NCERT Solutions for Applications of Integrals Exercise 8.2 Solutions

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Class 12 Maths NCERT Solutions Chapter 8 Applications of Integrals Exercise 8.2 is provided in the article. Class 12 Chapter 8 Applications of Integrals Exercises include questions on area between a curve, parabola, and ellipse. All questions under this chapter are solved using diagrams and an easy to understand method.

Download PDF: NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2

Also check other Exercise Solutions of Class 12 Maths Chapter 8 Applications of Integrals

p>Also Read:

Chapter 8 Applications of Integrals Important Topics
Trapezoidal Rule Formula	Area Under Curve Formula	Trapezoidal Rule
Applications of Integrals Important Questions	SImpsons Rule Formula	Surface Integral

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CBSE CLASS XII Related Questions

1.
Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).
View Solution
2.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :
- \( -6 \) unit/sec
- \( 0 \) unit/sec
- \( 18 \) unit/sec
- \( 6 \) unit/sec
View Solution
3.
Solve the differential equation \( y + \frac{d}{dx}(xy) = x(\sin x + \log x) \).
View Solution
4.
Find the sum of maximum and minimum values of \(f(x) = \sin x\,(1 + \cos x)\) in the interval \([0, \pi]\).
View Solution
5.
For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :
- \(\frac{1}{2}\)
- \(\frac{3}{5}\)
- \(\frac{1}{5}\)
- \(\frac{3}{10}\)
View Solution
6.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).
View Solution

View All

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NCERT Solutions for Applications of Integrals Exercise 8.2 Solutions

CBSE CLASS XII Related Questions

1.
Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).

2.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

3.
Solve the differential equation \( y + \frac{d}{dx}(xy) = x(\sin x + \log x) \).

4.
Find the sum of maximum and minimum values of \(f(x) = \sin x\,(1 + \cos x)\) in the interval \([0, \pi]\).

5.
For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :

6.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

Similar Mathematics Concepts

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NCERT Solutions for Applications of Integrals Exercise 8.2 Solutions

CBSE CLASS XII Related Questions

1.Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).

2.The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

3.Solve the differential equation \( y + \frac{d}{dx}(xy) = x(\sin x + \log x) \).

4.Find the sum of maximum and minimum values of \(f(x) = \sin x\,(1 + \cos x)\) in the interval \([0, \pi]\).

5.For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :

6.Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).

2.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

3.
Solve the differential equation \( y + \frac{d}{dx}(xy) = x(\sin x + \log x) \).

4.
Find the sum of maximum and minimum values of \(f(x) = \sin x\,(1 + \cos x)\) in the interval \([0, \pi]\).

5.
For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :

6.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).