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Application of Integral
Question 1: Find the area of the region bounded by the curve y^{2} = x & the lines x = 1, x = 4 & the xaxis.
ANSWER :  The area of the region bounded by the curve, y^{2} = x, the lines, x = 1 and x = 4, and the xaxis is the area ABCD.
Question 2: Find the area of the region bounded by y^{2} = 9x, x = 2, x = 4 & the xaxis in the first quadrant.
ANSWER :  The area of the region bounded by the curve, y^{2} = 9x, x = 2, and x = 4, and the xaxis is the area ABCD.
Question 3: Find the area of the region bounded by x^{2} = 4y, y = 2, y = 4 and the yaxis in the first quadrant.
ANSWER :  The area of the region bounded by the curve, x^{2} = 4y, y = 2, and y = 4, and the yaxis is the area ABCD.
Question 4: Find the area of the region bounded by the ellipse
ANSWER :  The given equation of the ellipse, , can be represented as
It can be observed that the ellipse is symmetrical about xaxis and yaxis.
∴ Area bounded by ellipse = 4 × Area of OAB
Therefore, area bounded by the ellipse = 4 × 3π = 12π units
Question 5: Find the area of the region bounded by the ellipse
ANSWER : The given equation of the ellipse can be represented as
It can be observed that the ellipse is symmetrical about xaxis and yaxis.
∴ Area bounded by ellipse = 4 × Area OAB
Therefore, area bounded by the ellipse =
Question 6: Find the area of the region in the first quadrant enclosed by xaxis, line and the circle
ANSWER : The area of the region bounded by the circle, , and the xaxis is the area OAB.
The point of intersection of the line and the circle in the first quadrant is .
Area OAB = Area ΔOCA Area ACB
Area of OAC
Area of ABC
Therefore, area enclosed by xaxis, the line, and the circle in the first quadrant =
Question 7: Find the area of the smaller part of the circle x^{2} + y^{2} = a^{2} cut off by the line
ANSWER : The area of the smaller part of the circle, x^{2} + y^{2} = a^{2}, cut off by the line, , is the area ABCDA.
It can be observed that the area ABCD is symmetrical about xaxis.
∴ Area ABCD = 2 × Area ABC
Therefore, the area of smaller part of the circle, x^{2} + y^{2} = a^{2}, cut off by the line, , is units.
Question 8: The area between x = y^{2} and x = 4 is divided into two equal parts by the line x = a, find the value of a.
ANSWER : 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 xaxis.
⇒ Area OED Area EFCD
From (1) and (2), we obtain
Therefore, the value of a is .
Question 9: Find the area of the region bounded by the parabola y = x^{2} and
ANSWER : The area bounded by the parabola, x^{2} = y,and the line,, can be represented as
The given area is symmetrical about yaxis.
∴ Area OACO = Area ODBO
The point of intersection of parabola, x^{2} = y, and line, y = x, is A (1, 1).
Area of OACO = Area ΔOAB – Area OBACO
⇒ Area of OACO = Area of ΔOAB – Area of OBACO
Therefore, required area = units
Question 10: Find the area bounded by the curve x^{2} = 4y and the line x = 4y – 2
ANSWER : The area bounded by the curve, x^{2} = 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 xaxis.
It can be observed that,
Area OBAO = Area OBCO Area OACO … (1)
Then, Area OBCO = Area OMBC – Area OMBO
Similarly, Area OACO = Area OLAC – Area OLAO
Therefore, required area =
Question 11: Find the area of the region bounded by the curve y^{2} = 4x and the line x = 3
ANSWER : The region bounded by the parabola, y^{2} = 4x, and the line, x = 3, is the area OACO.
The area OACO is symmetrical about xaxis.
∴ 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 x^{2} + y^{2} = 4 and the lines x = 0 and x = 2 is
A. π
B.
C.
D.
ANSWER : 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: Area of the region bounded by the curve y^{2} = 4x, yaxis and the line y = 3 is
A. 2
B.
C.
D.
ANSWER :  The area bounded by the curve, y^{2} = 4x, yaxis, and y = 3 is represented as
Thus, the correct answer is B.
Question 14: Find the area of the circle 4x^{2} + 4y^{2} = 9 which is interior to the parabola x^{2} = 4y
ANSWER : The required area is represented by the shaded area OBCDO.
Solving the given equation of circle, 4x^{2} + 4y^{2} = 9, and parabola, x^{2} = 4y, we obtain the point of
intersection as.
It can be observed that the required area is symmetrical about yaxis.
∴ Area OBCDO = 2 × Area OBCO
We draw BM perpendicular to OA.
Therefore, the coordinates of M are.
Therefore, Area OBCO = Area OMBCO – Area OMBO
Therefore, the required area OBCDO is units.
Question 15: Find the area bounded by curves (x – 1)^{2} + y^{2} = 1 and x^{2} + y ^{2} = 1
ANSWER : The area bounded by the curves, (x – 1)^{2} + y^{2} = 1 and x^{2} + y ^{2} = 1, is represented by the shaded area as
On solving the equations, (x – 1)^{2} + y^{2} = 1 and x^{2} + y ^{2} = 1, we obtain the point of intersection as Aand B.
It can be observed that the required area is symmetrical about xaxis.
∴ 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 .
Therefore, required area OBCAO = units.
Question 16: Find the area of the region bounded by the curves y = x^{2} + 2, y = x, x = 0 and x = 3
ANSWER : The area bounded by the curves, y = x^{2} + 2, y = x, x = 0, and x = 3, is represented by the shaded area OCBAO as
Then, Area OCBAO = Area ODBAO – Area ODCO
Question 17: Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).
ANSWER : BL and CM are drawn perpendicular to xaxis.
It can be observed in the following figure that,
Area (ΔACB) = Area (ALBA) Area (BLMCB) – Area (AMCA) … (1)
Equation of line segment AB is
Equation of line segment BC is
Equation of line segment AC is
Therefore, from equation (1), we obtain
Area (ΔABC) = (3 + 5 – 4) = 4 units
Question 18: Using integration find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.
ANSWER : 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 (ΔACB) = Area (OLBAO) –Area (OLCAO)
Question 19: Smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2 is
A. 2 (π – 2)
B. π – 2
C. 2π – 1
D. 2 (π 2)
ANSWER : The smaller area enclosed by the circle, x^{2} + y^{2} = 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 (ΔOAB)
Thus, the correct answer is B.
Question 20: Area lying between the curve y^{2} = 4x and y = 2x is
A.
B.
C.
D.
ANSWER : The area lying between the curve, y^{2} = 4x and y = 2x, is represented by the shaded area OBAO as
The points of intersection of these curves are O (0, 0) and A (1, 2).
We draw AC perpendicular to xaxis such that the coordinates of C are (1, 0).
∴ Area OBAO = Area (ΔOCA) – Area (OCABO)
Thus, the correct answer is B.
Question 21: Find the area under the given curves and given lines:
(i) y = x^{2}, x = 1, x = 2 and xaxis
(ii) y = x^{4}, x = 1, x = 5 and x –axis
ANSWER :  (i) The required area is represented by the shaded area ADCBA as
Question 22: Find the area between the curves y = x and y = x^{2}
ANSWER : The required area is represented by the shaded area OBAO as
The points of intersection of the curves, y = x and y = x^{2}, is A (1, 1).
We draw AC perpendicular to xaxis.
∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)
Question 23: Find the area of the region lying in the first quadrant and bounded by y = 4x^{2}, x = 0, y = 1 and y = 4
ANSWER : The area in the first quadrant bounded by y = 4x^{2}, x = 0, y = 1, and y = 4 is represented by the shaded area ABCDA as
Question 24: Sketch the graph of and evaluate
ANSWER : The given equation is
The corresponding values of x and y are given in the following table.
On plotting these points, we obtain the graph of as follows.
It is known that,
Question 25: Find the area bounded by the curve y = sin x between x = 0 and x = 2π
ANSWER : The graph of y = sin x can be drawn as
∴ Required area = Area OABO Area BCDB
Question 26: Find the area enclosed between the parabola y^{2} = 4ax and the line y = mx
ANSWER : The area enclosed between the parabola, y^{2} = 4ax, and the line, y = mx, is represented by the shaded area OABO as
The points of intersection of both the curves are (0, 0) and .
We draw AC perpendicular to xaxis.
∴ Area OABO = Area OCABO – Area (ΔOCA)
Question 27: Find the area enclosed by the parabola 4y = 3x^{2} and the line 2y = 3x + 12
ANSWER : The area enclosed between the parabola, 4y = 3x^{2}, 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 xaxis.
∴ Area OBAO = Area CDBA – (Area ODBO Area OACO)
Question 28:Find the area of the smaller region bounded by the ellipse and the line
ANSWER : 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)
Question 29:Find the area of the smaller region bounded by the ellipse & the line
ANSWER : 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)
Question 30: Find the area of the region enclosed by the parabola x^{2} = y, the line y = x + 2 and xaxis
ANSWER : The area of the region enclosed by the parabola, x^{2} = y, the line, y = x + 2, and xaxis is represented by the shaded region OABCO as
The point of intersection of the parabola, x^{2} = y, and the line, y = x + 2, is A (–1, 1).
∴ Area OABCO = Area (BCA) Area COAC
Question 31: 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 : 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 xaxis and yaxis.
∴ Area ADCB = 4 × Area OBAO
Question 32: Find the area bounded by curves
ANSWER : The area bounded by the curves, , is represented by the shaded region as
It can be observed that the required area is symmetrical about yaxis.
Question 33: 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 : The vertices of ΔABC are A (2, 0), B (4, 5), and C (6, 3).
Equation of line segment AB is
Equation of line segment BC is
Equation of line segment CA is
Area (ΔABC) = Area (ABLA) Area (BLMCB) – Area (ACMA)
Question 34: 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 : The given equations of lines are
2x + y = 4 … (1)
3x – 2y = 6 … (2)
And, x – 3y + 5 = 0 … (3)
The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on xaxis.
Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)
Question 35: Find the area of the region
ANSWER : 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 xaxis.
∴ Area OABCO = 2 × Area OBC
Area OBCO = Area OMC Area MBC
Therefore, the required area is units
Question 36: Area bounded by the curve y = x^{3}, the xaxis and the ordinates x = –2 and x = 1 is
A. – 9
B.
C.
D.
ANSWER : 
Thus, the correct answer is B.
Question 37:The area bounded by the curve, xaxis and the ordinates x = –1 & x = 1 is given by [Hint: y = x^{2} if x > 0 and y = –x^{2} if x < 0]
A. 0
B.
C.
D.
ANSWER : 
Thus, the correct answer is C.
Question 38: The area of the circle x^{2} + y^{2} = 16 exterior to the parabola y^{2} = 6x is
A.
B.
C.
D.
ANSWER : The given equations are
x^{2} + y^{2} = 16 … (1)
y^{2} = 6x … (2)
Area bounded by the circle and parabola
Area of circle = π (r)^{2}
= π (4)^{2}
= 16π units
Thus, the correct answer is C.
Question 39: The area bounded by the yaxis, y = cos x and y = sin x when
A.
B.
C.
D.
ANSWER : The given equations are
y = cos x … (1)
And, y = sin x … (2)
Required area = Area (ABLA) area (OBLO)
Integrating by parts, we obtain
Thus, the correct answer is B.
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