x = 2cos2 2x -1

Again using  cos 2θ = 2cos2 θ -1

cos 4x = 2Sq(2cos2 x -1) -1

It is the square of 2cos2 x -1

cos 4x = 2Sq(2cos2 x -1) -1

cos 4x = 2(4cos4 x +1 - 4cos2 x) -1

cos 4x = 8cos4 x +2 - 8cos2 x -1

cos 4x =  8cos4 x - 8cos2 x +1

Hence the Proof

## What is the value of sin3x?

To find the value of sin 3x ,  use this formula which contain sin (A+B)
therefore sin (A+B) = sin A cos B cos A sin B——-(1)
put A = 2x and B = x in (1)
then Sin 3x = sin 2x cos x + cos 2x sin x

## As we know that cos 2x = 1 - 2sin3 x and sin 2x = 2 sin x cos x

sin 3x = (2 sin x cos x) cos x + (1 - 2sin3 x ) sin x
sin 3 x = 2 sin x cos2 x + sin x -  2sin3 x

As we know that cos2 x = 1sin2 x

sin 3x= 2 sin x (1-sin3 x) + sin x - 2sin3 x
sin 3x = 2 sin x -2 sin3 x + sin x - 2sin3 x
sin 3x = 3 sin x - 4 sin3 x

Similarly we prove that cos 3x= 4 cos3 x - 3 cos x
For learning and memorising more trigonometric formulas

## Conclusion

In this post I have discussed trigonometric identities ,trigonometric identities problems, proving trigonometric identities formulas . If this post helped you little bit, then please share it with your friends to benefit them, comment your views on it to boost me and to do better, and also follow me on my Blog .We shell meet in next post till then Bye

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