Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.1 | Set 1

Chapter 20 of RD Sharma's Class 12 textbook focuses on Definite Integrals a fundamental concept in calculus. This chapter introduces the concept of definite integrals including their properties, evaluation, and applications. Understanding definite integrals is crucial for solving problems involving the areas under curves and physical quantities in mathematics.

Definite Integrals

The Definite integrals are used to compute the exact area under a curve between the two specific points on the x-axis. They are defined as the limit of a Riemann sum as the number of the subdivisions approaches infinity. The value of a definite integral is obtained using the Fundamental Theorem of the Calculus which connects differentiation and integration. This provides a powerful tool for evaluating and interpreting the accumulated quantities.

Evaluate the following definite integrals:

Question 1. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 2[√9 - √4 ] 

I = 2 (3 − 2)

I = 2 (1)

I = 2

Therefore, the value of is 2.

Question 2. 

Solution:

We have,

I = 

I = 

I = log (3 + 7) − log (−2 + 7)

I = log 10 − log 5

I = 

I = log 2

Therefore, the value of is log 2.

Question 3. 

Solution:

We have,

I = 

Let x = sin t, so we have, 

=> dx = cos t dt

Now, the lower limit is,

=> x = 0 

=> sin t = 0

=> t = 0

Also, the upper limit is,

=> x = 1/2

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I = 

I = 

I = 

I = 

I = 

I =  π/6 - 0

I = π/6​

Therefore, the value of is π/6.

Question 4. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = π/4

Therefore, the value of is π/4.

Question 5. 

Solution:

We have,

I = 

Let x² + 1 = t, so we have,

=> 2x dx = dt

=> x dx = dt/2

Now, the lower limit is, x = 2

=> t = x² + 1

=> t = (2)² + 1

=> t = 4 + 1

=> t = 5

Also, the upper limit is, x = 3

=> t = x² + 1

=> t = (3)² + 1

=> t = 9 + 1

=> t = 10

So, the equation becomes,

I = 

I = 

I = 

I = 1/2[log10 - log5] 

I = 1/2[log10/5]

I = 1/2[log2]

I = log√2

Therefore, the value of is log√2.

Question 6. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 1/ab[tan^-1∞ - tan^-10] 

I = 1/ab[π/2 - 0]  

I = 1/ab[π/2]  

I = π/2ab   

Therefore, the value of is π/2ab.

Question 7. 

Solution:

We have,

I = 

I = 

I = [tan^-11 - tan^-1(-1)]  

I = [π/4 - (-π/4)]   

I = [π/4 + π/4]  

I = 2π/4 

I = π/2  

Therefore, the value of is π/2.

Question 8. 

Solution:

We have,

I = 

I = 

I = -e^-∞ - (-e⁰) 

I = − 0 + 1

I = 1

Therefore, the value of is 1.

Question 9. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = [1 − 0] − [log(1 + 1) − log(0 + 1)]

I = 1 − [log2 − log1]

I = 1 - log2/1 

I = 1 − log 2

I = log e − log 2

I = loge/2 

Therefore, the value of  is loge/2.

Question 10. 

Solution:

We have,

I = 

I = 

I = 

I = [-cosπ/2 + cos0] + [sinπ/2 - sin0] 

I = [−0 + 1] + 1

I = 1 + 1

I = 2

Therefore, the value of is 2.

Question 11. 

Solution:

We have,

I = 

I = 

I = log(sinπ/2) - log(sinπ/4)

I = log1 - log1/√2 

I = 

I = log√2  

Therefore, the value of is log√2.

Question 12. 

Solution:

We have,

I = 

I = 

I = log(secπ/4 + tanπ/4 - log(sec0 + tan0) 

I = log(√2 + 1) - log(1 + 0) 

I = 

I = log(√2 + 1) 

Therefore, the value of is log(√2 + 1).

Question 13. 

Solution:

We have,

I = 

I = 

I = [log|cosecπ/4 - cotπ/4|] - [log|cosecπ/6 - cotπ/6|]

I = [log|√2 - 1|] - [log|2 - √3|] 

I = 

Therefore, the value of is .

Question 14. 

Solution: 

We have, 

I = 

Let x = cos 2t, so we have,

=> dx = –2 sin 2t dt

Now, the lower limit is,

=> x = 0

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is,

=> x = 1

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I = 

I = 

I = 

I = 

I = 

Let cos t = z, so we have,

=> – sin t dt = dz

=> sin t dt = – dz

Now, the lower limit is,

=> t = 0

=> z = cos t

=> z = cos 0

=> z = 1

Also, the upper limit is,

=> t = π/4

=> z = cos t

=> z = cos π/4

=> z = 1/√2

So, the equation becomes,

I = 

I = 

I = 

I = 

I = 

I = 

I = -4[(log1/√2 - 1/2(2)) - (log1 - 1/2)]

I = -4[(log1/√2 - 1/4) - (0 - 1/2)]

I = -4[log1/√2 - 1/4 - 0 + 1/2]

I = -4[-log√2 + 1/4]

I = 4log√2 - 1 

I = 4 × 1/2log2 - 1

I = 2log2 - 1 

Therefore, the value of is 2log2 - 1.

Question 15. 

Solution: 

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = [tan π – tan0] – [sec π – sec 0]

I = [0 – 0] – [–1 – 1]

I = 0 – (–2)

I = 2

Therefore, the value of is 2.

Question 16. 

Solution: 

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = [tan π/4 – tan(–π/4)] – [sec π/4 – sec (–π/4)]

I = [1 – (–1)] – [sec π/4 – sec (π/4)]

I = 2 – 0

I = 2

Therefore, the value of is 2.

Question 17. 

Solution: 

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 1/2[π/2 - 0] + 1/4[sinπ - sin0]

I = 1/2[π/2] + 1/4[0 - 0]

I = π/4 

Therefore, the value of  is π/4.

Question 18. 

Solution: 

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 1/12 [-1 - 0] + 3/4[1 - 0]

I = 3/4 - 1/12

I = (9 - 1)/12

I = 8/12

I = 2/3

Therefore, the value of  is 2/3.

Question 19. 

Solution: 

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 1/6[sinπ/2 - sin0] + 1/2[sinπ/6 - sin0]

I = 1/6[1 - 0] + 1/2[1/2 - 0]

I = 1/6 + 1/4

I = (4 + 6)/24 

I = 10/24 

I = 5/12 

Therefore, the value of  is 5/12.

Question 20. 

Solution: 

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 1/2[sinπ/2 - sin0] - 1/6[sin3π/2 - sin0]

I = 1/2[1 - 0] - 1/6[-1 - 0]

I = 1/2 - 1/6(-1)

I = 1/2 + 1/6 

I = (6 + 2)/12

I = 8/12 

I = 2/3

Therefore, the value of  is 2/3.

Question 21. 

Solution: 

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 2[-cotπ/2 + cot2π/3]

I = 2[-1/√3 - 0]

I = -2/√3 

Therefore, the value of  is -2/√3.

Question 22. 

Solution: 

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 1/4[π/2 + π/4 + 0 + 0 - 0 - 0 - 0 - 0]

I = 1/4[3π/4] 

I = 3π/16 

Therefore, the value of is 3π/16.

Summary

Exercise 20.1 in RD Sharma's Class 12 textbook focuses on definite integrals. This section typically covers:

• The concept of definite integrals
• Properties of definite integrals
• Evaluation of definite integrals using fundamental theorems of calculus
• Integration techniques applied to definite integrals