Advertisements
Advertisements
Question
The area of the region bounded by parabola y2 = x and the straight line 2y = x is ______.
Options
`4/3`sq.units
1 sq.units
`2/3`sq.units
`1/3`sq.units
Solution
The area of the region bounded by parabola y2 = x and the straight line 2y = x is `4/3`sq.units.
Explanation:
Given equation of parabola is y2 = x ......(i)
And equation of straight line is 2y = x ......(ii)
Solving equation (i) and (ii)
We get `(x/2)^2` = x
⇒ `x^2/4` = x
⇒ x2 = 4x
⇒ x(x – 4) = 0
∴ x = 0, 4
Required area = `int_0^4 sqrt(x) "d"x - int_0^4 x/2 "d"x`
= `2/3 [x^(3/2)]_0^4 - 1/2 * 1/2 [x^2]_0^4`
= `2/3 [(4)^(3/2) - 0] - 1/4 [(4)^2 - 0]`
= `2/3 xx 8 - 1/4 xx 16`
= `16/3 - 4`
= `4/3` sq.units
APPEARS IN
RELATED QUESTIONS
Find the area bounded by the curve y2 = 4ax, x-axis and the lines x = 0 and x = a.
Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y − 2.
Find the area of the sector of a circle bounded by the circle x2 + y2 = 16 and the line y = x in the ftrst quadrant.
Find the equation of a curve passing through the point (0, 2), given that the sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at that point by 5
Using definite integrals, find the area of the circle x2 + y2 = a2.
Find the area bounded by the curve y = cos x, x-axis and the ordinates x = 0 and x = 2π.
Find the area of the region in the first quadrant bounded by the parabola y = 4x2 and the lines x = 0, y = 1 and y = 4.
Find the area of the region bounded by x2 + 16y = 0 and its latusrectum.
Find the area of the region bounded by the curve \[a y^2 = x^3\], the y-axis and the lines y = a and y = 2a.
Using integration, find the area of the region bounded by the triangle whose vertices are (2, 1), (3, 4) and (5, 2).
Find the area of the region between the circles x2 + y2 = 4 and (x − 2)2 + y2 = 4.
Find the area of the region included between the parabola y2 = x and the line x + y = 2.
Using integration find the area of the region:
\[\left\{ \left( x, y \right) : \left| x - 1 \right| \leq y \leq \sqrt{5 - x^2} \right\}\]
Make a sketch of the region {(x, y) : 0 ≤ y ≤ x2 + 3; 0 ≤ y ≤ 2x + 3; 0 ≤ x ≤ 3} and find its area using integration.
If the area enclosed by the parabolas y2 = 16ax and x2 = 16ay, a > 0 is \[\frac{1024}{3}\] square units, find the value of a.
The area bounded by the curve y = 4x − x2 and the x-axis is __________ .
The area bounded by the curve y2 = 8x and x2 = 8y is ___________ .
Draw a rough sketch of the curve y2 = 4x and find the area of region enclosed by the curve and the line y = x.
Using the method of integration, find the area of the region bounded by the lines 3x − 2y + 1 = 0, 2x + 3y − 21 = 0 and x − 5y + 9 = 0
The area of the region bounded by the curve x = y2, y-axis and the line y = 3 and y = 4 is ______.
Find the area of the region bounded by the parabola y2 = 2px, x2 = 2py
The area of the region bounded by the curve x = 2y + 3 and the y lines. y = 1 and y = –1 is ______.
Let f(x) be a continuous function such that the area bounded by the curve y = f(x), x-axis and the lines x = 0 and x = a is `a^2/2 + a/2 sin a + pi/2 cos a`, then `f(pi/2)` =
Find the area of the region bounded by the curve `y^2 - x` and the line `x` = 1, `x` = 4 and the `x`-axis.
The area bounded by `y`-axis, `y = cosx` and `y = sinx, 0 ≤ x - (<pi)/2` is
Let T be the tangent to the ellipse E: x2 + 4y2 = 5 at the point P(1, 1). If the area of the region bounded by the tangent T, ellipse E, lines x = 1 and x = `sqrt(5)` is `sqrt(5)`α + β + γ `cos^-1(1/sqrt(5))`, then |α + β + γ| is equal to ______.
Let a and b respectively be the points of local maximum and local minimum of the function f(x) = 2x3 – 3x2 – 12x. If A is the total area of the region bounded by y = f(x), the x-axis and the lines x = a and x = b, then 4A is equal to ______.
Let g(x) = cosx2, f(x) = `sqrt(x)`, and α, β (α < β) be the roots of the quadratic equation 18x2 – 9πx + π2 = 0. Then the area (in sq. units) bounded by the curve y = (gof)(x) and the lines x = α, x = β and y = 0, is ______.
Find the area of the following region using integration ((x, y) : y2 ≤ 2x and y ≥ x – 4).
Using integration, find the area of the region bounded by y = mx (m > 0), x = 1, x = 2 and the X-axis.
