# RD Sharma Class 12 Ex 20.4 Solutions Chapter 20 Definite Integrals

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## RD Sharma Class 12 Ex 20.4 Solutions Chapter 20 Definite Integrals

Solution:

We know that

so,

we know,

if

I =

then

I =

2I =

l=π

Solution:

We know that

So,

if

I =

then

I =

2I =

2I=

2I =

2I=0

I=0

Solution:

We know

So,

if

I=

then

I=

So

2I=

2I =

2I=

2I=π/6

I=π/12

Solution:

We know

So,

if

I =

then,

I =

2I =

2I =

2I =

2I=π/6

I=π/12

### Question 5.

Solution:

We know

so,

if

then,

we know if

f(x) is even

f(x) is odd

Here, f(x) = tan2x which is even

hence,

I =

Solution:

We know

So,

if,  then

So,

Solution:

We know

Hence,

if,

then

so,

Solution:

We know

hence,

if

Then,

So,

### Question 9.

Solution:

if f(x) is even

if f(x) is odd

here,   is odd and

is even

Hence,

2

Solution:

if

then,

Solution:

let,

we know that,

hence,

Solution:

Let,

we know that ,

so,

then,

Solution:

We know that,

So,

then,

I

Solution:

We know that,

so,

then,

Solution:

We know that,

Let,

hence,

Solution:

### Question 1.

Solution:

We have,

Let

—— 1

So,,

———— 2

Hence, by adding 1 and 2 ..

Solution:

We have,

Let, I=—– 1

So,—— 2

Solution:

We have ,

Let,

So,

————— 2

Solution:

Let,—— 1

———- 2

Solution:

Let,———– (1)

So,

Solution:

———- 1

—– 2

Solution:

Let,

Let,

Now, x=0 ,, then

——————– 1

So,

————- 2

Solution:

Let,

then

———— (1)

——————- (2)

Solution:

Let,

### Question 10.

Solution:

Let,

Equating coefficients, we get

So,

Solution:

————- (1)

—————— (2)

### Question 12.

Solution:

Let ,————– 1

So,

Let,

As, x=0, t=1 ; x=π , t=-1

Hence,

Solution:

Let,

### Question 14.

Solution:

we have,

=

Since, f(x) = f(-x) , f(x) is an even function.

————– 1

—————- 2

Now, let

Putting 2x=t, we get

Solution:

Let,

### Question 16.

Solution:

We have , I=\int\limits_0^{π} \frac{x}{1+cos\alpha sinx}dx ———- 1

[Tex]=\int\limits_0^{π} \frac{(π-x)}{1+cos\alpha sinx}dx [/Tex]——- 2

substituting s

when x=0 , t=0 ; x=π ,

Solution:

Let,

Solution:

Solution:

Solution:

### Question 21.

Solution:

Now,

Let cosx=t

sinx dx=-dt

[Tex]I= \frac{π }{8}[\fracπ 4+\fracπ 4] dt[/Tex]

Solution:

—————— 1

———— 2

Let ,

### Question 23.

Solution:

Let,

Here, f(x)=-f(x)

Hence, f(x) is odd function

### Question 24.

Solution:

We have,  is an even function.

### Question 25.

Solution:

we have,

Since,

this is an odd function

### Question 26.

Solution:

we have,

sin2x is even function

Hence,

Solution:

Solution:

we have ,

Let,

Then,

### Question 29.

Solution:

Put cosx = t then -sinx dx = dt

### Question 30.

Solution:

Let

is an odd function

Solution:

### Question 32.

Solution:

Substitute π+x=u then dx=du

Solution:

Let,

### Question 34.

Solution:

Applying the property ,

Thus,

Solution:

Let,

### Question 36.

Solution:

[Tex][\because \int\limits_0^{2a} f(x)dx= 2\int\limits_0^{a} f(x)dx, f(2a-x)=f(x)][/Tex]

let tanx = v

dv = sec2xdx

### Question 37.

Solution:

Putthen

x=0 ⇒ t=0 and x=π ⇒

Solution:

we know,

Also here,

f(x) = f(2π -x)

So,

Solution:

then,

### Question 40. If f is an integrable function such that f(2a-x)=f(x), then prove that

Solution:

We have ,

Then,

Let , 2a-t =x then dx=-dt

if t=a ⇒x=a

if t=2a ⇒ x=0

[Tex]=2\int\limits_{0}^{a}f(x)dx[/Tex]

Hence Proved.

### Question 41. If, prove that

Solution:

We have,

Let 2a-t=x then dx=-dt

t=a , x=a ; t=2a , x=0

### (i)

Solution:

we have ,

clearly f(x2) is an even function .

So,

### (ii)

Solution:

clearly , xf(x2) is odd function .

So,

### Question 43. If f(x) is a continuous function defined on [0,2a] . Then, prove that

Solution:

We have from LHS,

substituting

we get,

### Question 44. If f(a+b-x) = f(x), then prove that

Solution:

——————[ Given that f(a+b-x) = f(x) ]

### Question 45. If f(x) is a continuous function defined on [-a,a], then prove that

Solution:

we have ,

Let, x=-t, then dx=-dt

x=-a ⇒ t=a

x=0 ⇒ t=0

Hence, Proved.

Solution:

### 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 .

### 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) = x− 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) = ex.

=> 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) = ex.

=> h =

=> nh = b − a

So, we get,

I =

=

=

=

=

=

=

= ea (eb-a −1)

= eb − ea

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

### 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.

Evaluate the following definite integrals as limits of sums:
Question 23.
Solution:

We have,
I =
We know,
, where h =
Here a = 0, b = 4 and f(x) = x + e2x.
=> h = 4/n
=> nh = 4
So, we get,
I =
=
=
=
=
=
Now if h −> 0, then n −> ∞. So, we have,
=
=
=
=
Therefore, the value ofas limit of sum is.
Question 24.
Solution:
We have,
I =
We know,
, where h =
Here a = 0, b = 2 and f(x) = x2 + x.
=> 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 25.
Solution:
We have,
I =
We know,
, where h =
Here a = 0, b = 2 and f(x) = x2 + 2x + 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 26.
Solution:
We have,
I =
We know,
, where h =
Here a = 0, b = 3 and f(x) = 2x2 + 3x + 5.
=> h = 3/n
=> nh = 3
So, we get,
I =
=
=
=
Now if h −> 0, then n −> ∞. So, we have,
=
=
= 15 + 18 +
=
Therefore, the value ofas limit of sum is.
Question 27.
Solution:
We have,
I =
We know,
, where h =
Here a = a, b = b and f(x) = x.
=> h =
=> nh = b − a
So, we get,
I =
=
=
=
Now if h −> 0, then n −> ∞. So, we have,
=
=
=
=
=
=
Therefore, the value ofas limit of sum is.
Question 28.
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 29.
Solution:
We have,
I =
We know,
, where h =
Here a = 2, b = 3 and f(x) = x2.
=> h = 1/n
=> nh = 1
So, we get,
I =
=
=
=
=
Now if h −> 0, then n −> ∞. So, we have,
=
=
=
=
=
Therefore, the value ofas limit of sum is.
Question 30.
Solution:
We have,
I =
We know,
, where h =
Here a = 1, b = 3 and f(x) = x2 + x.
=> 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 31.
Solution:
We have,
I =
We know,
, where h =
Here a = 0, b = 2 and f(x) = x2 − x.
=> 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 32.
Solution:
We have,
I =
We know,
, where h =
Here a = 1, b = 3 and f(x) = 2x2 + 5x.
=> h = 2/n
=> nh = 2
So, we get,
I =
=
=
=
Now if h −> 0, then n −> ∞. So, we have,
=
=
= 14 + 18 +
=
Therefore, the value ofas limit of sum is.
Question 33.
Solution:
We have,
I =
We know,
, where h =
Here a = 1, b = 3 and f(x) = 3x2 + 1.
=> h = 2/n
=> nh = 2
So, we get,
I =
=
=
=
Now if h −> 0, then n −> ∞. So, we have,
=
=
= 8 + 12 + 8
= 28
Therefore, the value ofas limit of sum is 28.

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