CHAPTER-8-APPLICATION OF INTEGRALS
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CHAPTER-8
APPLICATION OF INTEGRALS
EXRCISE-8.1
PAGE-365
Question 1:
Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4
and the x-axis.
Answer
Discussion
The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and
the x-axis is the area ABCD.
http://cbse.meritnation.com/study-online/solution/Math/CKKqCIjQgAg8OfG3zNWCTw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/CKKqCIjQgAg8OfG3zNWCTw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/CKKqCIjQgAg8OfG3zNWCTw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/CKKqCIjQgAg8OfG3zNWCTw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/CKKqCIjQgAg8OfG3zNWCTw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/CKKqCIjQgAg8OfG3zNWCTw!!#answer%23answer -
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Question 2:
Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the
first quadrant.
Answer
Discussion
The area of the region bounded by the curve, y2 = 9x, x = 2, and x = 4, and the x-
axis is the area ABCD.
http://cbse.meritnation.com/study-online/solution/Math/X1BGnc@LYOzsAp4sYqJTVg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/X1BGnc@LYOzsAp4sYqJTVg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/X1BGnc@LYOzsAp4sYqJTVg!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/X1BGnc@LYOzsAp4sYqJTVg!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/X1BGnc@LYOzsAp4sYqJTVg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/X1BGnc@LYOzsAp4sYqJTVg!!#discussion%23discussion -
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Question 3:
Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the
first quadrant.
Answer
Discussion
The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-
axis is the area ABCD.
http://cbse.meritnation.com/study-online/solution/Math/HxM7v9XtZjl$$7DybyUEHw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/HxM7v9XtZjl$$7DybyUEHw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/HxM7v9XtZjl$$7DybyUEHw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/HxM7v9XtZjl$$7DybyUEHw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/HxM7v9XtZjl$$7DybyUEHw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/HxM7v9XtZjl$$7DybyUEHw!!#discussion%23discussion -
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Question 4:
Find the area of the region bounded by the ellipse
Answer
Discussion
The given equation of the ellipse, , can be represented as
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It can be observed that the ellipse is symmetrical about x-axis and y-axis.
Area bounded by ellipse = 4 Area of OAB
Therefore, area bounded by the ellipse = 4 3 = 12 units
Question 5:
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Find the area of the region bounded by the ellipse
Answer
Discussion
The given equation of the ellipse can be represented as
It can be observed that the ellipse is symmetrical about x-axis and y-axis.
Area bounded by ellipse = 4 Area OAB
http://cbse.meritnation.com/study-online/solution/Math/n8dP7G4jA8KGoh0hc9NKhg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/n8dP7G4jA8KGoh0hc9NKhg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/n8dP7G4jA8KGoh0hc9NKhg!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/n8dP7G4jA8KGoh0hc9NKhg!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/n8dP7G4jA8KGoh0hc9NKhg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/n8dP7G4jA8KGoh0hc9NKhg!!#discussion%23discussion -
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Therefore, area bounded by the ellipse =
Question 6:
Find the area of the region in the first quadrant enclosed by x-axis, line
and the circle
Answer
Discussion
The area of the region bounded by the circle, , and the x-axis is
the area OAB.
http://cbse.meritnation.com/study-online/solution/Math/J4724pLu6njJO5Nhtsgruw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/J4724pLu6njJO5Nhtsgruw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/J4724pLu6njJO5Nhtsgruw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/J4724pLu6njJO5Nhtsgruw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/J4724pLu6njJO5Nhtsgruw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/J4724pLu6njJO5Nhtsgruw!!#discussion%23discussion -
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The point of intersection of the line and the circle in the first quadrant is .
Area OAB = Area OCA + Area ACB
Area of OAC
Area of ABC
Therefore, area enclosed by x-axis, the line , and the circle in the
first quadrant =
Question 7:
Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line
Answer
Discussion
http://cbse.meritnation.com/study-online/solution/Math/aO9MDuTkDw4cW6X0iuOHvA!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/aO9MDuTkDw4cW6X0iuOHvA!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/aO9MDuTkDw4cW6X0iuOHvA!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/aO9MDuTkDw4cW6X0iuOHvA!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/aO9MDuTkDw4cW6X0iuOHvA!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/aO9MDuTkDw4cW6X0iuOHvA!!#discussion%23discussion -
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The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, , is
the area ABCDA.
It can be observed that the area ABCD is symmetrical about x-axis.
Area ABCD = 2 Area ABC
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Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the
line, , is units.
Question 8:
The area between x = y2 and x = 4 is divided into two equal parts by the line x = a,
find the value ofa.
Answer
Discussion
The line, x = a, divides the area bounded by the parabola and x = 4 into two equal
parts.
Area OAD = Area ABCD
It can be observed that the given area is symmetrical about x-axis.
Area OED = Area EFCD
http://cbse.meritnation.com/study-online/solution/Math/YjRpnGFzd$DWSdcktjneVw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/YjRpnGFzd$DWSdcktjneVw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/YjRpnGFzd$DWSdcktjneVw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/YjRpnGFzd$DWSdcktjneVw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/YjRpnGFzd$DWSdcktjneVw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/YjRpnGFzd$DWSdcktjneVw!!#discussion%23discussion -
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From (1) and (2), we obtain
Therefore, the value ofa is .
Question 9:
Find the area of the region bounded by the parabola y = x2 and
Answer
Discussion
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The area bounded by the parabola, x2 = y,and the line, , can be represented as
The given area is symmetrical about y-axis.
Area OACO = Area ODBO
The point of intersection of parabola, x2 = y, and line, y = x, is A (1, 1).
Area of OACO = Area OAB Area OBACO
Area of OACO = Area of OAB Area of OBACO
Therefore, required area = units
Question 10:
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Find the area bounded by the curve x2 = 4y and the line x = 4y 2
Answer
Discussion
The area bounded by the curve, x2 = 4y, and line, x = 4y 2, is represented by the
shaded area OBAO.
Let A and B be the points of intersection of the line and parabola.
Coordinates of point .
Coordinates of point B are (2, 1).
We draw AL and BM perpendicular to x-axis.
It can be observed that,
Area OBAO = Area OBCO + Area OACO (1)
Then, Area OBCO = Area OMBC Area OMBO
http://cbse.meritnation.com/study-online/solution/Math/Jx4NGjCMWQXfmY6w4NrHnQ!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/Jx4NGjCMWQXfmY6w4NrHnQ!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/Jx4NGjCMWQXfmY6w4NrHnQ!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/Jx4NGjCMWQXfmY6w4NrHnQ!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/Jx4NGjCMWQXfmY6w4NrHnQ!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/Jx4NGjCMWQXfmY6w4NrHnQ!!#discussion%23discussion -
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Similarly, Area OACO = Area OLAC Area OLAO
Therefore, required area =
Question 11:
Find the area of the region bounded by the curve y2 = 4x and the line x = 3
Answer
Discussion
http://cbse.meritnation.com/study-online/solution/Math/R6OQEU8JV6WACzX9Zv0gag!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/R6OQEU8JV6WACzX9Zv0gag!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/R6OQEU8JV6WACzX9Zv0gag!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/R6OQEU8JV6WACzX9Zv0gag!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/R6OQEU8JV6WACzX9Zv0gag!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/R6OQEU8JV6WACzX9Zv0gag!!#discussion%23discussion -
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The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO.
The area OACO is symmetrical about x-axis.
Area of OACO = 2 (Area of OAB)
Therefore, the required area is units.
Question 12:
Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the
lines x = 0 and x = 2 is
A.
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B.
C.
D.
Answer
Discussion
The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant
is represented as
Thus, the correct answer is A.
Question 13:
http://cbse.meritnation.com/study-online/solution/Math/fwXr4J03AsERPhXkdeT6Qw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/fwXr4J03AsERPhXkdeT6Qw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/fwXr4J03AsERPhXkdeT6Qw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/fwXr4J03AsERPhXkdeT6Qw!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/fwXr4J03AsERPhXkdeT6Qw!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/fwXr4J03AsERPhXkdeT6Qw!!#discussion%23discussion -
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Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is
A. 2
B.
C.
D.
Answer
Discussion
The area bounded by the curve, y2 = 4x, y-axis, and y = 3 is represented as
http://cbse.meritnation.com/study-online/solution/Math/qva5f94PXDA32OWg$yeMgg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/qva5f94PXDA32OWg$yeMgg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/qva5f94PXDA32OWg$yeMgg!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/qva5f94PXDA32OWg$yeMgg!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/qva5f94PXDA32OWg$yeMgg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/qva5f94PXDA32OWg$yeMgg!!#discussion%23discussion -
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Thus, the correct answer is B.
EXERCISE-8.2
PAGE-371
Question 1:
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y
Answer
Discussion
The required area is represented by the shaded area OBCDO.
http://cbse.meritnation.com/study-online/solution/Math/7rf4liFPnF2nyhjDsl1lsg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/7rf4liFPnF2nyhjDsl1lsg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/7rf4liFPnF2nyhjDsl1lsg!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/7rf4liFPnF2nyhjDsl1lsg!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/7rf4liFPnF2nyhjDsl1lsg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/7rf4liFPnF2nyhjDsl1lsg!!#discussion%23discussion -
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Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we obtain
the point of intersection as .
It can be observed that the required area is symmetrical about y-axis.
Area OBCDO = 2 Area OBCO
We draw BM perpendicular to OA.
Therefore, the coordinates of M are .
Therefore, Area OBCO = Area OMBCO Area OMBO
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Therefore, the required area OBCDO
is units
Question 2:
Find the area bounded by curves (x 1)2 + y2 = 1 and x2 + y 2 = 1
Answer
Discussion
The area bounded by the curves, (x 1)2 + y2 = 1 and x2 + y 2 = 1, is represented by
the shaded area as
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On solving the equations, (x 1)2 + y2 = 1 and x2 + y 2 = 1, we obtain the point of
intersection as A and B .
It can be observed that the required area is symmetrical about x-axis.
Area OBCAO = 2 Area OCAO
We join AB, which intersects OC at M, such that AM is perpendicular to OC.
The coordinates of M are .
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Therefore, required area OBCAO = units
Question 3:
Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3
Answer
Discussion
The area bounded by the curves, y = x2 + 2, y = x, x = 0, and x = 3, is represented
by the shaded area OCBAO as
http://cbse.meritnation.com/study-online/solution/Math/BPr7bbngW18rrs7D8DOkQg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/BPr7bbngW18rrs7D8DOkQg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/BPr7bbngW18rrs7D8DOkQg!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/BPr7bbngW18rrs7D8DOkQg!!#discussion%23discussionhttp://cbse.meritnation.com/study-online/solution/Math/BPr7bbngW18rrs7D8DOkQg!!#answer%23answerhttp://cbse.meritnation.com/study-online/solution/Math/BPr7bbngW18rrs7D8DOkQg!!#discussion%23discussion -
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Then, Area OCBAO = Area ODBAO Area ODCO
Question 4:
Using integration finds the area of the region bounded by the triangle whose
vertices are (1, 0), (1, 3) and (3, 2).
Answer
Discussion
BL and CM are drawn perpendicular to x-axis.
It can be observed in the following figure that,
Area (ACB) = Area (ALBA) + Area (BLMCB) Area (AMCA) (1)
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Area (ABC) = (3 + 5 4) = 4 units
Question 5:
Using integration find the area of the triangular region whose sides have the
equations y = 2x +1, y= 3x + 1 and x = 4.
Answer
Discussion
The equations of sides of the triangle are y = 2x +1, y = 3x + 1, and x = 4.
On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13),and C (4, 9).
It can be observed that,
Area (ACB) = Area (OLBAO) Area (OLCAO)
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Question 6:
Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is
A. 2 ( 2)
B. 2
C. 2 1
D. 2 ( + 2)
Answer
Discussion
The smaller area enclosed by the circle, x2 + y2 = 4, and the line, x + y = 2, is
represented by the shaded area ACBA as
It can be observed that,
Area ACBA = Area OACBO Area (OAB)
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Thus, the correct answer is B.
Question 7:
Area lying between the curve y2 = 4x and y = 2x is
A.
B.
C.
D.
Answer
Discussion
The area lying between the curve, y2 = 4x and y = 2x, is represented by the shaded
area OBAO as
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The points of intersection of these curves are O (0, 0) and A (1, 2).
We draw AC perpendicular to x-axis such that the coordinates of C are (1, 0).
Area OBAO = Area (OCA) Area (OCABO)
Thus, the correct answer is B.
EXERCISE-MISC
PAGE-375
Question 1:
Find the area under the given curves and given lines:
-
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(i)y = x2, x = 1, x = 2 and x-axis
(ii)y = x4, x = 1, x = 5 and x axis
Answer
Discussion
i. The required area is represented by the shaded area ADCBA as
ii. The required area is represented by the shaded area ADCBA as
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Question 2:
Find the area between the curves y = x and y = x2
Answer
Discussion
The required area is represented by the shaded area OBAO as
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The points of intersection of the curves, y = x and y = x2, is A (1, 1).
We draw AC perpendicular to x-axis.
Area (OBAO) = Area (OCA) Area (OCABO) (1)
Question 3:
Find the area of the region lying in the first quadrant and bounded by y = 4x2, x =
0, y = 1 and y = 4
Answer
Discussion
The area in the first quadrant bounded by y = 4x2, x = 0, y = 1, and y = 4 is
represented by the shaded area ABCDA as
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Question 4:
Sketch the graph of and evaluate
Answer
Discussion
The given equation is
The corresponding values ofx and y are given in the following table.
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x 6 5 4 3 2 1 0
y 3 2 1 0 1 2 3
On plotting these points, we obtain the graph of as follows.
It is known that,
Question 5:
Find the area bounded by the curve y = sin x between x = 0 and x = 2
Answer
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Discussion
The graph ofy = sin x can be drawn as
Required area = Area OABO + Area BCDB
Question 6:
Find the area enclosed between the parabola y2 = 4ax and the line y = mx
Answer
Discussion
The area enclosed between the parabola, y2 = 4ax, and the line, y = mx, is
represented by the shaded area OABO as
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The points of intersection of both the curves are (0, 0) and .
We draw AC perpendicular to x-axis.
Area OABO = Area OCABO Area (OCA)
Question 7:
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Answer
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Discussion
The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 12, is
represented by the shaded area OBAO as
The points of intersection of the given curves are A (2, 3) and (4, 12).
We draw AC and BD perpendicular to x-axis.
Area OBAO = Area CDBA (Area ODBO + Area OACO)
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Question 8:
Find the area of the smaller region bounded by the ellipse and the
line
Answer
Discussion
The area of the smaller region bounded by the ellipse, , and the
line, , is represented by the shaded region BCAB as
Area BCAB = Area (OBCAO) Area (OBAO)
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Question 9:
Find the area of the smaller region bounded by the ellipse and the
line
Answer
Discussion
The area of the smaller region bounded by the ellipse, , and the
line, , is represented by the shaded region BCAB as
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Area BCAB = Area (OBCAO) Area (OBAO)
Question 9:
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Find the area of the smaller region bounded by the ellipse and the
line
Answer
Discussion
The area of the smaller region bounded by the ellipse, , and the
line, , is represented by the shaded region BCAB as
Area BCAB = Area (OBCAO) Area (OBAO)
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Question 10:
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-
axis
Answer
Discussion
The area of the region enclosed by the parabola, x2 = y, the line, y = x + 2, and x-
axis is represented by the shaded region OABCO as
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The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (1,
1).
Area OABCO = Area (BCA) + Area COAC
Question 11:
Using the method of integration find the area bounded by the curve
[Hint: the required region is bounded by lines x + y = 1, x y = 1, x + y = 1 and
x y = 11]
Answer
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Discussion
The area bounded by the curve, , is represented by the shaded region
ADCB as
The curve intersects the axes at points A (0, 1), B (1, 0), C (0, 1), and D (1, 0).
It can be observed that the given curve is symmetrical about x-axis and y-axis.
Area ADCB = 4 Area OBAO
Question 12:
Find the area bounded by curves
Answer
Discussion
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The area bounded by the curves, , is represented by the
shaded region as
It can be observed that the required area is symmetrical about y-axis.
Question 13:
Using the method of integration find the area of the triangle ABC, coordinates of
whose vertices are A (2, 0), B (4, 5) and C (6, 3)
Answer
Discussion
The vertices of ABC are A (2, 0), B (4, 5), and C (6, 3).
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Equation of line segment AB is
Equation of line segment BC is
Equation of line segment CA is
Area (ABC) = Area (ABLA) + Area (BLMCB) Area (ACMA)
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Question 14:
Using the method of integration find the area of the region bounded by lines:
2x + y = 4, 3x 2y = 6 and x 3y + 5 = 0
Answer
Discussion
The given equations of lines are
2x + y = 4 (1)
3x 2y = 6 (2)
And, x 3y + 5 = 0 (3)
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The area of the region bounded by the lines is the area of ABC. AL and CM are
the perpendiculars on x-axis.
Area (ABC) = Area (ALMCA) Area (ALB) Area (CMB)
Question 15:
Find the area of the region
Answer
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Discussion
The area bounded by the curves, , is represented as
The points of intersection of both the curves are .
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about x-axis.
Area OABCO = 2 Area OBC
Area OBCO = Area OMC + Area MBC
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Therefore, the required area is
units
Question 16:
Area bounded by the curve y = x3, the x-axis and the ordinates x = 2 and x = 1 is
A. 9
B.
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C.
D.
Answer
Discussion
Thus, the correct answer is B.
Question 17:
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The area bounded by the curve , x-axis and the ordinates x = 1 and x = 1 is
given by
[Hint:y = x2 ifx > 0 and y = x2 ifx < 0]
A. 0
B.
C.
D.
Answer
Discussion
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Area bounded by the circle and parabola
Area of circle = (r)2
= (4)2
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= 16 units
Thus, the correct answer is C.
Question 19:
The area bounded by the y-axis, y = cos x and y = sin x when
A.
B.
C.
D.
Answer
Discussion
The given equations are
y = cos x (1)
And, y = sin x (2)
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Required area = Area (ABLA) + area (OBLO)
Integrating by parts, we obtain
Thus, the correct answer is B.