RBSE Maths Class 12 Chapter 10: Definite Integral Important Questions and Solutions

RBSE Class 12 Maths Chapter 10 – Definite Integral Important questions and solutions are provided here. The important questions and solutions of the chapter which are given here are important from the examination point of view. The RBSE Class 12 important questions and solutions, available at BYJU’S, has a full coverage of all the concepts of Class 12 and are updated as per the new pattern.

Chapter 10 of RBSE Class 12 contains only three exercises, all of which have questions on finding the integral of given functions under the given limits. The first exercise has questions on finding the integral as the limit of the sum of the function. All the questions under this chapter involve some of the other formulas to derive into simple forms.

RBSE Maths Chapter 10: Exercise 10.1 Textbook Important Questions and Solutions

Question 1: Evaluate the following definite integral as a limit of sums.

35 (x – 2) dx

Solution:

We know that,

∫ab f(x) dx = limh→0 h[f(a + h) + f(a + 2h) + f(a + 3h) + … + f(a + nh)]

35 (x – 2) dx

Here,

a = 3, b = 5

f(x) = x – 2

nh = 5 – 3 = 2

35 (x – 2) dx = limh→0 h [f(3 + h) + f(3 + 2h) + f(3 + 3h) + … + f(3 + nh)]

= limh→0 h [(3 + h – 2) + (3 + 2h – 2) + (3 + 3h – 2) + … + (3 + nh – 2)]

= limh→0 h [1 + h + 1 + 2h + 1 + 3h + … + 1 + nh]

= limh→0 h [h(1 + 2 + 3 + … + n) + (1 + 1 + 1 + … + n times)]

= limh→0 h [hn(n + 1)/2] + n

= limh→0 h(hn2 + hn + 2n)/2

= limh→0 (h2n2 + h2n + 2nh)/2

= limh→0 [(nh)2 + 2(nh) + h2n]/2

= [(2)2 + 2(2) + 0]/2

= (4 + 4)/2

= 8/2

= 4

Question 2: Evaluate the following definite integral as a limit of sums

ab x2 dx

Solution:

We know that,

∫ab f(x) dx = limh→0 h[f(a + h) + f(a + 2h) + f(a + 3h) + … + f(a + nh)]

ab x2 dx

Here, a = a, b = b and nh = b – a

ab x2 dx = limh→0 h[f(a + h) + f(a + 2h) + f(a + 3h) + … + f(a + nh)]

= limh→0 h [(a + h)2 + (a + 2h)2 + … + (a + nh)2]

= limh→0 h [(a2 + h2 + 2ah) + (a2 + 22 h2 + 2a2h) + … + (a2 + n2h2 + 2anh)]

= limh→0 h [(a2 + a2 + … + a2) + h2(12 + 22 + … + n2) + … + 2ah(1 + 2 + … + n)]

= limh→0 h [a2 × n + h2 × {n(2n + 1)(n + 1)/6} + 2ah × {n(n + 1)/2}]

= limh→0 a2nh + h3 [(n)(n + 1)(2n + 1)/6] + ah2n(n + 1)

= limh→0 a2(b – a) + [(nh)(nh + h)(2nh + h)/6] + anh(nh + h)

= limh→0 a2(b – a) + [(b – a)(b – a + h){2(b – a) + h})/6] + a(b – a)(b – a + h)]

= a2(b – a) + a(b – a)(b – a) + (1/3)(b – a)(b – a)(b – a)

= (b – a)a2 + a(b – a)2 + (1/2)(b – a)3

= [(b – a)/3] [3a2 + 3a(b – a) + (b – a)2]

= [(b – a)/3] [3a2 + 3ab – 3a2 + b2 – 2ab + a2]

= [(b – a)/3] [b2 + ab + a2]

= (b3 – a3)/3

RBSE Maths Chapter 10: Exercise 10.2 Textbook Important Questions and Solutions

Question 3: Evaluate the definite integral ∫13 (2x + 1)3 dx.

Solution:

13 (2x + 1)3 dx

Let 2x + 1 = t

⇒ 2dx = dt

⇒ dx = dt/2

If x = 1, then t = 3.

If x = 3, then t = 7.

13 (2x + 1)3 dx = ∫37 t3 (dt/2)

= (1/2) [t4/4]37

= (1/8) [(7)4 – (3)4]

= (2401 – 81)/8

= 2320/8

= 290

Question 4: Evaluate: ∫13 [cos (log x)]/x dx

Solution:

13 [cos (log x)]/x dx

Let log x = t

⇒ (1/x) dx = dt

If x = 1, then t = log 1 = 0

If x = 3, then t = log 3

13 [cos (log x)]/x dx = ∫0 log 3 cos t dt

= [sin t]0 log 3

= sin (log 3) – sin 0

= sin (log 3)

RBSE Maths Chapter 10: Exercise 10.3 Textbook Important Questions and Solutions

Question 5: Evaluate the definite integral ∫-22 |2x + 3| dx.

Solution:

-22 |2x + 3| dx = ∫-2-3/2 -(2x + 3) dx + ∫-3/22 (2x + 3) dx

= -[(2x2/2) + 3x]-2-3/2 + [(2x2/2) + 3x]-3/22

= (x2 + 3x)-3/22 – (x2 + 3x)-2-3/2

= [22 + 3(2) – (-3/2)2 – 3(-3/2)] – [(-3/2)2 + 3(-3/2) – (-2)2 – 3(-2)]

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

= 4 + 6 – (9/4) + (9/2) – (9/4) + (9/2) + 4 – 6

= 8 + 9 – (9/2)

= 8 + (9/2)

= (16 + 9)/2

= 25/2

Question 6: Evaluate the definite integral ∫03 [x] dx when [.] is the greatest integer function.

Solution:

03 [x] dx = ∫01 [x] dx + ∫12 [x] dx + ∫23 [x] dx

= ∫01 0 dx + ∫12 1 dx + ∫23 2 dx

= 0 + (x)12 + (2x)23

= (2 – 1) + [2(3) – 2(2)]

= 1 + (6 – 4)

= 3

RBSE Maths Chapter 10: Additional Important Questions and Solutions

Question 1: If A(x) = ∫0x θ2 dθ, then value of A(3) is

(A) 9

(B) 27

(C) 3

(D) 81

Solution:

Correct answer: (A)

Given,

A(x) = ∫0x θ2

= [θ3/ 3]0x

= (1/3) [x3 – 0]

= x3/3

A(3) = (1/3) × (3)3

= (3)2

= 9

Question 2: Evaluate the definite integral ∫01 tan-1x dx.

Solution:

01 tan-1x dx = ∫01 tan-1x . 1 dx ….(i)

∫tan-1x . 1 dx = (tan-1x)x – ∫[1/(1 + x2)] x dx

= x tan-1x – (1/2) ∫2x/(1 + x2) dx ….(ii)

Let 1 + x2 = t

2xdx = dt

From (ii),

∫tan-1x . 1 dx = x tan-1x – (1/2) ∫ dt/t

= x tan-1x – (1/2) log t

01 tan-1x dx = [x tan-1x – (1/2) log (1 + x2)]01

= [1. tan-1(1) – (1/2) log (1 + 12)] – 0

= (π/4) – (1/2) log 2

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