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Miscellaneous Exercise - Chapter 8 Applications Of Integrals class 12 ncert solutions Maths - SaraNextGen [2024]


Question 1:

Find the area under the given curves and given lines:

(i) y = x2x = 1, x = 2 and x-axis

(ii) y = x4x = 1, x = 5 and x –axis

Answer:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8022/Chapter%208_html_m4ce42dbe1.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8022/Chapter%208_html_m7e5227a7.gif

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8022/Chapter%208_html_m4ad77a8d1.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8022/Chapter%208_html_30c596bd.gif

Question 2:

Find the area between the curves y = x and y = x2

Answer:

The required area is represented by the shaded area OBAO as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8024/Chapter%208_html_404ae4fa1.jpg

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)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8024/Chapter%208_html_m2d2081bd.gif

Question 3:

Find the area of the region lying in the first quadrant and bounded by y = 4x2x = 0, y = 1 and = 4

Answer:

The area in the first quadrant bounded by y = 4x2x = 0, y = 1, and = 4 is represented by the shaded area ABCDA as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8026/Chapter%208_html_2c0231111.jpg

 

Area of ABCDA = ∫14 x dy =∫14 y2 dy as, y = 4×2 =12∫14y dy

 =12×23y3/214

 =1343/2 – 13/2=138 – 1 =13×7=73 square units

Question 4:

Sketch the graph of https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8028/Chapter%208_html_38fd028a.gif and evaluatehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8028/Chapter%208_html_mfd541dc.gif

Answer:

The given equation is https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8028/Chapter%208_html_38fd028a.gif

The corresponding values of 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 https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8028/Chapter%208_html_38fd028a.gif  as follows.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8028/Chapter%208_html_m465503671.jpg

It is known that, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8028/Chapter%208_html_79eb9f8d.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8028/Chapter%208_html_m68a9de36.gif

Question 5:

Find the area bounded by the curve y = sin between x = 0 and x = 2π

Answer:

The graph of y = sin x can be drawn as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8029/Chapter%208_html_m732621b81.jpg

∴ Required area = Area OABO + Area BCDB

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8029/Chapter%208_html_425b6c24.gif

Question 6:

Find the area enclosed between the parabola y2 = 4ax and the line y mx

Answer:

The area enclosed between the parabola, y2 = 4ax, and the line, y mx, is represented by the shaded area OABO as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8031/Chapter%208_html_m4169bafd1.jpg

The points of intersection of both the curves are (0, 0) and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8031/Chapter%208_html_5687dc28.gif .

We draw AC perpendicular to x-axis.

∴ Area OABO = Area OCABO – Area (ΔOCA)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8031/Chapter%208_html_54356da2.gif

Question 7:

Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12

Answer:

The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 12, is represented by the shaded area OBAO as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8033/Chapter%208_html_m163540c91.jpg

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)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8033/Chapter%208_html_m562c66f1.gif

Question 8:

Find the area of the smaller region bounded by the ellipse https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8035/Chapter%208_html_m61425d53.gif and the line https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8035/Chapter%208_html_m520b1e80.gif

Answer:

The area of the smaller region bounded by the ellipse, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8035/Chapter%208_html_m61425d53.gif , and the line, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8035/Chapter%208_html_m520b1e80.gif , is represented by the shaded region BCAB as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8035/Chapter%208_html_m42a7cec2.jpg

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8035/Chapter%208_html_5d486ae4.gif

Question 9:

Find the area of the smaller region bounded by the ellipse https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8036/Chapter%208_html_m64d080ba.gif  and the line https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8036/Chapter%208_html_7f0bf713.gif

Answer:

The area of the smaller region bounded by the ellipse, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8036/Chapter%208_html_m64d080ba.gif , and the line, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8036/Chapter%208_html_7f0bf713.gif , is represented by the shaded region BCAB as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8036/Chapter%208_html_5ca67f6b.jpg

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8036/Chapter%208_html_m7a8d236e.gif

Question 10:

Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis

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

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/9923.png

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

Area of OACO = ∫-12x + 2 dx  –  ∫-12 x2 dx⇒Area of OACO = x22 + 2x-12 – 13×3-12⇒Area of OACO = 222+22 – -122+2-1 – 1323 – -13⇒Area of OACO = 2 + 4 – 12-2 – 138 + 1⇒Area of OACO = 6 + 32 – 3⇒Area of OACO = 3 + 32 = 92 square units

 

Question 11:

Using the method of integration find the area bounded by the curve https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8038/Chapter%208_html_m2a261334.gif

[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – = 11]

Answer:

The area bounded by the curve, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8038/Chapter%208_html_m2a261334.gif , is represented by the shaded region ADCB as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8038/Chapter%208_html_542ee9ce.jpg

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8038/Chapter%208_html_m4d783e9e.gif

Question 12:

Find the area bounded by curves https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8039/Chapter%208_html_m40cd809f.gif

Answer:

The area bounded by the curves, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8039/Chapter%208_html_m40cd809f.gif , is represented by the shaded region as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8039/Chapter%208_html_m666b3e21.jpg

It can be observed that the required area is symmetrical about y-axis.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8039/Chapter%208_html_m7e1414f1.gif

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).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8040/Chapter%208_html_56aece2a.jpg

Equation of line segment AB is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8040/Chapter%208_html_m4ab4b266.gif

Equation of line segment BC is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8040/Chapter%208_html_m18e3d5c1.gif

Equation of line segment CA is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8040/Chapter%208_html_m772ecb89.gif

Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8040/Chapter%208_html_190ffc4e.gif

Question 14:

Using the method of integration find the area of the region bounded by lines:

2x + y = 4, 3x – 2y = 6 and x – 3+ 5 = 0

Answer:

The given equations of lines are

2x + y = 4 … (1)

3x – 2y = 6 … (2)

And, x – 3+ 5 = 0 … (3)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8041/Chapter%208_html_m1271b377.jpg

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)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8041/Chapter%208_html_74af47c0.gif

Question 15:

Find the area of the region https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8042/Chapter%208_html_2fe78b05.gif

Answer:

The area bounded by the curves, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8042/Chapter%208_html_2fe78b05.gif , is represented as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8042/Chapter%208_html_7b683b2c.jpg

The points of intersection of both the curves arehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8042/Chapter%208_html_m5faa08aa.gif .

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8042/Chapter%208_html_m4cb17cda.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8042/Chapter%208_html_1b33a96b.gif

Therefore, the required area is https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8042/Chapter%208_html_56e204ba.gif  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. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8043/Chapter%208_html_1ece43cd.gif

C. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8043/Chapter%208_html_70049d87.gif

D. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8043/Chapter%208_html_cee7b9e.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8043/Chapter%208_html_m1c1637fc.jpg

Required Area =

∫-2 0ydx+∫01ydx

=∫-2 0x3dx+∫01x3dx=x44-20+x4401=0-164+14-0=-4+14=4+14=174 sq. unitsThus, the correct Answer is D.

Question 17:

The area bounded by the curvehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8044/Chapter%208_html_5beae877.gifx-axis and the ordinates x = –1 and x = 1 is given by

[Hint: y = x2 if x > 0 and y = –x2 if x < 0]

A. 0

B. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8044/Chapter%208_html_18a817bc.gif

C. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8044/Chapter%208_html_m64d8549d.gif

D. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8044/Chapter%208_html_37cb7ba8.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8044/Chapter%208_html_7d54987e.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8044/Chapter%208_html_1a3c9425.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8044/Chapter%208_html_158de1ed.gif

Thus, the correct Answer is C.

Question 18:

The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is

A. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8045/Chapter%208_html_m68cb08ac.gif

B. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8045/Chapter%208_html_2e193833.gif

C. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8045/Chapter%208_html_m773a9b81.gif

D. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8045/Chapter%208_html_2e193833.gif

Answer:

The given equations are

x2 + y2 = 16       … (1)

y2 = 6x               … (2)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8045/Chapter%208_html_338bfcf8.jpg

Area bounded by the circle and parabola

=2areaOADO+areaADBA=2∫026x dx+∫2416-x2 dx=2∫026x dx+2∫2416-x2 dx=26∫02x dx+2∫2416-x2 dx=26×23×3202+2×216-x2+162sin-1×424 =46322-0+20+8sin-11-23+8sin-112=1633+28×π2-23-8×π6=1633+24π-23-4π3=1633+8π-43-8π3=163+24π-43-8π3=16π+1233=434π+3 square units

Area of circle = π (r)2

= π (4)2

= 16π square units

∴ Required area=16π-434π+3=16π-16π3-433=32π3-433=438π-3 square units

Thus, the correct Answer is C.

Question 19:

The area bounded by the y-axis, y = cos x and y = sin x when https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8046/Chapter%208_html_m1d0eb305.gif

A. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8046/Chapter%208_html_m5be11f4.gif

B. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8046/Chapter%208_html_3627989f.gif

C. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8046/Chapter%208_html_m1f7d3eb7.gif

D. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8046/Chapter%208_html_m74e5c629.gif

Answer:

The given equations are

y = cos x … (1)

And, y = sin x … (2)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8046/Chapter%208_html_m64b27871.jpg

Required area = Area (ABLA) + area (OBLO)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8046/Chapter%208_html_m6bcaf60b.gif

Integrating by parts, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8046/Chapter%208_html_m11c3f3e7.gif

Thus, the correct Answer is B.

Also Read : Exercise-9.1-Chapter-9-Differential-Equations-class-12-ncert-solutions-Maths

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