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

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 = x ^{4}, x=1, x= 5 and x-axis**

Ans.

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

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

(ii) We are given the curve y = x^{4}, x=1, x= 5 and the lines x = 1, x =2

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)

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.

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 | -1 | 0 |

y | 3 | 2 | 1 | 0 | 1 | 2 | 3 |

The graph of the given equation is drawn as follows

**= – 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

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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