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• RD Sharma Class 12 Solutions for Maths

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

### 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 .

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