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
2y2 = 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
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