Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.5 | Set 1

Evaluate the following definite integrals as limits of sums:

Question 1. 

Solution:

We have,

I =

We know,, where h =

Here a = 0, b = 3 and f(x) = x + 4.

=> h = 3/n

=> nh = 3

So, we get,

I =

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

= 12 + 

=

Therefore, the value ofas limit of sum is.

Question 2. 

Solution:

We have,

I =

We know,

, where h =

Here a = 0, b = 2 and f(x) = x + 3.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

= 6 + 2

= 8

Therefore, the value ofas limit of sum is 8.

Question 3. 

Solution:

We have,

I =

We know,

, where h =

Here a = 1, b = 3 and f(x) = 3x − 2.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

= 2 + 6

= 8

Therefore, the value ofas limit of sum is 8.

Question 4. 

Solution:

We have,

I =

We know,

, where h =

Here a = −1, b = 1 and f(x) = x + 3.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

= 4 + 2

= 6

Therefore, the value ofas limit of sum is 6.

Question 5. 

Solution:

We have,

I =

We know,

, where h =

Here a = 0, b = 5 and f(x) = x + 1.

=> h = 5/n

=> nh = 5

So, we get,

I =

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

= 5 +

=

Therefore, the value ofas limit of sum is.

Question 6. 

Solution:

We have,

I =

We know,

, where

Here a = 1, b = 3 and f(x) = 2x + 3.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

= 10 + 4

= 14

Therefore, the value ofas limit of sum is 14.

Question 7. 

Solution:

We have,

I =

We know,

, where h =

Here a = 3, b = 5 and f(x) = 2 − x.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

= –2 – 2

= –4

Therefore, the value ofas limit of sum is –4.

Question 8. 

Solution:

We have,

I =

We know,

, where h =

Here a = 0, b = 2 and f(x) = x² + 1.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

=

=

=

Therefore, the value ofas limit of sum is.

Question 9. 

Solution:

We have,

I =

We know,

, where h =

Here a = 1, b = 2 and f(x) = x2.

=> h = 1/n

=> nh = 1

So, we get,

I =

=

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

= 1 + 1 +

= 1 + 1 +

=

Therefore, the value ofas limit of sum is.

Question 10. 

Solution:

We have,

I =

We know,

, where h =

Here a = 2, b = 3 and f(x) = 2x² + 1.

=> h = 1/n

=> nh = 1

So, we get,

I =

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

= 9 + 4 +

=

Therefore, the value ofas limit of sum is.

Question 11. 

Solution:

We have,

I =

We know,

, where h =

Here a = 1, b = 2 and f(x) = x² − 1.

=> h = 1/n

=> nh = 1

So, we get,

I =

=

=

=

Now if h −> 0, then n −> ∞. So, we have,

=

=

= 1 +

= 1 +

=

Therefore, the value of as limit of sum is .

Summary

This section likely focuses on:

• Properties of definite integrals
• Evaluation of definite integrals using fundamental theorems
• Application of substitution method in definite integrals
• Integration of trigonometric functions
• Solving definite integrals with absolute value functions