हिंदी

Find the area of the region bounded by y = x and y = x. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the area of the region bounded by y = `sqrt(x)` and y = x.

योग

उत्तर


We are given the equations of curve y = `sqrt(x)` and line y = x.

Solving y = `sqrt(x)`

⇒ y2 = x and y = x,

We get x2 = x

⇒ x2 – x = 0

⇒ x(x – 1) = 0

∴ x = 0, 1

Required area of the shaded region

= `int_0^1 sqrt(x)  "d"x - int_0^1 x  "d"x`

= `2/3 [x^(3/2)]_0^1 - 1/2 [x^2]_0^1`

= `2/3[(1)^(3/2) - 0] - 1/2 [(1)^2 - 0]`

= `2/3 - 1/2`

⇒ `(4 - 3)/6`

⇒ `1/6` sq.units

Hence, the required area = `1/6` sq.units

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Application Of Integrals - Exercise [पृष्ठ १७७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 8 Application Of Integrals
Exercise | Q 13 | पृष्ठ १७७

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find the area of the region bounded by the parabola y2 = 4ax and its latus rectum.


Find the area of the region bounded by the curve y = sinx, the lines x=-π/2 , x=π/2 and X-axis


Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y − 2.


Find the area of the region bounded by the curve x2 = 16y, lines y = 2, y = 6 and Y-axis lying in the first quadrant.


Find the area under the curve y = \[\sqrt{6x + 4}\] above x-axis from x = 0 to x = 2. Draw a sketch of curve also.


Find the area enclosed by the curve x = 3cost, y = 2sin t.


Find the area of the region bounded by x2 = 16y, y = 1, y = 4 and the y-axis in the first quadrant.

 

Using integration, find the area of the triangular region, the equations of whose sides are y = 2x + 1, y = 3x+ 1 and x = 4.


Find the area of the region bounded by \[y = \sqrt{x}\] and y = x.


Find the area of the region bounded by the curves y = x − 1 and (y − 1)2 = 4 (x + 1).


Find the area of the region bounded by y = | x − 1 | and y = 1.


Find the area enclosed by the curves y = | x − 1 | and y = −| x − 1 | + 1.


Find the area enclosed by the parabolas y = 4x − x2 and y = x2 − x.


Find the area of the figure bounded by the curves y = | x − 1 | and y = 3 −| x |.


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.


Find the area of the region bounded by the parabola y2 = 2x and the straight line x − y = 4.


Find the area of the region bounded by the parabolas y2 = 6x and x2 = 6y.


Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32 is ______.


The area of the region bounded by the ellipse `x^2/25 + y^2/16` = 1 is ______.


Area lying in the first quadrant and bounded by the circle `x^2 + y^2 = 4` and the lines `x + 0` and `x = 2`.


Find the area of the region bounded by `x^2 = 4y, y = 2, y = 4`, and the `y`-axis in the first quadrant.


The area of the region S = {(x, y): 3x2 ≤ 4y ≤ 6x + 24} is ______.


Area of figure bounded by straight lines x = 0, x = 2 and the curves y = 2x, y = 2x – x2 is ______.


Let f(x) be a non-negative continuous function such that the area bounded by the curve y = f(x), x-axis and the ordinates x = `π/4` and x = `β > π/4` is `(βsinβ + π/4 cos β + sqrt(2)β)`. Then `f(π/2)` is ______.


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 ______.


Using integration, find the area of the region bounded by line y = `sqrt(3)x`, the curve y = `sqrt(4 - x^2)` and Y-axis in first quadrant.


Find the area of the following region using integration ((x, y) : y2 ≤ 2x and y ≥ x – 4).


Find the area of the region bounded by the curve x2 = 4y and the line x = 4y – 2.


Sketch the region enclosed bounded by the curve, y = x |x| and the ordinates x = −1 and x = 1.


Hence find the area bounded by the curve, y = x |x| and the coordinates x = −1 and x = 1.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×