# NCERT Solutions for Miscellaneous Exercise Chapter 8 Class 12 - Application of Integrals

NCERT solutions for Class 12 Maths chapter 8 Miscellaneous exercise deals with the area finding of the various curves and intersection of two curves. Class 12 Maths chapter 8 miscellaneous exercise is a good source to practice good questions before the exam. NCERT book Class 12 Maths chapter 8 miscellaneous exercise is relatively easier than the chapter 7 exercise as it provides questions based on the application part. Class 12 Maths chapter 8 miscellaneous exercise solutions with other two exercises are sufficient to score good marks. Also NCERT Exemplar questions can be referred for wide understanding of the topic.

Application of Integrals Exercise 8.1

Application of Integrals Exercise 8.2

**Application of Integrals**** Class 12 Chapter 8**** Miscellaneous: Exercise**** **

** Question:1 ** Find the area under the given curves and given lines:

(i) and -axis

** Answer: **

The area bounded by the curve and -axis

The area of the required region = area of ABCD

Hence the area of shaded region is 7/3 units

** Question:1 ** Find the area under the given curves and given lines:

(ii) and -axis

** Answer: **

The area bounded by the curev and -axis

The area of the required region = area of ABCD

Hence the area of the shaded region is 624.8 units

** Question:2 ** Find the area between the curves and .

** Answer: **

the area between the curves and .

The curves intersect at A(1,1)

Draw a normal to AC to OC(x-axis)

therefore, the required area (OBAO)= area of (OCAO) - area of (OCABO)

Thus the area of shaded region is 1/6 units

** Question:3 ** Find the area of the region lying in the first quadrant and bounded by and .

** Answer: **

the area of the region lying in the first quadrant and bounded by and .

The required area (ABCD) =

The area of the shaded region is 7/3 units

** Question:4 ** Sketch the graph of and evaluate

** Answer: **

y=|x+3|

the given modulus function can be written as

x+3>0

x>-3

for x>-3

y=|x+3|=x+3

x+3<0

x<-3

For x<-3

y=|x+3|=-(x+3)

Integral to be evaluated is

** Question:5 ** Find the area bounded by the curve between and .

** Answer: **

The graph of y=sinx is as follows

We need to find the area of the shaded region

ar(OAB)+ar(BCD)

=2ar(OAB)

The bounded area is 4 units.

** Question:6 ** Find the area enclosed between the parabola and the line .

** Answer: **

We have to find the area of the shaded region OBA

The curves y=mx and y ^{ 2 } =4ax intersect at the following points

The required area is

** Question:7 ** Find the area enclosed by the parabola and the line .

** Answer: **

We have to find the area of the shaded region COB

The two curves intersect at points (2,3) and (4,12)

Required area is

** Question:8 ** Find the area of the smaller region bounded by the ellipse and the line .

** Answer: **

We have to find the area of the shaded region

The given ellipse and the given line intersect at following points

Since the shaded region lies above x axis we take y to be positive

The required area is

** Question:9 ** Find the area of the smaller region bounded by the ellipse and the line .

** Answer: **

The area of the shaded region ACB is to be found

The given ellipse and the line intersect at following points

Y will always be positive since the shaded region lies above x axis

The required area is

** Question:10 ** Find the area of the region enclosed by the parabola the line and the -axis.

** Answer: **

We have to find the area of the shaded region BAOB

O is(0,0)

The line and the parabola intersect in the second quadrant at (-1,1)

The line y=x+2 intersects the x axis at (-2,0)

The area of the region enclosed by the parabola the line and the -axis is 5/6 units.

** Question:11 ** Using the method of integration find the area bounded by the curve

[ ** Hint: ** The required region is bounded by lines and ]

** Answer: **

We need to find the area of the shaded region ABCD

ar(ABCD)=4ar(AOB)

Coordinates of points A and B are (0,1) and (1,0)

Equation of line through A and B is y=1-x

The area bounded by the curve is 2 units.

** Question:12 ** Find the area bounded by curves .

** Answer: **

We have to find the area of the shaded region

In the first quadrant

y=|x|=x

Area of the shaded region=2ar(OADO)

The area bounded by the curves is 1/3 units.

** Question:13 ** Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are

** Answer: **

Equation of line joining A and B is

Equation of line joining B and C is

Equation of line joining A and C is

ar(ABC)=ar(ABL)+ar(LBCM)-ar(ACM)

ar(ABC)=8+5-6=7

Therefore the area of the triangle ABC is 7 units.

** Question:14 ** Using the method of integration find the area of the region bounded by lines:

** Answer: **

We have to find the area of the shaded region ABC

ar(ABC)=ar(ACLM)-ar(ALB)-ar(BMC)

The lines intersect at points (1,2), (4,3) and (2,0)

Area of the region bounded by the lines is 3.5 units

** Question:15 ** Find the area of the region .

** Answer: **

We have to find the area of the shaded region OCBAO

Ar(OCBAO)=2ar(OCBO)

For the fist quadrant

In the first quadrant, the curves intersect at a point

Area of the unshaded region in the first quadrant is

The total area of the shaded region is-

= Area of half circle - area of the shaded region in the first quadrant

** Question:16 ** Choose the correct answer.

** ** Area bounded by the curve , the -axis and the ordinates and is

(A) (B) (C) (D)

** Answer: **

Hence the required area

** Therefore the correct answer is B. **

** Question:17 ** Choose the correct answer.

** T ** he area bounded by the curve , -axis and the ordinates and is given by

(A) (B) (C) (D)

[ ** Hint : ** if and if . ]

** Answer: **

The required area is

** Question:18 ** Choose the correct answer.

The area of the circle exterior to the parabola is

(A) (B) (C) (D)

** Answer: **

The area of the shaded region is to be found.

Required area =ar(DOC)+ar(DOA)

The region to the left of the y-axis is half of the circle with radius 4 units and centre origin.

Area of the shaded region to the left of y axis is ar(1) =

For the region to the right of y-axis and above x axis

The parabola and the circle in the first quadrant intersect at point

Remaining area is 2ar(2) is

Total area of shaded region is

** Question:19 ** Choose the correct answer ** ** The area bounded by the -axis, and when is

(A) (B) (C) (D)

** Answer: **

Given : and

Area of shaded region = area of BCDB + are of ADCA

Hence, the correct answer is B.

**More About NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise**

Miscellaneous exercise chapter 8 Class 12 chapter application of integrals is quite interesting and important from exam perspective. Class 12 Maths chapter 8 miscellaneous solutions mainly deals with application based questions like area finding between two curves etc. NCERT solutions for Class 12 Maths chapter 8 miscellaneous exercise will not take much time to solve provided chapter 7 integrals are done well before chapter 8.

**Also Read| **Application of Integrals Class 12 Notes

**Benefits of Ncert Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercises**

The Class 12th Maths chapter 8 exercise has some good questions which must be done before examination.

NCERT syllabus Class 12 Maths chapter 8 miscellaneous exercises can be done along with chapter 7 as basic concepts used are the same.

These Class 12 Maths chapter 8 miscellaneous exercises are also useful in Physics and chemistry problems.

**Also see-**

NCERT Exemplar Solutions Class 12 Maths Chapter 8

NCERT Solutions for Class 12 Maths Chapter 8

**NCERT Solutions Subject Wise**

NCERT Solutions Class 12 Chemistry

NCERT Solutions for Class 12 Physics

NCERT Solutions for Class 12 Biology

NCERT Solutions for Class 12 Mathematics

**Subject Wise NCERT Exemplar Solutions**

NCERT Exemplar Class 12 Maths

NCERT Exemplar Class 12 Physics

NCERT Exemplar Class 12 Chemistry

NCERT Exemplar Class 12 Biology

Happy learning!!!