Self Studies

Application of ...

TIME LEFT -
  • Question 1
    1 / -0

    The parabolas $$y^{2}=4x$$ and $$x^{2}=4y$$ divide the square region bounded by the lines $$x=4, y=4$$ and the coordinate axes. If $$S_{1}$$, $$S_{2}$$, $$S_{3}$$ are respectively the areas of these parts numbered from top to bottom; $$S_{1}: S_{2}: S_{3}$$ is

  • Question 2
    1 / -0

    Let y=f(x) be the given curve and $$x=a$$, $$x=b$$ be two ordinates then area bounded by the curve $$y=f(x)$$, the axis of x between the ordinates $$x=a$$ & $$x=b$$, is given by definite integral
    $$\int_{a}^{b}ydx$$ or $$\int_{a}^{b}f\left ( x \right )dx$$ and the area bounded by the curve $$x=f(y)$$, the axis of y & two abscissae $$y=c$$ & $$y=d$$ is given by $$\int_{c}^{d}xdy$$ or $$\int_{c}^{d}f\left ( x \right )dy$$. Again if we consider two curves $$y=f(x)$$, $$y=g(x)$$ where $$f\left ( x \right )\geq g\left ( x \right )$$ in the interval [a, b] where $$x=a$$ & $$x=b$$ are the points of intersection of these two curves Shown by the graph given
    Then area bounded by these two curves is given by
    $$\int_{a}^{b}\left [ f\left ( x \right )-g\left ( x \right ) \right ]dx$$
    On the basis of above information answer the following questions.

    The area bounded by parabolas $$y=x^{2}+2x+1$$ & $$y=x^{2}-2x+1$$ and the line $$\displaystyle y=\frac{1}{4}$$ is equal to

  • Question 3
    1 / -0

    If f(x) be an increasing function defined on [a, b] then
    max {f(t) such that $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x)  & min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a) and if f(x) be decreasing function defined on [a, b] then
    max {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a),
    min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x).
    On the basis of above information answer the following questions.
    $$\int_{0}^{\pi /2}min\left \{ \sin x, \cos x \right \}dx$$ equals

  • Question 4
    1 / -0

    The  area bounded by $${ y }^{ 2 }+8x=16$$ and $${ y }^{ 2 }-24x=48$$ is $$\displaystyle \frac { a\sqrt { 6 }  }{ c } $$, then $$a+c=$$

  • Question 5
    1 / -0

    The area enclosed between the curves $$y=x^3$$ and $$y=\sqrt{x}$$ is (in square units)

  • Question 6
    1 / -0

    Find the area bounded by $$\displaystyle y = \cos ^{-1}x,y=\sin ^{-1}x$$ and $$y-$$axis

  • Question 7
    1 / -0

    Consider two curves $$\displaystyle C_{1}:y=\frac{1}{x}$$ and $$\displaystyle C_{2}$$ : $$y = \displaystyle lnx$$ on the xy plane Let $$\displaystyle D_{1}$$ denotes the region surrounded by $$\displaystyle C_{1}$$, $$\displaystyle C_{2}$$ and the line $$x = 1$$ and $$\displaystyle D_{2}$$ denotes the region surrounded by $$\displaystyle C_{1}$$, $$\displaystyle C_{2}$$ and the line $$x = a$$ If $$\displaystyle D_{1}$$=$$\displaystyle D_{2}$$ then the value of 'a':

  • Question 8
    1 / -0

    Suppose $$y = f(x)$$ and $$y = g(x)$$ are two functions whose graphs intersect at three points $$(0, 4), (2, 2)$$ and $$(4, 0)$$ with $$f(x) > g(x)$$ for $$0 < x < 2$$ and $$f(x) < g(x)$$ for $$2 < x < 4 $$. if $$\displaystyle \int_{0}^{4}\left ( f(x)-g(x) \right )dx=10$$ and $$\displaystyle \int_{2}^{4}\left ( g(x)-f(x) \right )dx=5$$, the area between two curves for $$0 < x < 2$$, is:

  • Question 9
    1 / -0

    Area of the region enclosed between the curves $$\displaystyle x=y^{2}-1$$ and $$\displaystyle x = \left | y \right |\sqrt{1-y^{2}}$$ is

  • Question 10
    1 / -0

    Find the area of the region bounded by the curves $$\displaystyle x=\frac{1}{2},x=2,y=logx$$ and $$y=2^{x}$$

Submit Test
Self Studies
User
Question Analysis
  • Answered - 0

  • Unanswered - 10

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
Submit Test
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now