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Solving Exercises Involving Integrals

A PRC licensed teacher from Manila and a graduate of MaEd Mathematics.

SOLUTION TO SOME INTEGRATION EXERCISES


BASIC INTEGRATION (Finding Integral-Basic)

Course Objectives:

1) To help students develop skills in finding basic integrals.

2) To be able to apply the lessons learned in Algebra in solving integrals.

3) To help students develop patience while doing the exercises in finding the integrals.


SOLUTION TO SOME BASIC INTEGRATION EXERCISES

Introduction:

Finding integrals is one of the fundamental skills in Integral Calculus that should be developed in every student of Calculus. Skills in integration are very much needed for finding integrals has wide range of applications. Among its applications are vector-valued functions and finding areas under a curve. In Electrical Engineering, integration is used to determine the exact length of power cable needed to connect two substations, which are miles away from each other. Integration is also used to find the volume of an object with curved sides such as wine barrels. In Physics, integration is used to calculate the Centre of Mass, Centre of Gravity and Mass Moment of Inertia of a sports utility vehicle.

Finding integrals is one of the most challenging lessons in Calculus. In this lesson, I present several exercises involving integration complete with step by step progressive solution. Beginners will find it easy to do with these step by step procedures in solving these exercises. Calculus need not be really difficult, it is indeed fun.

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In this segment, I present several integration exercises with their corresponding solutions. The solutions presented here are comprehensive to help beginners learn the skills in finding integrals easily. With the aid of some principles in Algebra such as "Laws of Exponent", "Distributive Property of Multiplication", "Square and Cube of a Binomial", the solutions are developed. Hope you will find this one challenging.


This is the main pattern to be followed in these exercises:

Integral of X^n = X^(n+1)/ (n+1) + C


Since the derivative of a constant is zero, any constant may be added to an indefinite integral. This constant is known as the constant of integration.

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Exercise Number One: ∫ ( 2X³ + 7X² + X) dx

Solution: ∫ 2X³ dx + ∫ 7X² dx + ∫ X dx

2 ∫ X³ dx + 7 ∫ X² dx + ∫ X dx

We find the integral by adding one to the power of the variable X and the sum will also be the denominator of the term.

2 X³⁺¹/(3 + 1) + 7 X²⁺¹/(2 + 1) + X¹⁺¹/2 + C

2X⁴/4 + 7X³/3 + X²/2 + C

Answer: X⁴/2 + 7X³/3 + X²/2 + C


Exercise Number Two: ∫ ( 4X^⅓ + 2/X⁻⁴ + 8) dx

∫ 4X^⅓ dx + ∫ 2/X⁻⁴ dx + ∫ 8 dx

Solution: 4 ∫ X^⅓ dx = 4 X^(⅓ ⁺ ¹)/ (1/3 + 1) = 4X^(⁴/³)/(4/3)

= 4(X^(⁴/³))(3/4)

= 3X^(⁴/³)

2 ∫ X⁴ dx = 2X⁴⁺¹/(4+1) = 2X⁵/5

8 ∫ dx = 8X

Answer: 3X^(⁴/³) + 2X⁵/5 + 8X + C

2X⁵/5 + 3X^⁴/³ + 8X + C


Exercise Number Three: ∫ SQRT(X) ( (SQRT(X) + 3X)² dx

This expression can be rewritten as :

∫ (X^½ ) ( X^½ + 3X )² dx

Solution: Expand the binomial: ( X^½ + 3X )²

Remember Square of a binomial: (a + b)² = a² + 2ab + b²

(X^½ + 3X )² = (X^½)² + 2 (X^½) (3X) + (3X)²

= X + 6X^(¹⁺½) + 9X²

= X + 6X^(³/²) + 9X²

∫ X^½ ( X + 6X^³/² + 9X²) dx

∫ (X^³/² + 6X² + 9X^⁵/²) dx

∫ X^³/² dx + ∫ 6X² dx + ∫ 9X^⁵/² dx

∫ X^³/² dx = X^(³/² ⁺¹)/(3/2 + 1) = X^⁵/²/(5/2) = 2/5X^⁵/²

∫ 6X² dx = 6 ∫ X² dx = 6 X²⁺¹/3 = 2X³

∫ 9X^⁵/² dx = 9 ∫ X^⁵/² dx = 9X^⁵/²⁺¹/(5/2 + 1) = 9X^⁷/²/(7/2)

= (2/7) 9X^⁷/²

= 18/7 X^⁷/²

Answer : 2/5X^⁵/² + 2X³ + 18/7X^⁷/² + C

18/7X^⁷/² + 2X³ + 2/5X^⁵/² + C


Exercise Number Four: ∫ (X + 2)³/ (X^¹/³) dx

Solution : Expand the cube of a binomial first.

(a + b)³ = a³ + 3a²b + 3ab² + b³

Applying this formula to (X + 2)³:

X³ + (3)(2)X² + 3X (2)² + 2³ = X³ + 6X² + 12X + 8

The problem can be rewritten as : ∫ (X³ + 6X² + 12X + 8) /X^⅓ dx

∫ X³/X^⅓ dx + 6 ∫ X²/X^⅓ dx + 12 ∫ X/X^⅓ dx + ∫ 8/X^⅓ dx

Applying Laws of Exponent : Quotient of Powers

∫ X^⁽³⁻⅓⁾ dx + 6 ∫ X^⁽²⁻⅓⁾ dx + 12 ∫ X^⁽¹⁻⅓⁾ dx + 8 ∫ X^⁻⅓ dx

∫ X^⁸⁄³ dx + 6 ∫ X^⁵⁄³ dx + 12 ∫ X^²⁄³ dx + 8 ∫ X^⁻⅓ dx

(X^⁸⁄³ ⁺¹)/ (8/3 +1) + 6 (X^⁵⁄³⁺¹)/(5/3 + 1) + 12(X^²⁄³⁺¹)/(2/3 + 1) + 8(X^⁻¹⁄³⁺¹)/(-1/3 + 1) + C

= X^¹¹⁄³/(11/3) + 6 X^⁸⁄³/(8/3) + 12X^⁵⁄³/(5/3) + 8X^²⁄³/(2/3) + C

= 3/11 X^¹¹⁄³ + 6(3/8) X^⁸⁄³ + 12(3/5)X^⁵⁄³ + 8(3/2)X^²⁄³ + C

= 3/11 X^¹¹⁄³ + 18/8 X^⁸⁄³ + 36/5 X^⁵⁄³ + 24/2 X^²⁄³ + C

Answer = 3/11 X^¹¹⁄³ + 9/4 X^⁸⁄³ + 36/5 X^⁵⁄³ + 12 X^²⁄³ + C


Hope you enjoy this learning process well!!!


SOURCE : These exercises were taken from a problem set given to a class in Kalayaan College, Philippines.




























Comments

Cristina Santander (author) from Manila on May 12, 2019:

Thanks for visiting this hub, thanks for the suggestions as well. Stay blessed.

George Dimitriadis from Templestowe on May 12, 2019:

Hi.

You have covered some good examples of the indefinite integral.

At the start, perhaps you should explain the meaning of the C, the constant of integration.

Perhaps some terms comprising negative powers can be included.

Regards

George

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