Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.1 | Set 1

• Last Updated : 02 Jun, 2021

Question 1.

Solution:

We have,

I =

I =

I =

I =

I = 2[√9 – √4 ]

I = 2 (3 − 2)

I = 2 (1)

I = 2

Therefore, the value of is 2.

Question 2.

Solution:

We have,

I =

I =

I = log (3 + 7) − log (−2 + 7)

I = log 10 − log 5

I =

I = log 2

Therefore, the value of is log 2.

Question 3.

Solution:

We have,

I =

Let x = sin t, so we have,

=> dx = cos t dt

Now, the lower limit is,

=> x = 0

=> sin t = 0

=> t = 0

Also, the upper limit is,

=> x = 1/2

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I =

I =

I =

I =

I =

I =  π/6 – 0

I = π/6

Therefore, the value of is π/6.

Question 4.

Solution:

We have,

I =

I =

I =

I =

I = π/4

Therefore, the value of is π/4.

Question 5.

Solution:

We have,

I =

Let x2 + 1 = t, so we have,

=> 2x dx = dt

=> x dx = dt/2

Now, the lower limit is, x = 2

=> t = x2 + 1

=> t = (2)2 + 1

=> t = 4 + 1

=> t = 5

Also, the upper limit is, x = 3

=> t = x2 + 1

=> t = (3)2 + 1

=> t = 9 + 1

=> t = 10

So, the equation becomes,

I =

I =

I =

I = 1/2[log10 – log5]

I = 1/2[log10/5]

I = 1/2[log2]

I = log√2

Therefore, the value of is log√2.

Question 6.

Solution:

We have,

I =

I =

I =

I =

I =

I = 1/ab[tan-1∞ – tan-10]

I = 1/ab[π/2 – 0]

I = 1/ab[π/2]

I = π/2ab

Therefore, the value of is π/2ab.

Question 7.

Solution:

We have,

I =

I =

I = [tan-11 – tan-1(-1)]

I = [π/4 – (-π/4)]

I = [π/4 + π/4]

I = 2π/4

I = π/2

Therefore, the value of is π/2.

Question 8.

Solution:

We have,

I =

I =

I = -e – (-e0)

I = − 0 + 1

I = 1

Therefore, the value of is 1.

Question 9.

Solution:

We have,

I =

I =

I =

I =

I =

I = [1 − 0] − [log(1 + 1) − log(0 + 1)]

I = 1 − [log2 − log1]

I = 1 – log2/1

I = 1 − log 2

I = log e − log 2

I = loge/2

Therefore, the value of  is loge/2.

Question 10.

Solution:

We have,

I =

I =

I =

I = [-cosπ/2 + cos0] + [sinπ/2 – sin0]

I = [−0 + 1] + 1

I = 1 + 1

I = 2

Therefore, the value of is 2.

Question 11.

Solution:

We have,

I =

I =

I = log(sinπ/2) – log(sinπ/4)

I = log1 – log1/√2

I =

I = log√2

Therefore, the value of is log√2.

Question 12.

Solution:

We have,

I =

I =

I = log(secπ/4 + tanπ/4 – log(sec0 + tan0)

I = log(√2 + 1) – log(1 + 0)

I =

I = log(√2 + 1)

Therefore, the value of is log(√2 + 1).

Question 13.

Solution:

We have,

I =

I =

I = [log|cosecπ/4 – cotπ/4|] – [log|cosecπ/6 – cotπ/6|]

I = [log|√2 – 1|] – [log|2 – √3|]

I =

Therefore, the value of is .

Question 14.

Solution:

We have,

I =

Let x = cos 2t, so we have,

=> dx = –2 sin 2t dt

Now, the lower limit is,

=> x = 0

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is,

=> x = 1

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

Let cos t = z, so we have,

=> – sin t dt = dz

=> sin t dt = – dz

Now, the lower limit is,

=> t = 0

=> z = cos t

=> z = cos 0

=> z = 1

Also, the upper limit is,

=> t = π/4

=> z = cos t

=> z = cos π/4

=> z = 1/√2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I = -4[(log1/√2 – 1/2(2)) – (log1 – 1/2)]

I = -4[(log1/√2 – 1/4) – (0 – 1/2)]

I = -4[log1/√2 – 1/4 – 0 + 1/2]

I = -4[-log√2 + 1/4]

I = 4log√2 – 1

I = 4 × 1/2log2 – 1

I = 2log2 – 1

Therefore, the value of is 2log2 – 1.

Question 15.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = [tan π – tan0] – [sec π – sec 0]

I = [0 – 0] – [–1 – 1]

I = 0 – (–2)

I = 2

Therefore, the value of is 2.

Question 16.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = [tan π/4 – tan(–π/4)] – [sec π/4 – sec (–π/4)]

I = [1 – (–1)] – [sec π/4 – sec (π/4)]

I = 2 – 0

I = 2

Therefore, the value of is 2.

Question 17.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I = 1/2[π/2 – 0] + 1/4[sinπ – sin0]

I = 1/2[π/2] + 1/4[0 – 0]

I = π/4

Therefore, the value of  is π/4.

Question 18.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I = 1/12 [-1 – 0] + 3/4[1 – 0]

I = 3/4 – 1/12

I = (9 – 1)/12

I = 8/12

I = 2/3

Therefore, the value of  is 2/3.

Question 19.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I = 1/6[sinπ/2 – sin0] + 1/2[sinπ/6 – sin0]

I = 1/6[1 – 0] + 1/2[1/2 – 0]

I = 1/6 + 1/4

I = (4 + 6)/24

I = 10/24

I = 5/12

Therefore, the value of  is 5/12.

Question 20.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I = 1/2[sinπ/2 – sin0] – 1/6[sin3π/2 – sin0]

I = 1/2[1 – 0] – 1/6[-1 – 0]

I = 1/2 – 1/6(-1)

I = 1/2 + 1/6

I = (6 + 2)/12

I = 8/12

I = 2/3

Therefore, the value of  is 2/3.

Question 21.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = 2[-cotπ/2 + cot2π/3]

I = 2[-1/√3 – 0]

I = -2/√3

Therefore, the value of  is -2/√3.

Question 22.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I = 1/4[π/2 + π/4 + 0 + 0 – 0 – 0 – 0 – 0]

I = 1/4[3π/4]

I = 3π/16

Therefore, the value of is 3π/16.

My Personal Notes arrow_drop_up
Related Articles