मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Choose the correct alternative: Area of the region bounded by the curve x2 = 8y, the positive Y-axis lying in the first quadrant and the lines y = 4 and y = 9 is ______

Advertisements
Advertisements

प्रश्न

Choose the correct alternative:

Area of the region bounded by the curve x2 = 8y, the positive Y-axis lying in the first quadrant and the lines y = 4 and y = 9 is ______

पर्याय

  • `(76sqrt(2))/3` sq.units

  • `(76sqrt(2))/2` sq.units

  • `(38sqrt(2))/3` sq.units

  • `76sqrt(2)` sq.units

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

`(76sqrt(2))/3` sq.units

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1.7: Application of Definite Integration - Q.1 (A)

संबंधित प्रश्‍न

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


Find the area of the region lying in the first quadrant and bounded by y = 4x2x = 0, y = 1 and = 4


Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}


Find the area of the smaller region bounded by the ellipse \[\frac{x^2}{9} + \frac{y^2}{4} = 1\] and the line \[\frac{x}{3} + \frac{y}{2} = 1 .\]


Find the area of the region bounded by the following curves, the X-axis and the given lines: y = `sqrt(16 - x^2)`, x = 0, x = 4


Find the area of the region bounded by the following curves, the X-axis and the given lines:  2y = 5x + 7, x = 2, x = 8


Find the area of the region bounded by the following curves, the X-axis and the given lines:

y = x2 + 1, x = 0, x = 3


The area of the region bounded by y2 = 4x, the X-axis and the lines x = 1 and x = 4 is _______.


Area of the region bounded by x2 = 16y, y = 1 and y = 4 and the Y-axis, lying in the first quadrant is _______.


Using definite integration, area of the circle x2 + y2 = 49 is _______.


Fill in the blank :

The area of the region bounded by the curve x2 = y, the X-axis and the lines x = 3 and x = 9 is _______.


State whether the following is True or False :

The area bounded by the curve x = g (y), Y-axis and bounded between the lines y = c and y = d is given by `int_"c"^"d"x*dy = int_(y = "c")^(y = "d") "g"(y)*dy` 


Solve the following :

Find the area of the region bounded by the curve xy = c2, the X-axis, and the lines x = c, x = 2c.


Choose the correct alternative:

Area of the region bounded by the curve y = x3, x = 1, x = 4 and the X-axis is ______


Choose the correct alternative:

Using the definite integration area of the circle x2 + y2 = 16 is ______


The area bounded by the parabola x2 = 9y and the lines y = 4 and y = 9 in the first quadrant is ______


The area of the region bounded by the curve y2 = 4x, the X axis and the lines x = 1 and x = 4 is ______


Find the area of the region bounded by the curve 4y = 7x + 9, the X-axis and the lines x = 2 and x = 8


Find the area of the region bounded by the curve y = (x2 + 2)2, the X-axis and the lines x = 1 and x = 3


Find area of the region bounded by the parabola x2 = 4y, the Y-axis lying in the first quadrant and the lines y = 3


`int_0^log5 (e^xsqrt(e^x - 1))/(e^x + 3)` dx = ______ 


The area of the region bounded by the curve y = x IxI, X-axis and the ordinates x = 2, x = –2 is ______.


The equation of curve through the point (1, 0), if the slope of the tangent to t e curve at any point (x, y) is `(y - 1)/(x^2 + x)`, is


Equation of a common tangent to the circle, x2 + y2 – 6x = 0 and the parabola, y2 = 4x, is:


The area included between the parabolas y2 = 4a(x +a) and y2 = 4b(x – a), b > a > 0, is


The area of the region bounded by the curve y = x2, x = 0, x = 3, and the X-axis is ______.


The area (in sq.units) of the part of the circle x2 + y2 = 36, which is outside the parabola y2 = 9x, is ______.


Area bounded by y = sec2x, x = `π/6`, x = `π/3` and x-axis is ______.


The area (in sq. units) of the region {(x, y) : y2 ≥ 2x and x2 + y2 ≤ 4x, x ≥ 0, y ≥ 0} is ______.


If the area enclosed by y = f(x), X-axis, x = a, x = b and y = g(x), X-axis, x = a, x = b are equal, then f(x) = g(x).


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×