RD Sharma Solutions Class 12 Definite Integrals Exercise 20.3

RD Sharma Solutions for Class 12 Maths Exercise 20.3 Chapter 20 Definite Integrals are given here. The main aim is to help students understand and solve these problems. We, at BYJU’S, have prepared the RD Sharma Solutions for Class 12 Maths Chapter 20, wherein problems are solved step-by-step with detailed explanations. Students need to practise thoroughly to score well in the exam.

RD Sharma Solutions for Class 12 Maths Chapter 20 Exercise 3

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Access RD Sharma Solutions for Class 12 Maths Chapter 20 Exercise 3

EXERCISE 20.3

Question. 1(i)

Solution:

Then, applying limits, we get,

= [((16/2) + 6) – ((4/2) + 3)] + [((48/2) + 20) – ((12/2) + 10)]

= [(8 + 6) – (2 + 3)] + [(24 + 20) – (6 + 10)]

= [14 -5] + [44 – 16]

= 9 + 28

= 37

Question. 1(ii)

Solution:

Then, applying limits, we get,

= [- cos (π/2) + cos 0] + [3 – (π/2)] + [e^{9 – 3} – e^{3 – 3}]

= [0 + 1] + [3 – (π/2)] + [e⁶ – e⁰]

= 0 + 1 + 3 – (π/2) + e⁶ – e⁰

= 1 + 3 – (π/2) + e⁶ – 1

= 3 – π/2 + e⁶

Question. 1(iii)

Solution:

Then, applying limits, we get,

= [(((7 × 9)/2) + (3 × 3)) – (((7 × 1)/2) + (3 × 1))] + [(((8 × 16)/2) – ((8 × 9)/2))]

= [(63/2) + 9 – (7/2) – 3] + [64 – 36]

= 34 + 28

= 62

Question. 2

Solution:

Now, applying limits,

= – [((4/2) – 4) – ((16/2) – 8)] + [((16/2) + 8) – ((4/2) – 4)]

= – [(2 – 4) – (8 – 8)] + [(8 + 8) – (2 – 4)]

= – [- 2 – 0] + [16 – (- 2)]

= – [-2] + [16 + 2]

= 2 + 18

= 20

Question. 3

Solution:

Question. 4

Solution:

Question. 5

Solution:

Now, applying limits we get,

= – [((18/8) – (9/2)) – ((8/2) – 6)] + [((8/2) + 6) – ((18/8) – (9/2))]

= – [((9/4) – (9/2)) – ((4/1) – 6)] + [((4/1) + 6) – ((9/4) – (9/2))]

= – [((9/4) – (9/2)) – (- 2)] + [(10) + (9/4)]

= – [(- 9/4) + 2] + [10 + (9/4)]

= 9/4 – 2 + 10 + 9/4

= 18/4 + 8

= 9/2 + 8

= (9 + 16)/2

= 25/2