NCERT Solutions for Class 12 Maths Chapter 8 – Application of Integrals Ex 8.1
Page No 365:
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 = 9x, x = 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)
Page No 366:
Question 3:
Find the area of the region bounded by x2 = 4y, y = 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 = 4y, y = 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)
![](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,
, can be represented as
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7878/Chapter%208_html_m1e3b9c76.gif)
![](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)
![](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)
![](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
and the circle ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7990/Chapter%208_html_m3c3d4c8c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7990/Chapter%208_html_7426353d.gif)
![](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,
, 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_7c9e45e6.gif)
![](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)
![](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_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)
![](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,
, is the area ABCDA.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7992/Chapter%208_html_74a06138.gif)
![](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,
, is
units.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7992/Chapter%208_html_74a06138.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7992/Chapter%208_html_m5df8a683.gif)
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 y = x2 and ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7997/Chapter%208_html_m5e6049d1.gif)
![](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,
, can be represented as
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7997/Chapter%208_html_m5e6049d1.gif)
![](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, y = 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/Chapter8_html_7ae9a240_3743494708373738394.png)
![](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)
![](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 =
units
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/7997/Chapter%208_html_99d7320.gif)
Question 10:
Find the area bounded by the curve x2 = 4y and the line x = 4y – 2
Answer:
The area bounded by the curve, x2 = 4y, and line, x = 4y – 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)
![](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
units.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8001/Chapter%208_html_m55261ceb.gif)
Question 12:
Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is
A. π
B. ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/237/8003/Chapter%208_html_m4c4df2e.gif)
![](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)
![](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)
![](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 = 4x, y-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)
![](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)
![](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)
![](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 = 4x, y-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.