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

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

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

Let A be the required area.

Given equation of the curve is y = (x2 + 2)2 

∴ A = `int_1^3 y  "d"x`

= `int_1^3 (x^2 + 2)^2  "d"x`

= `int_0^3 (x^4 + 4x^2 + 4)`

= `[x^5/5 + 4(x^3/3) + 4x]_1^3`

= `[3^5/5 + 4(3^3/3) + 4(3)] - [1^5/5 + 4(1^3/3) + 4(1)]`

= `(243/5 + 36 + 12) - (1/5 + 4/3 + 4)`

= `483/5 - 83/15`

= `(1449 - 83)/15`

= `1366/15` sq.units

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1.7: Application of Definite Integration - Q.2

संबंधित प्रश्‍न

Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.


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


Find the area of the region bounded by the parabola y = x2 and y = |x| .


Find the area of the region bounded by the curve y2 = 4x and the line x = 3


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


Find the area under the given curve and given line:

y = x4, x = 1, x = 5 and x-axis


Find the area between the curves y = x and y = x2


Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4


Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12


Find the area of the smaller region bounded by the ellipse `x^2/9 + y^2/4` and the line `x/3 + y/2 = 1`


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 parabola y2 = 16x and the line x = 4. 


Find the area of the region. 

{(x,y) : 0 ≤ y ≤ x2 , 0 ≤ y ≤ x + 2 ,-1 ≤ x ≤ 3} .


Using integration find the area of the triangle formed by negative x-axis and tangent and normal to the circle `"x"^2 + "y"^2 = 9  "at" (-1,2sqrt2)`.


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

y = `sqrt(6x + 4), x = 0, x = 2`


Find the area of the region bounded by the following curve, the X-axis and the given line:

y = 2 – x2, x = –1, x = 1


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


If the curve, under consideration, is below the X-axis, then the area bounded by curve, X-axis and lines x = a, x = b is positive.


Solve the following :

Find the area of the region bounded by the curve y = x2 and the line y = 10.


Choose the correct alternative:

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


State whether the following statement is True or False:

The equation of the area of the circle is `x^2/"a"^2 + y^2/"b"^2` = 1


The area of the shaded region bounded by two curves y = f(x), and y = g(x) and X-axis is `int_"a"^"b" "f"(x) "d"x + int_"a"^"b" "g"(x)  "d"x`


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


The area enclosed between the curve y = loge(x + e) and the coordinate axes is ______.


The slope of a tangent to the curve y = 3x2 – x + 1 at (1, 3) is ______.


The area enclosed by the parabola x2 = 4y and its latus rectum is `8/(6m)` sq units. Then the value of m is ______.


What is the area of an arbitrary elementary vertical strip of height \[y\] and width \[dx\]?


Why is the area enclosed by \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] written as \[4\int_0^a y\,dx\]?


What is the area enclosed by the ellipse \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×