Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.5 | Set 2

Evaluate the following definite integrals as limits of sums:

Question 12. 

Solution:

We have,

I =

We know,

, where h =

Here a = 0, b = 2 and f(x) = x² + 4.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

=

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

=

=

= 8 +

= 8 +

=

Therefore, the value ofas limit of sum is.

Question 13. 

Solution:

We have,

I =

We know,

, where h =

Here a = 1, b = 4 and f(x) = x² − x.

=> h = 3/n

=> nh = 3

So, we get,

I =

=

=

=

=

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

=

=

=

=

Therefore, the value ofas limit of sum is.

Question 14. 

Solution:

We have,

I =

We know,

, where h =

Here a = 0, b = 1 and f(x) = 3x² + 5x.

=> h = 1/n

=> nh = 1

So, we get,

I =

=

=

=

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

=

=

= 1 +

=

Therefore, the value ofas limit of sum is.

Question 15. 

Solution:

We have,

I =

We know,

, where h =

Here a = 0, b = 2 and f(x) = e^x.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

=

=

= e² − 1

Therefore, the value ofas limit of sum is e² − 1.

Question 16. 

Solution:

We have,

I =

We know,

, where h =

Here a = a, b = b and f(x) = e^x.

=> h =

=> nh = b − a

So, we get,

I =

=

=

=

=

=

=

= e^a (e^b-a −1)

= e^b − e^a

Therefore, the value ofas limit of sum is e^b − e^a.

Question 17. 

Solution:

We have,

I =

We know,

, where h =

Here a = a, b = b and f(x) = cos x.

=> h =

=> nh = b − a

So, we get,

I =

=

=

=

=

=

=

=

= sin b − sin a

Therefore, the value ofas limit of sum is sin b − sin a.

Question 18. 

Solution:

We have,

I =

We know,

, where h =

Here a = 0, b =and f(x) = sin x.

=> h =

=> nh =

So, we get,

I =

=

=

=

=

=

= 1

Therefore, the value ofas limit of sum is 1.

Question 19. 

Solution:

We have,

I =

We know,

, where h =

Here a = 0, b =and f(x) = cos x.

=> h =

=> nh =

So, we get,

I =

=

=

=

=

= 1

Therefore, the value ofas limit of sum is 1.

Question 20. 

Solution:

We have,

I =

We know,

, where h =

Here a =1, b = 4 and f(x) = 3x² + 2x.

=> h = 3/n

=> nh = 3

So, we get,

I =

=

=

=

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

=

=

= 15 + 36 + 27

= 78

Therefore, the value ofas limit of sum is 78.

Question 21. 

Solution:

We have,

I =

We know,

, where h =

Here a =0, b = 2 and f(x) = 3x² − 2.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

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

=

=

= −4 + 8

= 4

Therefore, the value ofas limit of sum is 4.

Question 22. 

Solution:

We have,

I =

We know,

, where h =

Here a =0, b = 2 and f(x) = x² + 2.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

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

=

=

= 4 +

= 4 +

=

Therefore, the value ofas limit of sum is.