The equation of parabola is
y2 = x ...(1)
The equation of line is
x + y = 2 ...(2)
From (2), y = 2 - x ...(3)
Putting this value of y in (1), we get,
(2 - x)2 = x
or x2 - 4 x + 4 = x or x2 - 5 x + 4 = 0
∴ (x - 1) (x - 4) = 0
∴ x = 1, 4
∴ from (3), y = 1, - 2
∴ parabola (1) and line (2) intersect in the points A (1, 1), B (4, - 1)
Also line (2) meets x-axis in C (2,0)
Required area is shaded.
Area above x-axis = area AOL + area ALC
Area below x-axis = Area OBM - area CBM
Find the area bounded by the curve y = x2 and the line y = x.
OR
Find the area of the region {(x. y): x2 ≤ y ≤ x}.
Find the area of the region bounded by the line y = 3 x + 2, the x-axis and the ordinates x = - 1 and x = 1.
The equation of given line is
y = 3 x + 2 ...(1)
Consider the lines
x = -1 ...(2)
and x = 1 ...(3)
Line (1) meets x-axis where y = 0
putting y = 0 in (1), we get,
line (1) meets x-axis in
Let line (1) meet lines (2) and (3) in B and D respectively. From B, draw BC ⊥ x-axis and from D, draw DE ⊥ x-axis.
Required area = Area of region ACBA + area of region ADEA
The equation of curve is x2 = 4 y ...(1)
which is upward parabola with vertex O.
The equation of line is
x = 4 y - 2 ...(2)
Let us solve (1) and (2)
Putting x = 4y - 2 in (1), we get
From A, draw AM ⊥ x-axis and from B. draw BN ⊥ x-axis.
Required area = area AOB
= Area of trapezium BNMA - (area BNO + area OMA)
=