Find the area bounded by the curve  y = x2 and the line y = x.
OR
Find the area of the region {(x. y): x2 ≤ y ≤ x}.


The given region is {(x, y): x2 ≤ y ≤ x}
This region is the intersection of the following regions:
straight R subscript 1 space equals space open curly brackets open parentheses straight x comma space straight y right parenthesis close parentheses space colon space straight x squared space less or equal than space straight y close curly brackets
straight R subscript 2 space equals space open curly brackets left parenthesis straight x comma space straight y right parenthesis space colon space straight y space less or equal than straight x close curly brackets
Consider the equations
                   straight y space equals straight x squared                  ...(1)
  and            y = x                     ...(2)
From (1) and (2), we get
                          straight x space equals space straight x squared space space or space space straight x squared minus straight x space equals space 0
rightwards double arrow space space space space straight x left parenthesis straight x minus 1 right parenthesis space equals space 0 space space space space rightwards double arrow space space space space straight x space equals space 0 comma space space 1
∴    from (2), y = 0, 1
∴    curves (1) and (2) intersect in the points O (0, 0) and A (1, 1)
Required area = area of the shaded region
equals space integral subscript 0 superscript 1 straight x space dx space minus space integral subscript 0 superscript 1 straight x squared space dx space equals space open square brackets straight x squared over 2 close square brackets subscript 0 superscript 1 space minus space open square brackets straight x cubed over 3 close square brackets subscript 0 superscript 1
equals space open parentheses 1 half minus 0 close parentheses space minus space open parentheses 1 third minus 0 close parentheses space equals space 1 half minus 1 third space equals space fraction numerator 3 minus 2 over denominator 6 end fraction space equals space 1 over 6 space sq. space units.
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Find the area of the region included between the parabola straight y space equals 3 over 4 straight x squared space and space the space line space 3 straight x space minus space 2 straight y space plus space 12 space equals space 0

The equations of the given curves are
                 straight y equals space 3 over 4 straight x squared                          ...(1)
  and        3 straight x minus 2 straight y plus 12 space equals space 0             ...(2)
From (2), 2 straight y space equals space 3 straight x plus 12
therefore space space space straight y space equals space fraction numerator 3 straight x plus 12 over denominator 2 end fraction
Putting this value of y in (1), we get,
                                  fraction numerator 3 straight x plus 12 over denominator 2 end fraction space equals 3 over 4 straight x squared
therefore space space 6 straight x space plus 24 space equals space 3 straight x squared space space space space space rightwards double arrow space space straight x squared minus 2 straight x minus 8 space equals space 0 space space space space rightwards double arrow space space space space left parenthesis straight x plus 2 right parenthesis thin space left parenthesis straight x minus 4 right parenthesis space equals space 0
rightwards double arrow space space space space straight x space equals space minus 2.4
therefore space space space space straight y space equals space 3 comma space 12
therefore curves (1) and (2) intersect in points A (4, 12) and B(-2, 3).
From A. draw AM ⊥ x -axis and from B, draw BN ⊥ x-axis.
Required area = area of trapezium BNMA - (area BNO + area OMA)
equals space 1 half left parenthesis 3 plus 12 right parenthesis space cross times 6 space minus space integral subscript negative 2 end subscript superscript 4 3 over 4 straight x squared dx                          open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
equals space 45 space minus space 3 over 4 integral subscript negative 2 end subscript superscript 4 straight x squared dx space equals space 45 minus 3 over 4 open square brackets straight x cubed over 3 close square brackets subscript negative 2 end subscript superscript 4
equals space 45 minus 1 fourth open square brackets straight x cubed close square brackets subscript negative 2 end subscript superscript 4 space equals space 45 minus 1 fourth left square bracket 64 plus 8 right square bracket space equals space 27 space sq. space units. space


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Find the area of the region bounded by the line y = 3 x + 2, the x-axis and the ordinates x = - 1 and x = 1.


The equation of given line is
y = 3 x + 2    ...(1)
Consider the lines
x = -1    ...(2)
and    x = 1    ...(3)
Line (1) meets x-axis where y = 0

therefore  putting y = 0 in (1), we get,
       0 space equals space 3 straight x plus 2 space space space or space space straight x space space equals space minus 2 over 3
therefore line (1) meets x-axis in straight A open parentheses negative 2 over 3 comma space 0 close parentheses
Let line (1) meet lines (2) and (3) in B and D respectively. From B, draw BC ⊥ x-axis and from D, draw DE ⊥ x-axis.
Required area = Area of region ACBA + area of region ADEA
equals space open vertical bar integral subscript negative 1 end subscript superscript fraction numerator negative 2 over denominator 3 end fraction end superscript left parenthesis 3 straight x plus 2 right parenthesis space dx close vertical bar space plus space integral subscript negative 2 over 3 end subscript superscript 1 left parenthesis 3 straight x plus 2 right parenthesis space dx
equals space open vertical bar open square brackets fraction numerator 3 straight x squared over denominator 2 end fraction plus 2 straight x close square brackets subscript negative 1 end subscript superscript negative 2 over 3 end superscript close vertical bar space plus space open square brackets fraction numerator 3 straight x squared over denominator 2 end fraction plus 2 straight x close square brackets subscript negative 2 over 3 end subscript superscript 1
equals space open vertical bar open curly brackets 3 over 2 minus open parentheses negative 2 over 3 close parentheses squared plus 2 space open parentheses negative 2 over 3 close parentheses close curly brackets space minus space open curly brackets 3 over 2 left parenthesis negative 1 right parenthesis squared plus 2 left parenthesis negative 1 right parenthesis close curly brackets close vertical bar
space space space space space space space space space space space space space space space space space
                                                          plus open square brackets 3 over 2 left parenthesis 1 right parenthesis squared plus 2 left parenthesis 1 right parenthesis close square brackets space minus space open square brackets 3 over 2 open parentheses negative 2 over 3 close parentheses squared plus 2 open parentheses negative 2 over 3 close parentheses close square brackets
equals space open vertical bar open parentheses 2 over 3 minus 4 over 3 close parentheses minus open parentheses 3 over 2 minus 2 close parentheses close vertical bar plus open parentheses 3 over 2 plus 2 close parentheses minus open parentheses 2 over 3 minus 4 over 3 close parentheses space equals space open vertical bar negative 2 over 3 plus 1 half close vertical bar plus open parentheses 7 over 2 plus 2 over 3 close parentheses
equals space open vertical bar negative 1 over 6 close vertical bar plus 25 over 6 space equals space 1 over 6 plus 25 over 6 equals space 26 over 6 equals space 13 over 3 space sq. space units.

112 Views

Find the area bounded by the parabola x2 = 4 y and the straight line x = 4 y - 2.

The equation of curve is x2 = 4 y    ...(1)
which is upward parabola with vertex O.
The equation of line is
x = 4 y - 2    ...(2)
Let us solve (1) and (2)
Putting x = 4y - 2 in (1), we get
                  left parenthesis 4 space straight y space minus 2 right parenthesis squared space equals space 4 space straight y
therefore space space space 16 space straight y squared minus space 16 space straight y plus space 4 space equals space 4 straight y
therefore space space 16 space straight y squared minus 20 straight y space plus space 4 space equals space 0
or space 4 straight y squared minus 5 straight y plus 1 space equals space 0
therefore space space space space straight y space equals space fraction numerator 5 plus-or-minus square root of 25 minus 16 end root over denominator 8 end fraction space equals space fraction numerator 5 plus-or-minus 3 over denominator 8 end fraction space equals space 8 over 8 comma space 2 over 8
therefore space space space space straight y space equals space 1 comma space space 1 fourth
therefore space space from space left parenthesis 2 right parenthesis comma space space straight x space equals space 4 minus 2 comma space 1 minus 2 space equals space 2 comma space minus 1
therefore space space curve space left parenthesis 1 right parenthesis space and space line space left parenthesis 2 right parenthesis space intersect space in space two space points space straight A left parenthesis 2 comma space 1 right parenthesis space and space straight B space open parentheses negative 1 comma space 1 fourth close parentheses



From A, draw AM ⊥ x-axis and from B. draw BN ⊥ x-axis.
Required area = area AOB
= Area of trapezium BNMA - (area BNO + area OMA)
1 half open parentheses 1 plus 1 fourth close parentheses cross times 3 space minus space integral subscript negative 1 end subscript superscript 2 straight y space dx space equals space 1 half cross times 5 over 4 cross times 3 space minus space integral subscript negative 1 end subscript superscript 2 straight x squared over 4. dx              open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets
equals space 15 over 8 minus 1 fourth integral subscript negative 1 end subscript superscript 2 straight x squared space dx space equals space 15 over 8 minus 1 fourth open square brackets straight x cubed over 3 close square brackets subscript negative 1 end subscript superscript 2 space equals space 15 over 8 minus fraction numerator 1 over denominator 4 cross times 3 end fraction open square brackets straight x close square brackets subscript negative 1 end subscript superscript 2
equals space 15 over 8 minus 1 over 12 open square brackets left parenthesis 2 right parenthesis cubed minus left parenthesis negative 1 right parenthesis cubed close square brackets space equals space 15 over 8 minus 1 over 12 open square brackets 8 minus left parenthesis negative 1 right parenthesis close square brackets
equals space 15 over 8 minus 1 over 12 left parenthesis 8 plus 1 right parenthesis space equals space 15 over 8 minus 9 over 12 space equals space 9 over 8 space sq. space units. space

201 Views

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Find the area of the region included between the parabola y2 = x and the line x + y = 2.


The equation of parabola is
y2 = x    ...(1)
The equation of line is
x + y = 2    ...(2)
From (2), y = 2 - x    ...(3)
Putting this value of y in (1), we get,
(2 - x)= x
or x2 - 4 x + 4 = x or x2 - 5 x + 4 = 0



∴ (x - 1) (x - 4) = 0
∴ x = 1, 4
∴ from (3), y = 1, - 2
∴ parabola (1) and line (2) intersect in the points A (1, 1), B (4, - 1)
Also line (2) meets x-axis in C (2,0)
Required area is shaded.
Area above x-axis = area AOL + area ALC
    equals space integral subscript 0 superscript 1 square root of straight x dx plus integral subscript 1 superscript 2 left parenthesis 2 minus straight x right parenthesis space dx space equals space 2 over 3 open square brackets straight x to the power of 3 over 2 end exponent close square brackets subscript 0 superscript 1 plus open square brackets 2 space straight x space minus space straight x squared over 2 close square brackets subscript 1 superscript 2
equals space 2 over 3 left parenthesis 1 minus 0 right parenthesis space plus space open square brackets left parenthesis 4 minus 2 right parenthesis space minus space left parenthesis 2 minus 1 half right parenthesis close square brackets space equals space 2 over 3 plus 2 minus 3 over 2 equals 2 over 3 plus 1 half equals 7 over 6 space sq. space units
Area below x-axis = Area OBM - area CBM
                             equals space integral subscript 0 superscript 4 square root of straight x minus integral subscript 2 superscript 4 left parenthesis 2 minus straight x right parenthesis space dx space equals space 2 over 3 open square brackets straight x to the power of 3 divided by 2 end exponent close square brackets subscript 0 superscript 4 space minus space open square brackets 2 straight x minus straight x squared over 2 close square brackets subscript 2 superscript 4
equals space 2 over 3 left square bracket 4 to the power of 3 divided by 2 end exponent minus 0 right square bracket space minus space open square brackets left parenthesis 8 minus 8 right parenthesis space minus space left parenthesis 4 minus 2 right parenthesis close square brackets
equals space 16 over 3 minus left parenthesis negative 2 right parenthesis space equals space 16 over 3 minus 2 space space space space space space space space space space space space space space left square bracket because space space area space is space always space positive right square bracket
equals space 10 over 3 space sq. space units
therefore space space total space area space equals 7 over 6 plus 10 over 3 equals fraction numerator 7 plus 20 over denominator 6 end fraction equals 27 over 6 equals 9 over 2 space sq. space units. space


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