Here we provide RD Sharma Class 12 Ex 20.4 Solutions Chapter 20 Definite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 20.4 Solutions Chapter 20 Definite Integrals book pdf download. Now you will get step-by-step solutions to each questio

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 20 |

Exercise | 20.4 |

Category | RD Sharma Solutions |

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

**Evaluate of each of the following integrals (1-16):**

**Question 1.**

**Solution:**

We know that

so,

we know,

if

I =

then

I =

2I =

l=π

**Question 2.**

**Solution:**

We know that

So,

if

I =

then

I =

2I =

2I=

2I =

2I=0

I=0

**Question 3.**

**Solution:**

We know

So,

if

I=

then

I=

So

2I=

2I =

2I=

2I=π/6

I=π/12

**Question 4. **

**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) = tan^{2}x which is even

hence,

I =

**Question 6. **

**Solution:**

We know

So,

if, then

So,

**Question 7. **

**Solution:**

We know

Hence,

if,

then

so,

**Question 8. **

**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

**Question 10. **

**Solution:**

if

then,

**Question 11. **

**Solution:**

let,

we know that,

hence,

**Question 12. **

**Solution:**

Let,

we know that ,

so,

then,

**Question 13. **

**Solution:**

We know that,

So,

then,

I

**Question 14. **

**Solution:**

We know that,

so,

then,

**Question 15. **

**Solution:**

We know that,

Let,

hence,

**Question 16. If f(a+b-x)=f(x), then prove that **

**Solution:**

### Evaluate of each of the following integrals (1-46):

**Question 1. **

**Solution:**

We have,

Let

—— 1

So,,

———— 2

Hence, by adding 1 and 2 ..

**Question 2. ****,**

**Solution:**

We have,

Let, I=—– 1

So,—— 2

Adding 1 & 2 ——-

**Question 3. ****,**

**Solution:**

We have ,

Let,

So,

————— 2

Adding 1 and 2 ——–

**Question 4. **

**Solution:**

Let,—— 1

———- 2

Adding 1 & 2 —–

**Question 5. **

**Solution:**

Let,———– (1)

So,

**Question 6. **

**Solution:**

———- 1

—– 2

Adding 1 & 2 ——-

**Question 7. **

**Solution:**

Let,

Let,

Now, x=0 ,, then

——————– 1

So,

————- 2

Adding (1) and (2) —————-

**Question 8. **

**Solution:**

Let,

then

———— (1)

——————- (2)

Adding (1) & (2) ————

**Question 9. **

**Solution:**

Let,

**Question 10. **

**Solution:**

Let,

Equating coefficients, we get

So,

**Question 11. **

**Solution:**

————- (1)

—————— (2)

Adding (1) & (2) —————-

**Question 12. **

**Solution:**

Let ,————– 1

So,

Let,

As, x=0, t=1 ; x=π , t=-1

Hence,

**Question 13. **

**Solution:**

Let,

**Question 14. **

**Solution:**

we have,

=

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

————– 1

—————- 2

Adding 1 and 2 —————-

Now, let

Putting 2x=t, we get

**Question 15. **

**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

Adding 1 and 2 —-

substituting s

when x=0 , t=0 ; x=π ,

**Question 17. **

**Solution:**

Let,

**Question 18. **

**Solution:**

**Question 19. **

**Solution:**

**Question 20. **

**Solution:**

**Question 21. **

**Solution:**

Now,

Let cosx=t

sinx dx=-dt

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

**Question 22. **

**Solution:**

—————— 1

———— 2

Adding 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,

sin^{2}x is even function

Hence,

**Question 27. **

**Solution:**

**Question 28. **

**Solution:**

we have ,

Let,

Then,

**Question 29. **

**Solution:**

Put cosx = t then -sinx dx = dt

**Question 30. **

**Solution:**

Let

is an odd function

**Question 31. **

**Solution:**

**Question 32. **

**Solution:**

Substitute π+x=u then dx=du

**Question 33. **

**Solution:**

Let,

**Question 34. **

**Solution:**

Applying the property ,

Thus,

–

**Question 35. **

**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 = sec^{2}xdx

**Question 37. **

**Solution:**

Putthen

x=0 ⇒ t=0 and x=π ⇒

**Question 38. **

**Solution:**

we know,

Also here,

f(x) = f(2π -x)

So,

**Question 39. **

**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

**Question 42. If f is an integrable function, show that**

**(i)**

**Solution:**

we have ,

clearly f(x^{2}) is an even function .

So,

**(ii)**

**Solution:**

clearly , xf(x^{2}) 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.

**Question 46. Prove that:**

**Solution:**

**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) = x

^{2}+ 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) = 2x

^{2}+ 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) = x

^{2}− 1.=> h = 1/n

=> nh = 1

So, we get,

I =

=

=

=

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

=

=

= 1 +

= 1 +

=

Therefore, the value ofas limit of sum is.

**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) = x

^{2}+ 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

^{2 }− 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) = 3x

^{2}+ 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 =

=

=

=

=

=

= e

^{2}− 1

Therefore, the value ofas limit of sum is e^{2}− 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) = 3x

^{2}+ 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) = 3x

^{2}− 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) = x

^{2}+ 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 + e^{2x}.

=> 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) = x^{2}+ 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) = x^{2}+ 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) = 2x^{2}+ 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) = x^{2}.

=> 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) = x^{2}+ 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) = x^{2}− 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) = 2x^{2}+ 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) = 3x^{2}+ 1.

=> h = 2/n

=> nh = 2

So, we get,

I =

=

=

=

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

=

=

= 8 + 12 + 8

= 28Therefore, the value ofas limit of sum is 28.

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