For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

# Integral Calculus

I greet you this day,
First: read the notes. Second: view the videos. Third: solve the solved examples and word problems. Fourth: check your solutions with my thoroughly-explained solutions. Fifth: check your answers with the calculators as applicable.
Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome.
If you are my student, please do not contact me here. Contact me via the school's system. Thank you for visiting!!!

Samuel Dominic Chukwuemeka (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

## Objectives

Students will:

(1.) Discuss the concept of the antiderivative of a functions.

(2.) Determine the antiderivative of functions using the Power Rule.

(3.) Determine the antiderivative of functions using the Methods of Integration.

(4.) Integrate exponential functions.

(5.) Integrate logarithmic functions.

(6.) Integrate trigonometric functions.

(7.) Solve applied problems involving the antiderivative of functions.

## Power Rule

Prerequisite Topic: Exponents

The Power Rule of Integrals states that:

$If\:\: \dfrac{dy}{dx} = x^n \\[3ex] dy = x^n dx \\[3ex] then\:\: \displaystyle\int dy = \displaystyle\int x^n dx \\[3ex] y = \dfrac{x^{n + 1}}{n + 1} + C \:\:where\:\: n \ne -1 \\[5ex] If\:\: \dfrac{dy}{dx} = ax^n \:\:where\:\: a \:\:is\:\:a\:\:constant \\[3ex] then\:\: \displaystyle\int dy = \displaystyle\int ax^n dx \\[3ex] y = a\displaystyle\int x^n = \dfrac{ax^{n + 1}}{n + 1} + C \:\:where\:\: n \ne -1$

## Integration by Algebraic Substitution

Some functions are integrated using the Algebraic Substitution method
How do you know those kind of functions?
How do you know when to use Algebraic Substitution?

So, typically; when you are given a function that:
(1.) are of certain forms (we shall discuss these forms)
(2.) are not easy to integrate right away/directly
(3.) may involve some cancellation(s) when you differentiate some part(s) of it and make some substitution(s)
Then, you probably need to use Algebraic Substitution.

Steps in Using Algebraic Substitution
Given a function to integrate:
The function has an independent variable say $x$

(1.) Use a variable to make an "appropriate" substitution of some part of that function
Assume the variable is $p$

(2.) Find the derivative of that variable $wrt$ (with respect to) the independent variable
In other words, find $\dfrac{dp}{dx}$

(3.) Find $\dfrac{dx}{dp}$

(4.) Find $dx$ in terms of $dp$.
In other words, make $dx$ the subject of the formula.

(5.) Substitute appropriately for: $p$ for some part of the function and $dx$ in the main function
You will then have a simpler function in $p$ and $dp$ only.
If you do not have a simpler function in $p$, then try another method.

(6.) Integrate the function (substituted function) $wrt$ $p$

(7.) Substitute back.
You were given the original function in $x$
Your answer should also be a function in $x$

Let us discuss the forms of functions that are integrated by Algebraic Substitution

$\color{red}{(1.)\:\: \displaystyle\int f(x)f'(x) dx} \\[3ex] Let\:\: p = f(x) \\[3ex] \dfrac{dp}{dx} = f'(x) \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{f'(x)} \\[5ex] dx = \dfrac{dp}{f'(x)} \\[5ex] Substitute \\[3ex] \rightarrow \displaystyle\int f(x)f'(x) dx = \displaystyle\int p * f'(x) * \dfrac{dp}{f'(x)} \\[5ex] \displaystyle\int p dp \\[3ex] = \dfrac{p^2}{2} + C \\[5ex] Substitute\:\:back \\[3ex] = \dfrac{f^2(x)}{2} + C \\[5ex] \therefore \color{red}{\displaystyle\int f(x)f'(x) dx = \dfrac{f^2(x)}{2} + C} \\[5ex] Similarly:\:\: \color{red}{\displaystyle\int -f(x)f'(x) dx = \dfrac{-f^2(x)}{2} + C} \\[5ex] \color{red}{(2.)\:\: \displaystyle\int \dfrac{f'(x)}{f(x)} dx} \\[5ex] Let\:\: p = f(x) \\[3ex] \dfrac{dp}{dx} = f'(x) \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{f'(x)} \\[5ex] dx = \dfrac{dp}{f'(x)} \\[5ex] Substitute \\[3ex] \rightarrow \displaystyle\int \dfrac{f'(x)}{f(x)} dx = \displaystyle\int \dfrac{f'(x)}{p} * \dfrac{dp}{f'(x)} \\[5ex] = \displaystyle\int \dfrac{dp}{p} \\[5ex] = \ln p + C \\[3ex] Substitute\:\:back \\[3ex] = \ln f(x) + C \\[3ex] \therefore \color{red}{\displaystyle\int \dfrac{f'(x)}{f(x)} dx = \ln f(x) + C} \\[5ex] Similarly:\:\: \color{red}{\displaystyle\int \dfrac{-f'(x)}{f(x)} dx = -\ln f(x) + C} \\[5ex] \color{red}{(3.)\:\: \displaystyle\int e^{kx} dx ...k \:\:is\:\:a\:\:constant} \\[3ex] Let\:\: p = kx \\[3ex] \dfrac{dp}{dx} = k \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{k} \\[5ex] dx = \dfrac{dp}{k} \\[5ex] Substitute \\[3ex] \rightarrow \displaystyle\int e^{kx} dx = \displaystyle\int e^{p} * \dfrac{dp}{k} \\[5ex] = \dfrac{1}{k} \displaystyle\int e^p dp \\[5ex] = \dfrac{1}{k} * e^p + C \\[5ex] Substitute\:\:back \\[3ex] = \dfrac{1}{k} * e^{kx} + C \\[5ex] = \dfrac{e^{kx}}{k} + C \\[5ex] \therefore \color{red}{\displaystyle\int e^{kx} dx = \dfrac{e^{kx}}{k} + C} \\[5ex] \color{red}{(4.)\:\: \displaystyle\int e^{-kx} dx ...k \:\:is\:\:a\:\:constant} \\[3ex] Let\:\: p = -kx \\[3ex] \dfrac{dp}{dx} = -k \\[5ex] \dfrac{dx}{dp} = -\dfrac{1}{k} \\[5ex] dx = -\dfrac{dp}{k} \\[5ex] Substitute \\[3ex] \rightarrow \displaystyle\int e^{kx} dx = \displaystyle\int e^{p} * -\dfrac{dp}{k} \\[5ex] = -\dfrac{1}{k} \displaystyle\int e^p dp \\[5ex] = -\dfrac{1}{k} * e^p + C \\[5ex] Substitute\:\:back \\[3ex] = -\dfrac{1}{k} * e^{kx} + C \\[5ex] = -\dfrac{e^{kx}}{k} + C \\[5ex] \therefore \color{red}{\displaystyle\int e^{-kx} dx = -\dfrac{e^{kx}}{k} + C} \\[5ex] \color{red}{(5.)\:\: \displaystyle\int (ax + b)^n dx ...a, b, n \:\:are\:\:constants} \\[3ex] Let\:\: p = ax + b \\[3ex] \dfrac{dp}{dx} = a \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{a} \\[5ex] dx = \dfrac{dp}{a} \\[5ex] Substitute \\[3ex] \rightarrow \displaystyle\int (ax + b)^n dx = \displaystyle\int p^n * \dfrac{dp}{a} \\[5ex] = \dfrac{1}{a} \displaystyle\int p^n dp \\[5ex] = \dfrac{1}{a} * \dfrac{p^{n + 1}}{n + 1} + C \\[5ex] Substitute\:\:back \\[3ex] = \dfrac{1}{a} * \dfrac{(ax + b)^{n + 1}}{n + 1} + C \\[5ex] = \dfrac{(ax + b)^{n + 1}}{a(n + 1)} + C \\[5ex] \therefore \color{red}{\displaystyle\int (ax + b)^n dx = \dfrac{(ax + b)^{n + 1}}{a(n + 1)} + C} \\[5ex] Similarly:\:\: \color{red}{\displaystyle\int (ax - b)^n dx = \dfrac{(ax - b)^{n + 1}}{a(n + 1)} + C} \\[5ex] \color{red}{(6.)\:\: \displaystyle\int (-ax + b)^n dx ...a, b, n \:\:are\:\:constants} \\[3ex] Let\:\: p = -ax + b \\[3ex] \dfrac{dp}{dx} = -a \\[5ex] \dfrac{dx}{dp} = -\dfrac{1}{a} \\[5ex] dx = -\dfrac{dp}{a} \\[5ex] Substitute \\[3ex] \rightarrow \displaystyle\int (ax + b)^n dx = \displaystyle\int p^n * -\dfrac{dp}{a} \\[5ex] = -\dfrac{1}{a} \displaystyle\int p^n dp \\[5ex] = -\dfrac{1}{a} * \dfrac{p^{n + 1}}{n + 1} + C \\[5ex] Substitute\:\:back \\[3ex] = -\dfrac{1}{a} * \dfrac{(ax + b)^{n + 1}}{n + 1} + C \\[5ex] = -\dfrac{(ax + b)^{n + 1}}{a(n + 1)} + C \\[5ex] \therefore \color{red}{\displaystyle\int (-ax + b)^n dx = -\dfrac{(ax + b)^{n + 1}}{a(n + 1)} + C} \\[5ex] Similarly:\:\: \color{red}{\displaystyle\int (-ax - b)^n dx = -\dfrac{(ax - b)^{n + 1}}{a(n + 1)} + C} \\[5ex]$

## Integration by Parts

### References

Chukwuemeka, S.D (2016, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.samuelchukwuemeka.com

Stroud, K. A., & Booth, D. J. (2001). Engineering Mathematics ($5^{th}$ ed.). Basingstoke: Palgrave.

Tan, S. T. (2004). Applied Calculus for the Managerial, Life, and Social Sciences ($5^{th}$ ed.). Pacific Grove, CA: Brooks/Cole Publishing Company.

Waner, S., & Costenoble, S. (2013). Applied Calculus ($6^{th}$ ed.). Boston, MA: Cengage Learning.