Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.3 | Set 1

Evaluate the following integrals:

Question 1(i). 

Solution:

We have,

I = 

I = 

Using additive property, we get

I = 

I = 

I = 8 + 6 – 2 – 3 + 24 + 20 – 6 – 10

I = 37

Question 1(ii). 

Solution:

We have,

I = 

I = 

Using additive property, we get

I = 

I = 

I = 

I = 0 + 1 + 3 – π/2 + e⁶ – e⁰

I = 3 – π/2 + e⁶

Question 1(iii). 

Solution:

We have,

I = 

I = 

Using additive property, we get

I = 

I = 

I = 

I = 63/2 + 9 – 7/2 – 3 + 64 – 36

I = 56/2 + 34

I = 62

Question 2. 

Solution:

We have,

I = 

We know that, 

So, we get

I = 

I = 

I = 

I = –2 + 4 – 8 – 8 + 8 + 8 – 2 + 4

I = 20

Question 3. 

Solution:

We have,

I = 

We know that, 

So, we get

I = 

I = 

I = 0 + 2 + 8 – 0

I = 10

Question 4. 

Solution:

We have,

I = 

We know that,

So, we get

I = 

I = 

I = –1/4 + 1/2 + 1 – 1 + 1 + 1 – 1/4 + 1/2

I = 5/2

Question 5. 

Solution:

We have,

I = 

We know that,

So, we get

I = 

I = 

I = –9/4 + 9/2 + 4 – 6 + 4 + 6 – 9/4 + 9/2

I = 25/2

Question 6. 

Solution:

We have,

I = 

We know that,

So we get,

I = 

I = 

I = 

I = 1/3 – 3/2 + 2 – [8/3 – 6 + 4 – 1/3 + 3/2 – 2]

I = 1/3 – 3/2 + 2 – 8/3 + 6 – 2 + 1/3 – 3/2

I = 1

Question 7. 

Solution:

We have,

I = 

We know that,

So we get,

I = 

I = 

I = –1/6 + 1/3 – 0 + 27/2 + 3 – 1/6 – 1/3

I = 65/6

Question 8. 

Solution:

We have,

I = 

We know that,

So, we get 

I = 

I = 

I = 

I = –2 + 4 + 18 - 12 + 18 + 12 – 2 + 4

I = 40

Question 9. 

Solution:

We have,

I = 

We know that,

So we get,

I = 

I = 

I = 

I = -1/2 + 1 + 2 - 2 + 2 + 2 - 1/2 + 1

I = 5

Question 10. 

Solution:

We have,

I = 

We know that,

So we get,

I = 

I = 

I = 

I = - 2 - 6 + 1/2 + 3

I = - 5 + 1/2

I = (-10 + 1)/2

I = -9/2

Question 11. 

Solution:

We have,

I = 

We know that,

So we get,

I = 

I = 

I = 

I = 1/2 - 0 - 0 + 1/2

I = 1

Question 12. 

Solution:

We have,

I = 

We know that,

So we get,

I = 

I = 

I = 

I = 1 + 1 + 1 – (–1)

I = 1 + 1 + 1 + 1

I = 4

Question 13. 

Solution:

We have,

I = 

We know that,

So we get,

I = 

I = 

I = 

I = 1 - 1/√2 - 1/√2 + 1

I = 2 - 2/√2

I = 2 – √2

Question 14. 

Solution:

We have,

I = 

We know that,

So we get,

I = 

I = 

I = 

I = - 25/2 + 25 + 2 - 10 + 32 - 40 - 25/2 + 25

I = 9

Summary

This exercise usually covers:

• Evaluation of definite integrals using fundamental theorem of calculus
• Application of integration techniques to definite integrals
• Properties of definite integrals
• Integrals involving algebraic and trigonometric functions
• Evaluation of symmetric integrals