Self Studies

Mathematics Test - 19

Result Self Studies

Mathematics Test - 19
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    5 / -1
    The area bounded by the curve x = f(y), the y-axis and the two lines y = a and y = b is equal to:
    Solution

    Concept:

    The area (A) under the curve y = f(x) between x = a & x = b is given by, 

    A = abf(x) dx

    Calculation:

    Here, the curve x = f(y) and lines y = a and y = b

    ∴Area = abf(y)dy(function is f(y))

    abx dy                   

    (∵ f(y) = x)

    Hence, option (3) is correct. 

  • Question 2
    5 / -1
    Find the area under the curve y = 2e4x from x = 0 to x = 4 as the end points
    Solution

    Concept:

    The area of the curve y = f(x) is given by:

    A = x1x2f(x)dx

    where x1 and x2 are the endpoints between which the area required.

     

    Calculation:

    The f(x) = y = 2e4x 

    Given the end points x1 = 0, x2 = 4

    Area of the curve (A) = |042e4x dx|

    ⇒ A = |[2e4x4]04|

    ⇒ A = 12|[e4x]04|

    ⇒ A = 12|[e16e0]|

    ⇒ A = 12(e161)

     

    Additional Information

    Integral property:

    • ∫ xn dx = xn+1n+1+ C ; n ≠ -1
    • 1xdx=lnx + C
    • ∫ edx = ex+ C
    • ∫ adx = (ax/ln a) + C ; a > 0,  a ≠ 1
    • ∫ sin x dx = - cos x + C
    • ∫ cos x dx = sin x + C 
  • Question 3
    5 / -1
    What is the area of the region bounded by the lines y = x, y = 0 and x = 4?
    Solution

    Concept:

    Area of triangle = 12×base×height

    Calculations:

    Given lines are y = x, y = 0 and x = 4

    To find the area of region bounded by line y = x, y = 0 and x = 4, fist draw a graph of the lines.



    First find the point of intersection.

    when y = 0 , x = 0

    when x = 4, y = 4

    So, point of intersection is (0, 0), and (4, 4).

    By these lines y = x, y = 0 and x = 4, we get bounded region as triangle.

    Area of triangle = 12×base×height

    ⇒Area of triangle = 12×4×4

    ⇒Area of triangle = 8 units

    Hence,the area of the region bounded by the lines y = x, y = 0 and x = 4 is 8 units

  • Question 4
    5 / -1
    Area bounded by the curve y = 2x – x2 and the line x + y = 0 is -
    Solution

    Formula used:

    Area of the curve y = f(x) bounded b/w x = a to x = b is  given by

    A=abf(x)dx

    Calculation:

    We have 

    f(x) = 2x - x   -----(1)

    y = -x              -----(2)

    From equation (1)

    ⇒ -x = 2x - x2

    ⇒ x2 - 3x = 0

    ⇒ x(x - 3) = 0

    x = 0 and 3

    Put these values in equation (2)

    y = 0 and -3

    So, parabola f(x) = 2x - x2 & line y = -x intersect at (0, 0) & (3, -3).

     

    Hence, the Required area (shaded region)

    A = 03[(2xx2)(x)]dx

    ⇒ A =03(3xx2)dx=[3x22x33]03

     =272273=92 sq units.

  • Question 5
    5 / -1
    Find the area between the lines y = |x - 2| and y = 5 - |x + 1|
    Solution

    Concept:

    The area between the curves y1 = f(x) and y2 = g(x) is given by:

    Area enclosed = x1x2(y1y2)dx

    Where x1 and x2 are the intersections of curves y1 and y2 

    Calculation:

    The shaded area has to be calculated

    Curve 1: y = |x - 2|

    ⇒ y = 2 - x for x < 2

    ⇒ y = x - 2 for x ≥ 2

    Curve 2: y = 5 - |x + 1|

    ⇒ y = 5 + x + 1 = x + 6 for x < -1

    ⇒ y = 5 - x - 1 = 4 - x for x ≥ -1

     

    Here, the enclosed region is a rectangle as the opposite lines are parallel.

    From the graph,

    Length of the smaller side = √[(5-4)2+(-1+2)2] = √2 

    Length of the larger side = √[(2- (-2))2+(0- 4)2] = 4√2

    ⇒ A =√2 × 4√2 

    ⇒ A = 8 sq. units

    Alternate Method

    Calculation:

    The shaded area has to be calculated

    Curve 1: y = |x - 2|

    ⇒ y = 2 - x for x < 2

    ⇒ y = x - 2 for x ≥ 2

    Curve 2: y = 5 - |x + 1|

    ⇒ y = 5 + x + 1 = x + 6 for x < -1

    ⇒ y = 5 - x - 1 = 4 - x for x ≥ -1

    Here, the enclosed region is a rectangle as the opposite lines are parallel.

    From the graph,

    Length of the smaller side = √[(5-4)2+(-1+2)2] = √2 

    Length of the larger side = √[(2- (-2))2+(0- 4)2] = 4√2

    ⇒ A =√2 × 4√2 

    ⇒ A = 8 sq. units

  • Question 6
    5 / -1
    Find the area of region bounded by the curve y2 = x and the line x = 1, x = 4 and the x-axis
    Solution

    Given:

    Area bounded by curve y2 = x and the line x = 1, x = 4 and x-axis

    Formula used:

    xn.dx=xn+1n+1; n can't be - 1

    Area under a curve y = f(x) is given by x.dx

    Calculation:

    From the figure, area is in the curve y2 = x, from x = 1 to x = 4

    y = x1/2

    The area is given by:

    14x1/2.dx

    ⇒ |x3/23/2|, x from 1 to 4; By putting values of x:

    ⇒ |8/(3/2) - 1/(3/2)|

    ⇒ 14/3

    ∴ Area of the region = 14/3

  • Question 7
    5 / -1

    The area bounded by the circle x2 + y2 = 8, the parabola x2 = 2y and the line y = x in y ≥ o

    I. Has area = (2π+43) sq units

    I. Has area = (2π+23) sq units

    III. The point of intersection of circle, parabola and line in 1st quadrant is (2, 2).

    Solution

    Formula used : 

    a2x2dx=12xa2x2+a22sin1xa+C

    xn dx=xn+1n+1+C

    Calculations :

    Given that,

    x2 + y2 = 8            -----(1)

    ⇒ y = √(8 - x2)      -----(2)

    Also, the given equation of a line is y = x. Hence, from equation (1)

    ⇒ x2 + x2 = 8

    ⇒ x = ± 2

    Now using equation (2), we get y = 2

    Hence, we get the intersection of line and circle at (2, 2)

    Since we have parabola x2 = 2y. Again using the equation (1)

    2y + y2 = 8 

    ⇒ y2 + 2y - 8 = 0

    ⇒ (y + 4)(y - 2) = 0

    ⇒ y = 2 & -4

    Put y = 2 in parabola x2 = 2y

    ⇒ x2 = 2 × 2 = 4

    ⇒ x = ± 2

    Hence, we got the intersection of circle and parabola at (-2, 2) and (2, 2).

     

    solution

    Required area = area of circle - area of parabola - area of line

    ⇒ 228x22012x202xdx

    ⇒ 2028x2dx[x36]20[x22]02

    By using the above formula, 

    ⇒ 2[x28x2+4sin1x22]432

    ⇒ 2[2+4π2]103

    ∴ The area bounded is 23+2π sq units and point of intersection is (2, 2).

  • Question 8
    5 / -1
    Find the Area of the region (in square unit) bounded by the curve y = x – 2 and x = 0 to x= 4.
    Solution

    Concept used:

    The area between the curves y1 = f(x) and y2 = g(x) is given by:

    Area enclosed = x1x2|y2y1|dx

    Where x1 and x2 are the intersections of curves y1 and y2

    Calculation:

    In figure ΔABC and ΔAOD is similar

    So, Area of the region = 2 × (Area of ΔABC)

    For the area of ΔABC

    24(x2)dx

    [x222x]24     

    4222(4)(2222×2)

    = 8 - 8 - 2 + 4

    = 2 sq. unit

    So, Area of the region = 2 × (Area of ΔABC) = 2 × 2 = 4 sq. unit

    Alternate Method

    Using the triange formula = 1/2 × base × height

    in the figure = 1/2 × (2 × 2)

    Area of ΔABC = 2

    Total bounded area = 2 × 2 = 4 sq. unit 

  • Question 9
    5 / -1
    The area of the region bounded by the curve x = 9y2 and y-axis is 
    Solution

    Concept:

    a2x2=x2a2x2+a22sin1xa

     

    Calculations:

    Given, the curve is  x=9y2

     

    Now, equation of Y-axis x = 0, 

    ⇒ y2 = 9 

    ⇒ y = -3, 3

    Area bounded by  the curve y = 9x2 and y-axis 

    =203ydx

    =2039x2dx

    =2[x29x2+92sin1x3]03

    =2[x29x2+92sin1x3]03

    = 2[9π40]

    = 4.5 π sq. units 

    Hence, option 4 is correct.

  • Question 10
    5 / -1
    Find the area between the curve y = 2a sin x and the positive x-axis from x = 0 to π.
    Solution

    Concept:

    Area under a curve:

    • The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral |abf(x) dx|, for curves which are entirely on the same side of the x-axis in the given range.
    • If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.

     

    Calculation:

    The graph of the curve y = 2a sin x is presented 

    ∴ The required area = |0π2asinx dx| = 2a [cosx]0π =  4a.

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
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