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Exercise 8.1 - Chapter 8 Applications Of Integrals class 12 ncert solutions Maths - SaraNextGen [2024]


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:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7870/Chapter%208_html_1df1fb491.jpg

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.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7870/Chapter%208_html_m7ba02c3f.gif

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:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7872/Chapter%208_html_m3eb074131.jpg

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7872/Chapter%208_html_214ceb8d.gif

Question 3:

Find the area of the region bounded by x2 = 4yy = 2, y = 4 and the y-axis in the first quadrant.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7875/Chapter%208_html_m34460b581.jpg

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7875/Chapter%208_html_m1658b8f8.gif

Question 4:

Find the area of the region bounded by the ellipse https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7878/Chapter%208_html_m1e3b9c76.gif

Answer:

The given equation of the ellipse, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7878/Chapter%208_html_m1e3b9c76.gif , can be represented as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7878/Chapter%208_html_m395b2d61.jpg

It can be observed that the ellipse is symmetrical about x-axis and y-axis.

∴ Area bounded by ellipse = 4 × Area of OAB

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7878/Chapter%208_html_m58c77053.gif

Therefore, area bounded by the ellipse = 4 × 3π = 12π units

Question 5:

Find the area of the region bounded by the ellipse https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7987/Chapter%208_html_4835ed99.gif

Answer:

The given equation of the ellipse can be represented as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7987/Chapter%208_html_m35b62b981.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7987/Chapter%208_html_3f2641f4.gif

It can be observed that the ellipse is symmetrical about x-axis and y-axis.

∴ Area bounded by ellipse = 4 × Area OAB

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7987/Chapter%208_html_m604372c1.gif

Therefore, area bounded by the ellipse = https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7987/Chapter%208_html_m12f1f12b.gif

Question 6:

Find the area of the region in the first quadrant enclosed by x-axis, line https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7990/Chapter%208_html_7426353d.gif and the circle https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7990/Chapter%208_html_m3c3d4c8c.gif

Answer:

The area of the region bounded by the circle, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7990/Chapter%208_html_7c9e45e6.gif , and the x-axis is the area OAB.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7990/Chapter%208_html_4c03911.jpg

The point of intersection of the line and the circle in the first quadrant is https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7990/Chapter%208_html_3026407d.gif .

Area OAB = Area ΔOCA + Area ACB

Area of OAC https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7990/Chapter%208_html_66e4f0ff.gif

Area of ABC https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7990/Chapter%208_html_m5bc50ca2.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7990/Chapter%208_html_b768c8f.gif

Therefore, required area enclosed =

32 + π3 – 32 = π3 square units

Question 7:

Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7992/Chapter%208_html_74a06138.gif

Answer:

The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7992/Chapter%208_html_74a06138.gif , is the area ABCDA.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7992/Chapter%208_html_m3e61e30e1.jpg

It can be observed that the area ABCD is symmetrical about x-axis.

∴ Area ABCD = 2 × Area ABC

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7992/Chapter%208_html_m5c3d19e8.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7992/Chapter%208_html_4b049957.gif

Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7992/Chapter%208_html_74a06138.gif , is https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7992/Chapter%208_html_m5df8a683.gif  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 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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7995/Chapter%208_html_5a0529a1.jpg

It can be observed that the given area is symmetrical about x-axis.

⇒ Area OED = Area EFCD

https://img-nm.mnimgs.com/img/study_content/editlive_ncert/77/2012_02_15_17_52_44/mathmlequation2248380751147929950.png

 

From (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7995/Chapter%208_html_m2aa1a2fa.gif

Therefore, the value of a is https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7995/Chapter%208_html_7fc3b5b5.gif .

Question 9:

Find the area of the region bounded by the parabola x2 and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7997/Chapter%208_html_m5e6049d1.gif

Answer:

The area bounded by the parabola, x2 = y,and the line,https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7997/Chapter%208_html_m5e6049d1.gif , can be represented as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7997/Chapter%208_html_m4c527e161.jpg

The given area is symmetrical about y-axis.

∴ Area OACO = Area ODBO

The point of intersection of parabola, x2 = y, and line, x, is A (1, 1).

Area of OACO = Area ΔOAM – Area OMACO

Area of ΔOAM https://img-nm.mnimgs.com/img/study_content/editlive_ncert/58/2013_03_25_11_57_06/mathmlequation7793862169195928565.png   https://img-nm.mnimgs.com/img/study_content/editlive_ncert/58/2013_03_25_11_57_06/Chapter8_html_7ae9a240_3743494708373738394.png

Area of OMACO https://img-nm.mnimgs.com/img/study_content/editlive_ncert/58/2013_03_25_11_54_03/Chapter8_html_7ae9a240_4422415849257538710.png

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7997/Chapter%208_html_m28f6abc6.gif

Therefore, required area = https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7997/Chapter%208_html_99d7320.gif units

Question 10:

Find the area bounded by the curve x2 = 4y and the line x = 4– 2

Answer:

The area bounded by the curve, x2 = 4y, and line, x = 4– 2, is represented by the shaded area OBAO.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7999/Chapter%208_html_m732412a01.jpg

Let A and B be the points of intersection of the line and parabola.

Coordinates of point https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7999/Chapter%208_html_madb2133.gif .

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7999/Chapter%208_html_684de512.gif

Similarly, Area OACO = Area OLAC – Area OLAO

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7999/Chapter%208_html_400a11a2.gif

Therefore, required area = https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7999/Chapter%208_html_m485eb8da.gif

Question 11:

Find the area of the region bounded by the curve y2 = 4x and the line x = 3

Answer:

The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8001/Chapter%208_html_m3ed1eb9a1.jpg

The area OACO is symmetrical about x-axis.

∴ Area of OACO = 2 (Area of OAB)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8001/Chapter%208_html_m232a2eb1.gif

Therefore, the required area is https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8001/Chapter%208_html_m55261ceb.gif units.

Question 12:

Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and = 2 is

A. π

B. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8003/Chapter%208_html_m4c4df2e.gif

C. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8003/Chapter%208_html_m4e8d241e.gif

D. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8003/Chapter%208_html_m6201c3cd.gif

Answer:

The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8003/Chapter%208_html_m649c57df1.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8003/Chapter%208_html_6a728a22.gif

Thus, the correct Answer is A.

Question 13:

Area of the region bounded by the curve y2 = 4xy-axis and the line y = 3 is

A. 2

B. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8004/Chapter%208_html_3fb85c49.gif

C. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8004/Chapter%208_html_m7a4cb6cb.gif

D. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8004/Chapter%208_html_597d40a8.gif

Answer:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8004/Chapter%208_html_2f79e71b1.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8004/Chapter%208_html_29709a98.gif

Thus, the correct Answer is B.

Also Read : Exercise-8.2-Chapter-8-Applications-Of-Integrals-class-12-ncert-solutions-Maths

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