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Applications of the Integrals Test - 2

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Applications of the Integrals Test - 2
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  • Question 1
    1 / -0.25

    Solution

    Area of standard ellipse is given by :πab.

  • Question 2
    1 / -0.25

    The area bounded by the curve y = 4x - x2 and the x- axis is equal to

    Solution

    For x - axis, y = 0,

    Therefore, 4x - x2  = 0

    Therefore, x = 0 or x = 4

    Required area :

  • Question 3
    1 / -0.25

    The area of the smaller segment cut off from the circle  x2 +y2  = 9by x = 1 is

    Solution


  • Question 4
    1 / -0.25

    For which of the following values of m , is the area of the region bounded by the curve y = x - x2  and the line y = mx equal to  9/2  ?

    Solution


  • Question 5
    1 / -0.25

    The area of the region bounded by the parabola  (y −2)2  = x −1,the tangent to yhe parabola at the point (2 , 3) and the x –axis is equal to

    Solution


    Therefore, tangent at (2, 3) is y –3 = ½(x –2). i.e. x –2y +4 = 0 . therefore required area is  

  • Question 6
    1 / -0.25

    The area of the loop between the curve y = a sin x and the x –axis and x = 0 , x = π. is

    Solution


  • Question 7
    1 / -0.25

    The area bounded by the curve  y2  = x,line y = 4 and y –axis is equal to

    Solution

  • Question 8
    1 / -0.25

    The area enclosed by the curve  xy2 = a2 (a  − x) and the y –axis is

    Solution


  • Question 9
    1 / -0.25

    The area bounded by the curve y = x[x], the x –axis and the ordinates x = 1 and x = -1 is given by

    Solution


    Graph of y = x [x]

  • Question 10
    1 / -0.25

    The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and  x = π/2 is equal to

    Solution


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