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Page No 365:
Question 1:
Find the area of the region bounded by the curve y^{2} = x and the lines x = 1, x = 4 and the x-axis.
Answer:
The area of the region bounded by the curve, y^{2} = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.
Page No 365:
Question 2:
Find the area of the region bounded by y^{2} = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Answer:
The area of the region bounded by the curve, y^{2} = 9x, x = 2, and x = 4, and the x-axis is the area ABCD.
Page No 366:
Question 3:
Find the area of the region bounded by x^{2} = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Answer:
The area of the region bounded by the curve, x^{2} = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.
Page No 366:
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 x-axis and y-axis.
∴ Area bounded by ellipse = 4 × Area of OAB
Therefore, area bounded by the ellipse = 4 × 3π = 12π units
Page No 366:
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 x-axis and y-axis.
∴ Area bounded by ellipse = 4 × Area OAB
Therefore, area bounded by the ellipse =
Page No 366:
Question 6:
Find the area of the region in the first quadrant enclosed by x-axis, line and the circle
Answer:
The area of the region bounded by the circle, , and the x-axis 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, required area enclosed = $\frac{\sqrt{3}}{2}+\frac{\mathrm{\pi}}{3}-\frac{\sqrt{3}}{2}=\frac{\mathrm{\pi}}{3}\mathrm{square}\mathrm{units}$
Page No 366:
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 x-axis.
∴ 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.
Page No 366:
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 x-axis.
⇒ Area OED = Area EFCD
From (1) and (2), we obtain
Therefore, the value of a is .
Page No 366:
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 y-axis.
∴ 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 ΔOAM – Area OMACO
Area of ΔOAM
Area of OMACO
⇒ Area of OACO = Area of ΔOAM – Area of OMACO
Therefore, required area = units
Page No 366:
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 x-axis.
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 =
Page No 366:
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 x-axis.
∴ Area of OACO = 2 (Area of OAB)
Therefore, the required area is units.
Page No 366:
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.
Page No 366:
Question 13:
Area of the region bounded by the curve y^{2} = 4x, y-axis and the line y = 3 is
A. 2
B.
C.
D.
Answer:
The area bounded by the curve, y^{2} = 4x, y-axis, and y = 3 is represented as
Thus, the correct answer is B.
Page No 371:
Question 1:
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 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
$={\int}_{0}^{\sqrt{2}}\sqrt{\frac{(9-4{x}^{2})}{4}}dx-{\int}_{0}^{\sqrt{2}}\frac{{x}^{2}}{4}dx$
$={\int}_{0}^{\sqrt{2}}\sqrt{{\left(\frac{3}{2}\right)}^{2}-{x}^{2}}dx-\frac{1}{4}{\int}_{0}^{\sqrt{2}}{x}^{2}dx$
$={\overline{)\left(\frac{x}{2}\sqrt{{\left(\frac{3}{2}\right)}^{2}-{x}^{2}}+\frac{9}{8}{\mathrm{sin}}^{-1}\frac{2x}{3}\right)}}_{0}^{\sqrt{2}}-{\overline{)\frac{1}{4}\left(\frac{{x}^{3}}{3}\right)}}_{0}^{\sqrt{2}}$
$=\frac{\sqrt{2}}{4}+\frac{9}{8}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}-\frac{1}{12}{\left(\sqrt{2}\right)}^{3}$
$=\frac{1}{2\sqrt{2}}+\frac{9}{8}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}-\frac{1}{3\sqrt{2}}$
$=\frac{1}{6\sqrt{2}}+\frac{9}{8}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}$
$=\frac{1}{2}\left[\frac{\sqrt{2}}{6}+\frac{9}{4}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}\right]$
Therefore, the required area OBCDO is units
Page No 371:
Question 2:
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 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 .
Therefore, required area OBCAO = units
Page No 371:
Question 3:
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
Page No 371:
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:
BL and CM are drawn perpendicular to x-axis.
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
Page No 371:
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:
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)
Page No 372:
Question 6:
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.
Page No 372:
Question 7:
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 x-axis such that the coordinates of C are (1, 0).
∴ Area OBAO = Area (OCABO) – Area (ΔOCA)
square units
Thus, the correct answer is B.
Page No 375:
Question 1:
Find the area under the given curves and given lines:
(i) y = x^{2}, x = 1, x = 2 and x-axis
(ii) y = x^{4}, x = 1, x = 5 and x –axis
Answer:
The required area is represented by the shaded area ADCBA as
The required area is represented by the shaded area ADCBA as
Page No 375:
Question 2:
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 x-axis.
∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)
Page No 375:
Question 3:
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
$\mathrm{Area}\mathrm{of}\mathrm{ABCDA}={\int}_{1}^{4}xdy\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}={\int}_{1}^{4}\frac{\sqrt{y}}{2}dy\left[\mathrm{as},y=4{x}^{2}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\int}_{1}^{4}\sqrt{y}dy\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \frac{2}{3}{\left[{y}^{3/2}\right]}_{1}^{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[{\left(4\right)}^{3/2}-{\left(1\right)}^{3/2}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(8-1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\times 7\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{7}{3}\mathrm{square}\mathrm{units}\phantom{\rule{0ex}{0ex}}$
Page No 375:
Question 4:
Sketch the graph of and evaluate
Answer:
The given equation is
The corresponding values of x and y are given in the following table.
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,
Page No 375:
Question 5:
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
Page No 375:
Question 6:
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 x-axis.
∴ Area OABO = Area OCABO – Area (ΔOCA)
Page No 375:
Question 7:
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 x-axis.
∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)
Page No 375:
Question 8:
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)
Page No 375:
Question 9:
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)
Page No 375:
Question 10:
Find the area of the region enclosed by the parabola x^{2} = y, the line y = x + 2 and x-axis
Answer:
The area of the region enclosed by the parabola, x^{2} = y, the line, y = x + 2, and x-axis is represented by the shaded region OACO as
The point of intersection of the parabola, x^{2} = y, and the line, y = x + 2, is A (–1, 1) and C(2, 4).
$\mathrm{Area}\mathrm{of}\mathrm{OACO}={\int}_{-1}^{2}\left(x+2\right)dx-{\int}_{-1}^{2}{x}^{2}dx\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{OACO}={\left[\frac{{x}^{2}}{2}+2x\right]}_{-1}^{2}-\frac{1}{3}{\left[{x}^{3}\right]}_{-1}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{OACO}=\left[\left\{\frac{{\left(2\right)}^{2}}{2}+2\left(2\right)\right\}-\left\{\frac{{\left(-1\right)}^{2}}{2}+2\left(-1\right)\right\}\right]-\frac{1}{3}\left[{\left(2\right)}^{3}-{\left(-1\right)}^{3}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{OACO}=\left[2+4-\left(\frac{1}{2}-2\right)\right]-\frac{1}{3}\left(8+1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{OACO}=6+\frac{3}{2}-3\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{OACO}=3+\frac{3}{2}=\frac{9}{2}\mathrm{square}\mathrm{units}$
Page No 375:
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:
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
Page No 376:
Question 12:
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 y-axis.
Page No 376:
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:
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)
Page No 376:
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:
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 x-axis.
Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)
Page No 376:
Question 15:
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 x-axis.
∴ Area OABCO = 2 × Area OBC
Area OBCO = Area OMC + Area MBC
Therefore, the required area is units
Page No 376:
Question 16:
Area bounded by the curve y = x^{3}, the x-axis and the ordinates x = –2 and x = 1 is
A. – 9
B.
C.
D.
Answer:
Required Area = $\left|{\mathit{\int}}_{\mathit{-}\mathit{2}}^{\mathit{}\mathit{0}}y\mathit{d}x\right|+{\int}_{0}^{1}y\mathit{d}x$
$=\left|{\mathit{\int}}_{\mathit{-}\mathit{2}}^{\mathit{}\mathit{0}}{x}^{3}\mathit{d}x\right|+{\int}_{0}^{1}{x}^{3}\mathit{d}x\phantom{\rule{0ex}{0ex}}=\left|{\left[\frac{{x}^{4}}{4}\right]}_{-2}^{0}\right|+{\left[\frac{{x}^{4}}{4}\right]}_{0}^{1}\phantom{\rule{0ex}{0ex}}=\left|\left[0-\frac{16}{4}\right]\right|+\left[\frac{1}{4}-0\right]\phantom{\rule{0ex}{0ex}}=\left|-4\right|+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=4+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=\frac{17}{4}sq.units$
Thus, the correct answer is D.
Page No 376:
Question 17:
The area bounded by the curve, x-axis and the ordinates x = –1 and 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.
Page No 376:
Question 18:
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
$=2\left[\mathrm{area}\left(\mathrm{OADO}\right)+\mathrm{area}\left(\mathrm{ADBA}\right)\right]\phantom{\rule{0ex}{0ex}}=2\left[{\int}_{0}^{2}\sqrt{6x}dx+{\int}_{2}^{4}\sqrt{16-{x}^{2}}dx\right]\phantom{\rule{0ex}{0ex}}=2{\int}_{0}^{2}\sqrt{6x}dx+2{\int}_{2}^{4}\sqrt{16-{x}^{2}}dx\phantom{\rule{0ex}{0ex}}=2\sqrt{6}{\int}_{0}^{2}\sqrt{x}dx+2{\int}_{2}^{4}\sqrt{16-{x}^{2}}dx\phantom{\rule{0ex}{0ex}}=2\sqrt{6}\times \frac{2}{3}{\left[{x}^{\frac{3}{2}}\right]}_{0}^{2}+2{\left[\frac{x}{2}\sqrt{16-{x}^{2}}+\frac{16}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{4}\right)\right]}_{2}^{4}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{6}}{3}\left(2\sqrt{2}-0\right)+2\left[\left\{0+8{\mathrm{sin}}^{-1}\left(1\right)\right\}-\left\{2\sqrt{3}+8{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)\right\}\right]\phantom{\rule{0ex}{0ex}}=\frac{16\sqrt{3}}{3}+2\left[8\times \frac{\mathrm{\pi}}{2}-2\sqrt{3}-8\times \frac{\mathrm{\pi}}{6}\right]\phantom{\rule{0ex}{0ex}}=\frac{16\sqrt{3}}{3}+2\left(4\mathrm{\pi}-2\sqrt{3}-\frac{4\mathrm{\pi}}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{16\sqrt{3}}{3}+8\mathrm{\pi}-4\sqrt{3}-\frac{8\mathrm{\pi}}{3}\phantom{\rule{0ex}{0ex}}=\frac{16\sqrt{3}+24\mathrm{\pi}-4\sqrt{3}-8\mathrm{\pi}}{3}\phantom{\rule{0ex}{0ex}}=\frac{16\mathrm{\pi}+12\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\left[4\mathrm{\pi}+\sqrt{3}\right]\mathrm{square}\mathrm{units}$
Area of circle = π (r)^{2}
= π (4)^{2}
= 16π square units
$\therefore \mathrm{Required}\mathrm{area}=16\mathrm{\pi}-\frac{4}{3}\left(4\mathrm{\pi}+\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=16\mathrm{\pi}-\frac{16\mathrm{\pi}}{3}-\frac{4\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}=\frac{32\mathrm{\pi}}{3}-\frac{4\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\left[8\mathrm{\pi}-\sqrt{3}\right]\mathrm{square}\mathrm{units}$
Thus, the correct answer is C.
Page No 376:
Question 19:
The area bounded by the y-axis, 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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