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

Last Updated : 20 May, 2021

### 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 âˆ’> 0, then n âˆ’> âˆž. 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 âˆ’ x.

=> h = 3/n

=> nh = 3

So, we get,

I =

=

=

=

=

Now if h âˆ’> 0, then n âˆ’> âˆž. 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 âˆ’> 0, then n âˆ’> âˆž. 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) = ex.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

=

=

= e2 âˆ’ 1

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

### Question 16.

Solution:

We have,

I =

We know,

, where h =

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

=> h =

=> nh = b âˆ’ a

So, we get,

I =

=

=

=

=

=

=

= ea (eb-a âˆ’1)

= eb âˆ’ ea

Therefore, the value ofas limit of sum is eb âˆ’ ea.

### Question 17.

Solution:

We have,

I =

We know,

, where h =

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

=> h =

=> nh = b âˆ’ a

So, we get,

I =

=

=

=

=

=

=

=

= sin b âˆ’ sin a

Therefore, the value ofas limit of sum is sin b âˆ’ 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 âˆ’> 0, then n âˆ’> âˆž. 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 âˆ’ 2.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

Now if h âˆ’> 0, then n âˆ’> âˆž. So, we have,

=

=

= âˆ’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 âˆ’> 0, then n âˆ’> âˆž. So, we have,

=

=

= 4 +

= 4 +

=

Therefore, the value ofas limit of sum is.

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