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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 = x4, 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 = x4, 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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