NCERT Solutions for Class 12 Science Math Chapter 2 - Application Of Integrals

Answer:

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.

Answer:

The area of the region bounded by the curve, y² = 9x, x = 2, and x = 4, and the x-axis is the area ABCD.

Answer:

The area of the region bounded by the curve, x² = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.

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

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 =

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 = $\frac{\sqrt{3}}{2} + \frac{\mathrm{\pi }}{3} - \frac{\sqrt{3}}{2} = \frac{\mathrm{\pi }}{3} \mathrm{square} \mathrm{units}$

Answer:

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

Therefore, the area of smaller part of the circle, x² + y² = a², cut off by the line, , is units.

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 .

Answer:

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 ΔOAM – Area OMACO

Area of ΔOAM   

Area of OMACO

⇒ Area of OACO = Area of ΔOAM – Area of OMACO

Therefore, required area = units

Answer:

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

Similarly, Area OACO = Area OLAC – Area OLAO

Therefore, required area =

Answer:

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.

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.

Answer:

The area bounded by the curve, y² = 4x, y-axis, and y = 3 is represented as

Thus, the correct answer is B.

Answer:

The required area is represented by the shaded area OBCDO.

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
$={\int }_0^{\sqrt{2}}\sqrt{\frac{(9-4x^2)}{4}}dx-{\int }_0^{\sqrt{2}}\frac{x^2}{4}dx$
$={\int }_0^{\sqrt{2}}\sqrt{{\left(\frac{3}{2}\right)}^2-x^2}dx-\frac{1}{4}{\int }_0^{\sqrt{2}}{x}^{2}dx$
$={\overline{)\left(\frac{x}{2}\sqrt{{\left(\frac{3}{2}\right)}^2-x^2}+\frac{9}{8}{\sin }^{-1}\frac{2x}{3}\right)}}_0^{\sqrt{2}}-{\overline{)\frac{1}{4}\left(\frac{x^3}{3}\right)}}_0^{\sqrt{2}}$
$=\frac{\sqrt{2}}{4}+\frac{9}{8}{\sin }^{-1}\frac{2\sqrt{2}}{3}-\frac{1}{12}{\left(\sqrt{2}\right)}^3$
$=\frac{1}{2\sqrt{2}}+\frac{9}{8}{\sin }^{-1}\frac{2\sqrt{2}}{3}-\frac{1}{3\sqrt{2}}$
$=\frac{1}{6\sqrt{2}}+\frac{9}{8}{\sin }^{-1}\frac{2\sqrt{2}}{3}$
$=\frac{1}{2}\left[\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}\right]$

Therefore, the required area OBCDO is units

Answer:

The area bounded by the curves, (x – 1)² + y² = 1 and x² + y ² = 1, is represented by the shaded area as

On solving the equations, (x – 1)² + y² = 1 and x² + 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

Answer:

The area bounded by the curves, y = x² + 2, y = x, x = 0, and x = 3, is represented by the shaded area OCBAO as

Then, Area OCBAO = Area ODBAO – Area ODCO

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

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)

Answer:

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 (ΔOAB)

Thus, the correct answer is B.

Answer:

The area lying between the curve, y² = 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.

Answer:

1. The required area is represented by the shaded area ADCBA as

2. The required area is represented by the shaded area ADCBA as

Answer:

The required area is represented by the shaded area OBAO as

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 (ΔOCA) – Area (OCABO) … (1)

Answer:

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

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{ABCDA} = {\int }_1^4 x dy \\ ={\int }_1^4 \frac{\sqrt{y}}{2} dy \left[\mathrm{as}, y = 4x^2\right] \\ =\frac{1}{2}{\int }_1^4\sqrt{y} dy \\ =\frac{1}{2}\times \frac{2}{3}{\left[y^{3/2}\right]}_1^4 \\ =\frac{1}{3}\left[{\left(4\right)}^{3/2} - {\left(1\right)}^{3/2}\right] \\ =\frac{1}{3}\left(8 - 1\right) \\ =\frac{1}{3}\times 7 \\ =\frac{7}{3} \mathrm{square} \mathrm{units} \end{gathered}$$

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,

Answer:

The graph of y = sin x can be drawn as

∴ Required area = Area OABO + Area BCDB

Answer:

The area enclosed between the parabola, y² = 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)

Answer:

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)

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)

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)

Answer:

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 OACO as

The point of intersection of the parabola, x² = y, and the line, y = x + 2, is A (–1, 1) and C(2, 4).

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{OACO} = {\int }_{-1}^2\left(x + 2\right) dx - {\int }_{-1}^2 x^2 dx \\ \Rightarrow \mathrm{Area} \mathrm{of} \mathrm{OACO} = {\left[\frac{x^2}{2} + 2x\right]}_{-1}^2 - \frac{1}{3}{\left[x^3\right]}_{-1}^2 \\ \Rightarrow \mathrm{Area} \mathrm{of} \mathrm{OACO} = \left[\left\{\frac{{\left(2\right)}^2}{2}+2\left(2\right)\right\} - \left\{\frac{{\left(-1\right)}^2}{2}+2\left(-1\right)\right\}\right] - \frac{1}{3}\left[{\left(2\right)}^3 - {\left(-1\right)}^3\right] \\ \Rightarrow \mathrm{Area} \mathrm{of} \mathrm{OACO} = \left[2 + 4 - \left(\frac{1}{2}-2\right)\right] - \frac{1}{3}\left(8 + 1\right) \\ \Rightarrow \mathrm{Area} \mathrm{of} \mathrm{OACO} = 6 + \frac{3}{2} - 3 \\ \Rightarrow \mathrm{Area} \mathrm{of} \mathrm{OACO} = 3 + \frac{3}{2} = \frac{9}{2} \mathrm{square} \mathrm{units} \end{gathered}$$

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

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.

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)

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)

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 sq.units

Answer:

Required Area = $\left|{\mathit{\int }}_{\mathit{-}\mathit{2}}^{\mathit{ }\mathit{0}}y\mathit{d}x\right|+{\int }_0^{1}y\mathit{d}x$

                       $$\begin{gathered} =\left|{\mathit{\int }}_{\mathit{-}\mathit{2}}^{\mathit{ }\mathit{0}}{x}^3\mathit{d}x\right|+{\int }_0^1{x}^3\mathit{d}x \\ =\left|{\left[\frac{x^4}{4}\right]}_{-2}^0\right|+{\left[\frac{x^4}{4}\right]}_0^1 \\ =\left|\left[0-\frac{16}{4}\right]\right|+\left[\frac{1}{4}-0\right] \\ =\left|-4\right|+\frac{1}{4} \\ =4+\frac{1}{4} \\ =\frac{17}{4} sq. units \end{gathered}$$

Thus, the correct answer is D.

Answer:

Thus, the correct answer is C.

Answer:

The given equations are

x² + y² = 16       … (1)

y² = 6x               … (2)

Area bounded by the circle and parabola

$$\begin{gathered} =2\left[\mathrm{area}\left(\mathrm{OADO}\right)+\mathrm{area}\left(\mathrm{ADBA}\right)\right] \\ =2\left[{\int }_0^2\sqrt{6x} dx+{\int }_2^4\sqrt{16-x^2} dx\right] \\ =2{\int }_0^2\sqrt{6x} dx+2{\int }_2^4\sqrt{16-x^2} dx \\ =2\sqrt{6}{\int }_0^2\sqrt{x} dx+2{\int }_2^4\sqrt{16-x^2} dx \\ =2\sqrt{6}\times \frac{2}{3}{\left[x^{\frac{3}{2}}\right]}_0^2+2{\left[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\left(\frac{x}{4}\right)\right]}_2^4 \\ =\frac{4\sqrt{6}}{3}\left(2\sqrt{2}-0\right)+2\left[\left\{0+8{\sin }^{-1}\left(1\right)\right\}-\left\{2\sqrt{3}+8{\sin }^{-1}\left(\frac{1}{2}\right)\right\}\right] \\ =\frac{16\sqrt{3}}{3}+2\left[8\times \frac{\mathrm{\pi }}{2}-2\sqrt{3}-8\times \frac{\mathrm{\pi }}{6}\right] \\ =\frac{16\sqrt{3}}{3}+2\left(4\mathrm{\pi }-2\sqrt{3}-\frac{4\mathrm{\pi }}{3}\right) \\ =\frac{16\sqrt{3}}{3}+8\mathrm{\pi }-4\sqrt{3}-\frac{8\mathrm{\pi }}{3} \\ =\frac{16\sqrt{3}+24\mathrm{\pi }-4\sqrt{3}-8\mathrm{\pi }}{3} \\ =\frac{16\mathrm{\pi }+12\sqrt{3}}{3} \\ =\frac{4}{3}\left[4\mathrm{\pi }+\sqrt{3}\right] \mathrm{square} \mathrm{units} \end{gathered}$$

Area of circle = π (r)²

= π (4)²

= 16π square units

$$\begin{gathered} \therefore \mathrm{Required} \mathrm{area}=16\mathrm{\pi }-\frac{4}{3}\left(4\mathrm{\pi }+\sqrt{3}\right) \\ =16\mathrm{\pi }-\frac{16\mathrm{\pi }}{3}-\frac{4\sqrt{3}}{3} \\ =\frac{32\mathrm{\pi }}{3}-\frac{4\sqrt{3}}{3} \\ =\frac{4}{3}\left[8\mathrm{\pi }-\sqrt{3}\right] \mathrm{square} \mathrm{units} \end{gathered}$$

Thus, the correct answer is C.

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.