AREA BOUNDED BY THE GIVEN LINE X AXIS AND ORDINATES

Example 1 :

Find the area of the region bounded by the line

x-y = 1 

x - axis x = 2 and x = 4 

Solution :

Within the limit x  =  2 and x  =  4, we find area under the given curve x - y  =  1.

The area lies above the x-axis, so the required area is

=  integral a to b y dx

=  4 square units.

Example 2 :

Find the area of the region bounded by the line

x-y = 1 

x - axis , x = - 2 and x = 0

Solution :

Required area  =  integral a to b ∫-y dx

So, the required area is 4 square units.

Example 3 :

Find the area of the region by the line

x-2y-12  =  0

and  y - axis, y = 2 and y = 5

Solution :

The area lies on the right side of y-axis. So,

=  (25+60) - (4+24)

=  85-28

=  57

So, the required area is 57 square units.

Example 4 :

Find the area of the region bounded by the line

y = x - 5

and the x - axis between the ordinates x = 3 and x = 7.

Solution :

=  [-(25/2) + 25] + [(9/2)-15] + [(49/2)-35] - [(25/2)-25]

=  -12.5+25+4.5-15+24.5-35-12.5+25

=  -12.5-15-35-12.5+25+4.5+24.5+25

=  -75+54

=  21 square units.

Example 5 :

Find the area of the region bounded by

x2  =  36y 

y - axis , y = 2 and y = 4

Solution :

So, area of the shaded region is 8(4-√2) square units.

Example 6 :

Find the area included between the parabola

y2  =  4ax

and its latus rectum.

Solution :

Example 7 :

Find the area of the region bounded by the ellipse

(x2/9) + (y2/5)  =  1

between the two latus rectum.

Solution :

a2  =  9, b2 = 5

e  =  √[1-(b2/a2)]

e  =  √[1-(5/9)]

=  √[(9-5)/9]

=  √(4/9)

e  =  2/3

Equation of latus rectum x = ± ae

a  =  3, e  =  2/3

ae  =  3 (2/3)

ae  =  2

Equation of latus rectum x = ± 2

Required area  =  integral a to b ∫ y dx

(x2/9) + (y2/5)  =  1

(y2/5)  =  1 - (x2/9)

y2/5  =  (9 - x2)/9

y2  =  (5/9) (9-x2)

y  =  √(5/9) (9-x2)

y  =  √5/3 √(9-x2)

By using the limits x = 0 and x = 2,  we can find area above the x-axis.

To find the total shaded area we have to multiply the above area by 4.

Example 8 :

Find the area of the region bounded by the parabola

y2  =  4x

and the line

2x-y = 4

Solution : 

To find the point of intersection we have to solve both equations.

x  =  y2/4   ------ (1)

x  =  (4+y)/2 ------ (2)

(1)  =  (2)

y2/4  =  (4 + y)/2

2y =  4(4 + y)

2 y2  =  16 + 4 y

2y2-4y-16  =  0

now we are going to divide the whole equation by 2,

y2-2y-8  =  0

(y-4) (y+2)  =  0

y-4  =  0          y+2  =  0

y  =  4              y  =  -2

Point of intersection of two curves are (0, 4) (0, -2).

=  [(32+16)/4 - (64/12) ] - [ (-16 + 4)/4 - (-8/12) ]

=  [(48/4) - (16/3)] - [(-12/4) - (-8/12)]

=  [12-(16/3)] - [-3 + (2/3)]

=  [(36-16)/3)] - [(-9+2)/3]

=  [20/3] - [-7/3]

=  (20/3) + (7/3)

=  (20+7)/3

=  27/3

=  9 square units 

Apart from the stuff given on, if you need any other stuff in math, please use our google custom search here. 

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. How to Solve Age Problems with Ratio

    Mar 28, 24 02:01 AM

    How to Solve Age Problems with Ratio

    Read More

  2. AP Calculus BC Integration of Rational Functions by Partical Fractions

    Mar 26, 24 11:25 PM

    AP Calculus BC Integration of Rational Functions by Partical Fractions (Part - 1)

    Read More

  3. SAT Math Practice

    Mar 26, 24 08:53 PM

    satmathquestions1.png
    SAT Math Practice - Different Topics - Concept - Formulas - Example problems with step by step explanation

    Read More