NCERT Solutions of class 12 maths miscellaneous exercise of chapter 8-Application of Integrals - Future Study Point

NCERT Solutions of class 12 maths miscellaneous exercise of chapter 8-Application of Integrals

MISC CH.8

NCERT Solutions of class 12 maths Miscellaneous exercise -Application of Integrals

MISC CH.8

The NCERT solutions of miscellaneous exercise of chapter 8-Application of Integrals are a summary of all the methods used in all the exercises of chapter 8-Application of Integrals, therefore in this respect, NCERT solutions of miscellaneous exercise are very important for students.All the questions of miscellaneous exercise are solved by the expert by a step-by-step method so each student will understand the solution clearly.

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NCERT Solutions of class 12 maths Miscellaneous exercise -Application of Integrals

Exercise 8.1-Application of Integration

Exercise 8.2-Application of Integration

Exercise 8.3-Application of Integration

NCERT Solutions of Maths for Class 12 Miscellaneous exercise -Application of Integrals

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Q1. Find the area under the given curve and given lines

(i) y = x² , x = 1, x =2 and x-axis

(ii) y = x4, x=1, x= 5 and x-axis

Ans.

(i) We are given the curve y = x²  and the lines  x = 1, x =2

q1 chapter 8 class 12 maths miscellaneous -applicatopn of integrals

 

 

 

 

 

 

 

 

 

Area covered by the curve,x-axis and the given lines = arABCD

(ii) We are given the curve y = x4, x=1, x= 5 and the lines  x = 1, x =2

Miscellaneous ex. chapter 8 mayhs class 12

 

 

 

 

 

 

 

 

Area covered by the curve, x-axis and the given lines = arABCD

Q2. Find the area between the curves y = x and y = x²

Ans. We are given the equations of curve y = x² and y = x

Putting y = x, in y = x²

x = x² ⇒ x= 0 and x =1,again putting x = 0 and x= 1 in one of the equations, we get y=0 and y=1

Therefore both of curve and the line intersect each other at origin (0,0) and (1,1)

Exercise 8.4 application of integrals

 

 

 

 

 

 

 

 

Area covered by the line and the curve , area OCB = ar ΔOAB – ar OABCO

ar OABCO = 1/3 sq.unit

OA = 1 unit and AB = 1 unit

Q3. Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y =1 and y = 4.

Ans. The given curve is y = 4x² and the lines are x = 0, y =1 and y = 4.

Class 12 chapter 8 miscaneous Q3

 

 

 

 

 

 

 

 

The area covered by the curve y = 4x² and the lines are x = 0(y-axis), y =1, and y = 4 is ar ABCD

From the equation of curve, y = 4x², we have   

 

Q4. Sketch the graph of      and evaluate   

rom the equation of curve, y = 4x², we have   

 

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Q4. Sketch the graph of      and evaluate   

Ans.Solutions of the given equation 

x-6-5-4-3-2-10
y3210123

The graph of the given equation is drawn as follows

Q4 miscellaneous

 

 

 

 

 

 

 

 

 

=  – 9 + 18 = 9

Q5. Find the area bounded by the curve y = sinx between x = 0 and x = 2π.

Ans.The graph of the curve y = sinx between x = 0 and x = 2π, is shown below

sinx

 

 

 

 

 

 

 

The area bounded by the curve y = sinx, x = 0 and x = 2π

=Area of OAB + area of BCD

= 2 + 2 = 4 sq.unit

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