There are many practical applications of Definite Integration.

Definite integrals can be used to determine the mass of an object if its density function is known. We can also find work by integrating a force function, and the force exerted on an object submerged in a liquid. The most important application of Definite Integration is finding the area under the curve.

Let f be a continuous function defined on the closed interval [a, b] and F be an antiderivative of f then

  ∫ b a f(x)  dx = [F (x)] b a = F(b) - F(a)

It is very useful because it gives us a method of calculating the definite integral more easily. There is no need to keep integration constant C because if we consider F(x)+C instead of F(x). 

[Integrals Class 12] There are many practical applications of Definite - Case Based Questions (MCQ)

part 2 - Question 4 - Case Based Questions (MCQ) - Serial order wise - Chapter 7 Class 12 Integrals

Question 1

3 2 x 2   dx is equal to:

(a) 7/3

(b) 9 

c) 19/3 

(d) 1/3

This question is inspired from Question 30 - CBSE Class 12 - Sample Paper for 2020 Boards

part 3 - Question 4 - Case Based Questions (MCQ) - Serial order wise - Chapter 7 Class 12 Integrals

Question 2

√3 1 dx/1 + x 2  is equal to:

(a) π/3     

(b) 2π/3 

(c) π/6 

(d) π/12

part 4 - Question 4 - Case Based Questions (MCQ) - Serial order wise - Chapter 7 Class 12 Integrals part 5 - Question 4 - Case Based Questions (MCQ) - Serial order wise - Chapter 7 Class 12 Integrals

Question 3

1 -1  (x + 1)  dx is equal to:

(a) −1 

(b) 2 

(c) 1 

(d) 3

part 6 - Question 4 - Case Based Questions (MCQ) - Serial order wise - Chapter 7 Class 12 Integrals

Question 4

3 1/x  dx is equal to:

(a) 3/2 

(b) 1/2 

(c) log(3/2)

(d) log(2)

 

part 7 - Question 4 - Case Based Questions (MCQ) - Serial order wise - Chapter 7 Class 12 Integrals

Question 5

5 e x   dx is equal to:

(a) 1 

(b) e 5 - 1

(c) e 

(d) e 5 - e 4

part 8 - Question 4 - Case Based Questions (MCQ) - Serial order wise - Chapter 7 Class 12 Integrals

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Transcript

Question 4 There are many practical applications of Definite Integration. Definite integrals can be used to determine the mass of an object if its density function is known. We can also find work by integrating a force function, and the force exerted on an object submerged in a liquid. The most important application of Definite Integration is finding the area under the curve. Let 𝑓 be a continuous function defined on the closed interval [a, b] and F be an antiderivative of 𝑓 then ∫_π‘Ž^𝑏▒〖𝑓(π‘₯)γ€— 𝑑π‘₯=γ€–[𝐹 (π‘₯)]γ€—_π‘Ž^𝑏=𝐹(𝑏)βˆ’πΉ(π‘Ž) It is very useful because it gives us a method of calculating the definite integral more easily. There is no need to keep integration constant C because if we consider 𝐹(π‘₯)+𝐢 instead of 𝐹(π‘₯). We get ∫_π‘Ž^𝑏▒〖𝑓(π‘₯)γ€— 𝑑π‘₯=γ€–[𝐹 (π‘₯)+𝐢]γ€—_π‘Ž^𝑏 =𝐹(𝑏)+πΆβˆ’πΉ(π‘Ž)βˆ’πΆ =𝐹(𝑏)βˆ’πΉ(π‘Ž) Based on the above information, answer the following questions. Question 1 ∫_2^3β–’π‘₯^2 𝑑π‘₯ is equal to: (a) 7/3 (b) 9 (c) 19/3 (d) 1/3 ∫_2^3β–’π‘₯^2 𝑑π‘₯ = [𝒙^πŸ‘/πŸ‘]_𝟐^πŸ‘ =[3^3/3βˆ’2^3/3] =[(27 βˆ’ 8)/3] = πŸπŸ—/πŸ‘ So, the correct answer is (c) Question 2 ∫_1^(√3)▒𝑑π‘₯/(1 + π‘₯^2 ) is equal to: (a) πœ‹/3 (b) 2πœ‹/3 (c) πœ‹/6 (d) πœ‹/12 ∫_1^(√3)▒𝑑π‘₯/(1 + π‘₯^2 ) = [1/1 π‘‘π‘Žπ‘›^(βˆ’1) π‘₯]_1^√3 = [𝒕𝒂𝒏^(βˆ’πŸ) 𝒙]_𝟏^√3 = π‘‘π‘Žπ‘›^(βˆ’1) √3βˆ’π‘‘π‘Žπ‘›^(βˆ’1) 1 = πœ‹/3βˆ’πœ‹/4 = (4πœ‹ βˆ’ 3πœ‹)/12 = 𝝅/𝟏𝟐 So, the correct answer is (d) Question 3 ∫_(βˆ’1)^1β–’γ€–(π‘₯+1)γ€— 𝑑π‘₯ is equal to: (a) βˆ’1 (b) 2 (c) 1 (d) 3 ∫_(βˆ’1)^1β–’γ€–(π‘₯+1)γ€— 𝑑π‘₯ = [𝒙^𝟐+𝒙]_(βˆ’πŸ)^𝟏 = [(1^2+1)βˆ’((βˆ’1)^2+(βˆ’1)) ] = [(1+1)βˆ’(1βˆ’1)] = [2βˆ’0] = 𝟐 So, the correct answer is (b) Question 4 ∫_2^3β–’1/π‘₯ 𝑑π‘₯ is equal to: (a) 3/2 (b) 1/2 (c) log(3/2) (d) log(2) ∫_2^3β–’1/π‘₯ 𝑑π‘₯ = [π’π’π’ˆ 𝒙]_𝟐^πŸ‘ = [log⁑3βˆ’log⁑2 ] = π₯π¨π β‘γ€–πŸ‘/πŸγ€— So, the correct answer is (c) Question 5 ∫_4^5▒𝑒^π‘₯ 𝑑π‘₯ is equal to: (a) 1 (b) 𝑒^5βˆ’1 (c) 𝑒 (d) 𝑒^5βˆ’π‘’^4 ∫_4^5▒𝑒^π‘₯ 𝑑π‘₯ = [𝒆^𝒙 ]_πŸ’^πŸ“ = [𝑒^5βˆ’π‘’^4 ] So, the correct answer is (d)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo