हिंदी

The area of the region bounded by the curve y = sinx between the ordinates x = 0, x = π2 and the x-axis is ______. - Mathematics

Advertisements
Advertisements

प्रश्न

The area of the region bounded by the curve y = sinx between the ordinates x = 0, x = `pi/2` and the x-axis is ______.

विकल्प

  • 2 sq.units

  • 4 sq.units

  • 3 sq.units

  • 1 sq.unit

MCQ
रिक्त स्थान भरें

उत्तर

The area of the region bounded by the curve y = sinx between the ordinates x = 0, x = `pi/2` and the x-axis is 1 sq.unit.

Explanation:

= `int_0^(pi/2) sin x  "d"x`

= `- [cos x]_0^(pi/2)`

= `-[cos  pi/2 - cos 0]`

= `-[0 - 1]`

= 1 sq.unit

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Application Of Integrals - Exercise [पृष्ठ १७८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 8 Application Of Integrals
Exercise | Q 30 | पृष्ठ १७८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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.


Draw a rough sketch of the curve and find the area of the region bounded by curve y2 = 8x and the line x =2.


Draw the rough sketch of y2 + 1 = x, x ≤ 2. Find the area enclosed by the curve and the line x = 2.


Draw a rough sketch of the graph of the curve \[\frac{x^2}{4} + \frac{y^2}{9} = 1\]  and evaluate the area of the region under the curve and above the x-axis.


Draw a rough sketch of the curve \[y = \frac{x}{\pi} + 2 \sin^2 x\] and find the area between the x-axis, the curve and the ordinates x = 0 and x = π.


Find the area of the region bounded by x2 + 16y = 0 and its latusrectum.


Find the area of the region common to the parabolas 4y2 = 9x and 3x2 = 16y.


Draw a rough sketch of the region {(x, y) : y2 ≤ 3x, 3x2 + 3y2 ≤ 16} and find the area enclosed by the region using method of integration.


Prove that the area in the first quadrant enclosed by the x-axis, the line x = \[\sqrt{3}y\] and the circle x2 + y2 = 4 is π/3.


Prove that the area common to the two parabolas y = 2x2 and y = x2 + 4 is \[\frac{32}{3}\] sq. units.


Find the area bounded by the parabola y2 = 4x and the line y = 2x − 4 By using horizontal strips.


The area bounded by the parabola y2 = 4ax and x2 = 4ay is ___________ .


The area of the region (in square units) bounded by the curve x2 = 4y, line x = 2 and x-axis is


Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3, is


Sketch the graphs of the curves y2 = x and y2 = 4 – 3x and find the area enclosed between them. 


Using integration, find the area of the region bounded by the parabola y= 4x and the circle 4x2 + 4y2 = 9.


Find the area of the region bounded by the curve y2 = 4x, x2 = 4y.


Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.


The area of the region bounded by the curve y = `sqrt(16 - x^2)` and x-axis is ______.


Using integration, find the area of the region `{(x, y): 0 ≤ y ≤ sqrt(3)x, x^2 + y^2 ≤ 4}`


If a and c are positive real numbers and the ellipse `x^2/(4c^2) + y^2/c^2` = 1 has four distinct points in common with the circle `x^2 + y^2 = 9a^2`, then


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


Find the area of the region bounded by the ellipse `x^2/4 + y^2/9` = 1.


The area bounded by the curve `y = x^3`, the `x`-axis and ordinates `x` = – 2 and `x` = 1


The area bounded by the curve `y = x|x|`, `x`-axis and the ordinate `x` = – 1 and `x` = 1 is given by


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 ______.


Find the area of the smaller region bounded by the curves `x^2/25 + y^2/16` = 1 and `x/5 + y/4` = 1, using integration.


Using integration, find the area bounded by the curve y2 = 4ax and the line x = a.


Using integration, find the area of the region bounded by the curve y2 = 4x and x2 = 4y.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×