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Class 12 Science (PCM)

Mathematics Part 2

Integrals

Integrals

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Study Action Plan (Sorted by Difficulty)
Continuity Requirement in Definite Integrals
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Definite Integral as Limit of Sum
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Definite Integral Properties: Limits and Splitting
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Definite Integral Symmetry Property
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Substitution Pattern Recognition
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Integration by Parts
Accuracy: 0%
Total Attempted3
Total Correct0
Total Incorrect3
Difficulty Level:
High
Standard Integrals
Accuracy: 0%
Total Attempted3
Total Correct0
Total Incorrect3
Difficulty Level:
High
Integrals of Particular Algebraic Forms
Accuracy: 17%
Total Attempted6
Total Correct1
Total Incorrect5
Difficulty Level:
High
Definite Integrals over Symmetric Intervals
Accuracy: 25%
Total Attempted8
Total Correct2
Total Incorrect6
Difficulty Level:
High
Choosing First Function in Integration by Parts
Accuracy: 29%
Total Attempted7
Total Correct2
Total Incorrect5
Difficulty Level:
High
Fundamental Theorem of Calculus
Accuracy: 33%
Total Attempted12
Total Correct4
Total Incorrect8
Difficulty Level:
High
Integral Notation and Terminology
Accuracy: 33%
Total Attempted3
Total Correct1
Total Incorrect2
Difficulty Level:
High
Linear Numerator over Quadratic Denominator
Accuracy: 33%
Total Attempted3
Total Correct1
Total Incorrect2
Difficulty Level:
High
Indefinite Integral and Constant of Integration
Accuracy: 35%
Total Attempted17
Total Correct6
Total Incorrect11
Difficulty Level:
High
Even and Odd Function Definite Integrals
Accuracy: 50%
Total Attempted4
Total Correct2
Total Incorrect2
Difficulty Level:
Medium
Integration by Partial Fractions
Accuracy: 50%
Total Attempted8
Total Correct4
Total Incorrect4
Difficulty Level:
Medium
Special Integration by Parts Forms
Accuracy: 50%
Total Attempted4
Total Correct2
Total Incorrect2
Difficulty Level:
Medium
Properties of Indefinite Integrals
Accuracy: 56%
Total Attempted9
Total Correct5
Total Incorrect4
Difficulty Level:
Medium
Completing Square for Quadratic Denominators
Accuracy: 57%
Total Attempted7
Total Correct4
Total Incorrect3
Difficulty Level:
Medium
Area Function and Applications
Accuracy: 67%
Total Attempted3
Total Correct2
Total Incorrect1
Difficulty Level:
Medium
Integration Using Trigonometric Identities
Accuracy: 67%
Total Attempted3
Total Correct2
Total Incorrect1
Difficulty Level:
Medium
Partial Fraction Decomposition Cases
Accuracy: 71%
Total Attempted7
Total Correct5
Total Incorrect2
Difficulty Level:
Medium
Definite Integrals and Properties
Accuracy: 75%
Total Attempted4
Total Correct3
Total Incorrect1
Difficulty Level:
Low
Evaluation of Definite Integrals Using Antiderivatives
Accuracy: 75%
Total Attempted4
Total Correct3
Total Incorrect1
Difficulty Level:
Low
Integration by Substitution
Accuracy: 75%
Total Attempted4
Total Correct3
Total Incorrect1
Difficulty Level:
Low
Integration as Inverse of Differentiation
Accuracy: 80%
Total Attempted10
Total Correct8
Total Incorrect2
Difficulty Level:
Low
Standard Exponential and Logarithmic Integrals
Accuracy: 80%
Total Attempted5
Total Correct4
Total Incorrect1
Difficulty Level:
Low
Standard Inverse Trigonometric Integral Forms
Accuracy: 100%
Total Attempted8
Total Correct8
Total Incorrect0
Difficulty Level:
Low
Standard Trigonometric Integrals
Accuracy: 100%
Total Attempted4
Total Correct4
Total Incorrect0
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Understand the concept of integration as the inverse process of differentiation.
  • Identify and differentiate between indefinite and definite integrals.
  • Apply the Fundamental Theorem of Calculus to evaluate definite integrals.
  • Utilize various techniques of integration, including substitution, integration by parts, and partial fractions.
  • Solve practical problems involving areas under curves using definite integrals.
  • Recognize and apply standard integral formulas for common functions.
  • Analyze the relationship between derivatives and integrals in the context of calculus.

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Revision Notes & Summary

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Chapter 7: Integrals

7.1 Introduction

  • Differential Calculus focuses on derivatives, while Integral Calculus focuses on areas under curves.
  • Key Problems in Integral Calculus:
    • Finding a function from its derivative (anti-derivatives).
    • Calculating the area bounded by a function's graph.
  • Types of Integrals:
    • Indefinite Integrals
    • Definite Integrals

7.2 Integration as an Inverse Process of Differentiation

  • Integration is the reverse of differentiation.
  • Examples of anti-derivatives:
    • If f′(x)=extcos(x)f'(x) = ext{cos}(x), then f(x)=extsin(x)+Cf(x) = ext{sin}(x) + C
    • If f′(x)=x2f'(x) = x^2, then f(x)=x33+Cf(x) = \frac{x^3}{3} + C
    • If f′(x)=exf'(x) = e^x, then f(x)=ex+Cf(x) = e^x + C

7.3 Methods of Integration

  • Integration Techniques:
    1. Integration by Substitution
    2. Integration using Partial Fractions
    3. Integration by Parts

7.3.1 Integration by Substitution

  • Change the variable to simplify the integral.

7.3.2 Integration using Trigonometric Identities

  • Use known identities to find integrals involving trigonometric functions.

7.4 Integrals of Some Particular Functions

  • Standard Integral Formulas:
    1. ∫dxx2−a2=12alog⁡∣x−ax+a∣+C\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C
    2. ∫dxa2−x2=12alog⁡∣a+xa−x∣+C\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C
    3. ∫dxx2+a2=1atan⁡−1xa+C\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C

7.5 Definite Integral

  • Denoted by ∫abf(x) dx\int_a^b f(x) \, dx
  • Represents the area under the curve from aa to bb.
  • Value is given by F(b)−F(a)F(b) - F(a) if FF is an anti-derivative of ff.

7.6 Fundamental Theorem of Calculus

  • First Fundamental Theorem: If ff is continuous on [a,b][a, b], then A′(x)=f(x)A'(x) = f(x).

7.7 Integration by Parts

  • Formula: ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du

7.8 Common Integral Formulas

  • Basic Integrals:
    • ∫ex dx=ex+C\int e^x \, dx = e^x + C
    • ∫ dx=x+C\int \, dx = x + C
    • ∫sin⁡x dx=−cos⁡x+C\int \sin x \, dx = -\cos x + C
    • ∫cos⁡x dx=sin⁡x+C\int \cos x \, dx = \sin x + C

Exercises

  • Evaluate integrals and apply the methods discussed in this chapter.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

x2+1x^2 + 1

B.

tan⁡−1x\tan^{-1}x

C.

log⁡x\log x

D.

All of the above
Correct Answer: D

Solution:

A good substitution is usually a function whose derivative also appears in the integrand, such as x2+1x^2+1, tan⁡−1x\tan^{-1}x, log⁡x\log x, sin⁡x\sin x or cos⁡x\cos x.

Chapter Concept:

Substitution Pattern Recognition

A.

Change of variable

B.

Reversal of limits

C.

Interval splitting

D.

Symmetry
Correct Answer: B

Solution:

The property of reversal of limits states that reversing the limits of integration changes the sign of the definite integral.

Chapter Concept:

Definite Integrals and Properties

A.

Express the numerator as a linear combination of the derivative of the denominator and a constant.

B.

Use partial fraction decomposition directly on the numerator.

C.

Differentiate the denominator and equate it to the numerator.

D.

Integrate the numerator and denominator separately.
Correct Answer: A

Solution:

The correct approach is to express the numerator as a linear combination of the derivative of the denominator and a constant, i.e., A(2ax+b)+BA(2ax+b) + B, to simplify the integration process.

Chapter Concept:

Linear Numerator over Quadratic Denominator

A.

Let u=x2u = x^2, then du=2x dxdu = 2x \, dx

B.

Let u=sin⁡(x2)u = \sin(x^2), then du=cos⁡(x2) dxdu = \cos(x^2) \, dx

C.

Let u=xu = x, then du=dxdu = dx

D.

Let u=cos⁡(x2)u = \cos(x^2), then du=−sin⁡(x2) dxdu = -\sin(x^2) \, dx
Correct Answer: A

Solution:

The substitution u=x2u = x^2 simplifies the integral to 12∫sin⁡(u) du\frac{1}{2} \int \sin(u) \, du, which is easier to evaluate.

Chapter Concept:

Integration by Substitution

A.

∫dx1−x2\int \frac{dx}{\sqrt{1-x^2}}

B.

∫dx1+x2\int \frac{dx}{1+x^2}

C.

∫dxx2−1\int \frac{dx}{\sqrt{x^2-1}}

D.

∫dxx2+a2\int \frac{dx}{x^2+a^2}
Correct Answer: A

Solution:

The integral ∫dx1−x2\int \frac{dx}{\sqrt{1-x^2}} is the standard form that leads to the inverse sine function, sin⁡−1(x)+C\sin^{-1}(x) + C.

Chapter Concept:

Standard Inverse Trigonometric Integral Forms

A.

8

B.

16

C.

4

D.

12
Correct Answer: B

Solution:

Using the symmetry property ∫abf(x) dx=∫abf(a+b−x) dx\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx, we have f(x)=x2f(x) = x^2 and f(2−x)=(2−x)2f(2-x) = (2-x)^2. Thus, ∫02(x2+(2−x)2) dx=∫02(x2+x2) dx=2∫02x2 dx=2[x33]02=2×83=163\int_{0}^{2} (x^2 + (2-x)^2) \, dx = \int_{0}^{2} (x^2 + x^2) \, dx = 2 \int_{0}^{2} x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_{0}^{2} = 2 \times \frac{8}{3} = \frac{16}{3}. Therefore, the correct answer is 16.

Chapter Concept:

Definite Integral Symmetry Property

A.

A(x)A(x) is an increasing function for all xx.

B.

A(x)A(x) is a decreasing function for all xx.

C.

A(x)A(x) is constant for all xx.

D.

A(x)A(x) has a maximum at x=0x=0.
Correct Answer: A

Solution:

Since e−t2>0e^{-t^2} > 0 for all tt, A′(x)=e−x2>0A'(x) = e^{-x^2} > 0. Thus, A(x)A(x) is an increasing function for all xx.

Chapter Concept:

Area Function and Applications

A.

π2\frac{\pi}{2}

B.

π\pi

C.

π4\frac{\pi}{4}

D.

π3\frac{\pi}{3}
Correct Answer: A

Solution:

Completing the square for the denominator gives x2+4x+5=(x+2)2+1x^2 + 4x + 5 = (x+2)^2 + 1. The integral becomes ∫−111(x+2)2+1 dx\int_{-1}^{1} \frac{1}{(x+2)^2 + 1} \, dx, which evaluates to π2\frac{\pi}{2} after substitution and integration.

Chapter Concept:

Completing Square for Quadratic Denominators

A.

The integral evaluates to 8.

B.

The integral evaluates to 10.

C.

The integral evaluates to 12.

D.

The integral evaluates to 14.
Correct Answer: C

Solution:

The integral can be evaluated by finding the antiderivative of the integrand. The antiderivative of x2+2x+1x^2 + 2x + 1 is x33+x2+x\frac{x^3}{3} + x^2 + x. Evaluating this from 0 to 2 gives: (233+22+2)−(033+02+0)=83+4+2=243+83=323=12.\left(\frac{2^3}{3} + 2^2 + 2\right) - \left(\frac{0^3}{3} + 0^2 + 0\right) = \frac{8}{3} + 4 + 2 = \frac{24}{3} + \frac{8}{3} = \frac{32}{3} = 12.

Chapter Concept:

Definite Integral as Limit of Sum

A.

The constant of differentiation

B.

The constant of integration

C.

The variable of integration

D.

The integrand
Correct Answer: B

Solution:

The symbol CC represents the constant of integration, which accounts for the family of antiderivatives.

Chapter Concept:

Integral Notation and Terminology

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True or False

Correct Answer: True

Solution:

The integral of sec⁡2x\sec^2 x with respect to xx is indeed tan⁡x+C\tan x + C, where CC is the constant of integration.

Chapter Concept :

Standard Trigonometric Integrals

Correct Answer: True

Solution:

The property of linearity of indefinite integrals states that the integral of a sum is the sum of the integrals. This is expressed as ∫[f(x)+g(x)] dx=∫f(x) dx+∫g(x) dx\int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx.

Chapter Concept :

Properties of Indefinite Integrals

Correct Answer: True

Solution:

The definite integral is defined as having a unique value, which can be evaluated using the antiderivative of the function over the interval [a, b].

Chapter Concept :

Definite Integrals and Properties

Correct Answer: False

Solution:

Integration by partial fractions is used for decomposing proper rational functions into simpler fractions. Improper rational functions are first reduced by division.

Chapter Concept :

Integration by Partial Fractions

Correct Answer: True

Solution:

In integral calculus, the notation ∫f(x) dx\int f(x) \, dx is used to denote the indefinite integral of the function f(x)f(x) with respect to the variable xx. The function f(x)f(x) inside the integral is referred to as the integrand.

Chapter Concept :

Integral Notation and Terminology

Correct Answer: True

Solution:

Integration by substitution is indeed a technique used to simplify integrals by changing the variable of integration, often transforming the integral into a standard form.

Chapter Concept :

Integration by Substitution

Correct Answer: True

Solution:

The Fundamental Theorem of Calculus connects the concept of the antiderivative with the definite integral, stating that the integral of a function over an interval can be computed using its antiderivative.

Chapter Concept :

Fundamental Theorem of Calculus

Correct Answer: False

Solution:

Improper rational functions are first reduced by division before applying integration by partial fractions.

Chapter Concept :

Integration by Partial Fractions

Correct Answer: True

Solution:

The integral of sec⁡2x\sec^2 x is indeed tan⁡x+C\tan x + C because the derivative of tan⁡x\tan x is sec⁡2x\sec^2 x.

Chapter Concept :

Standard Integrals

Correct Answer: True

Solution:

According to the Fundamental Theorem of Calculus, if F′(x)=f(x)F'(x) = f(x), then the definite integral ∫abf(x) dx\int_{a}^{b} f(x) \, dx is calculated as F(b)−F(a)F(b) - F(a), where F(x)F(x) is an antiderivative of f(x)f(x).

Chapter Concept :

Fundamental Theorem of Calculus

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