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

Prerequisite Topics

Demonstration of some proficiency in these topics is important.
The topics are:
(1.) Factoring
(2.) Partial Fractions
(3.) Exponents and Logarithms
(4.) Trigonometry
(5.) Differential Calculus

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 several 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 \\[5ex] dy = x^n dx \\[3ex] then\:\: \displaystyle\int dy = \displaystyle\int x^n dx \\[5ex] y = \dfrac{x^{n + 1}}{n + 1} + C \:\:where\:\: n \ne -1 \\[5ex] If\:\: \dfrac{dy}{dx} = ax^n \:\:where\:\: a \:\:is\:\:a\:\:constant \\[5ex] then\:\: \displaystyle\int dy = \displaystyle\int ax^n dx \\[5ex] y = a\displaystyle\int x^n = \dfrac{ax^{n + 1}}{n + 1} + C \:\:where\:\: n \ne -1 \\[5ex] $ Let us do some examples
Ask students to tell you the question numbers applicable to the concept you just taught them.

Teacher: What happens if n = -1?
Student: It would make the expression undefined.
Because you cannot divide anything by zero.
Teacher: That is correct.
Student: How do we integrate such a function?
Teacher: It is a special integral
Remember when we had to find the derivative of $\ln x$
Student: Yes, the derivative is $\dfrac{1}{x}$
Teacher: This implies that the antiderivative (integral) of $\dfrac{1}{x}$
Student: is $\ln x$
Teacher: That is correct.

Integration by Algebraic Substitution

Algebraic Substitution method of Integration requires some form of algebraic substitution to simplify the function before finding the integral.
Some functions are integrated using the Algebraic Substitution method
How do we know those kind of functions?
How do we know when to use Algebraic Substitution?

Typically, when we 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 some forms of functions that are integrated by Algebraic Substitution

$ (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} \\[7ex] (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} \\[7ex] (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} \\[7ex] (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} \\[7ex] (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} \\[7ex] (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] $ Please NOTE:
(1.) Some functions may not look exactly like the ones discussed above, but may look similar.
In that case, we will need to do some operations before finding the integral of those functions using algebraic substitution.

(2.) Some functions may not look similar to the ones we discussed, but may still be integrated using algebraic substitution.

Let us do some examples
Ask students to tell you the question numbers applicable to the concept you just taught them.

Integration by Trigonometric Substitution

Trigonometric Substitution method of Integration requires some form of trigonometric substitution to simplify the function before determining the integral.
Provided the simplified form of the function does not have a variable as the numerator, functions of these types are integrated using trigonometric substitution.

$ (1.)\;\; \displaystyle\int \dfrac{dx}{a^2 + x^2} \;\;\;use\;\;the\;\;substitution:\;\;x = a\tan\theta \\[5ex] (2.)\;\; \displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} \;\;\;use\;\;the\;\;substitution:\;\;x = a\sin\theta \\[5ex] (3.)\;\; \displaystyle\int \sqrt{a^2 - x^2} \;\;\;use\;\;the\;\;substitution:\;\;x = a\sin\theta \\[5ex] $ Let us discuss each function.

$ (1.)\:\: \displaystyle\int \dfrac{dx}{a^2 + x^2} \\[5ex] x = a\tan\theta \\[3ex] a^2 + x^2 \\[3ex] = a^2 + (a\tan\theta)^2 \\[3ex] = a^2 + a^2\tan^2\theta \\[3ex] = a^2(1 + \tan^2\theta) \\[3ex] 1 + \tan^2\theta = \sec^2\theta ... Pythagorean\;\;Identity \\[3ex] = a^2\sec^2\theta \\[5ex] x = a\tan\theta \\[3ex] \dfrac{dx}{d\theta} = a\sec^2\theta \\[5ex] dx = a\sec^2\theta d\theta \\[5ex] x = a\tan\theta \\[3ex] a\tan\theta = x \\[3ex] \tan\theta = \dfrac{x}{a} \\[5ex] \theta = \tan^{-1}\left(\dfrac{x}{a}\right) \\[5ex] \implies \\[3ex] \displaystyle\int \dfrac{dx}{a^2 + x^2} dx \\[5ex] = \displaystyle\int \dfrac{a\sec^2\theta}{a^2\sec^2\theta} d\theta \\[5ex] = \displaystyle\int \dfrac{d\theta}{a} \\[5ex] = \dfrac{1}{a} \displaystyle\int d\theta \\[5ex] = \dfrac{1}{a} * \theta + C \\[5ex] = \dfrac{1}{a} * \tan^{-1}\left(\dfrac{x}{a}\right) + C \\[5ex] = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right) + C \\[5ex] \therefore \color{red}{\displaystyle\int \dfrac{dx}{a^2 + x^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right) + C} \\[7ex] $ $ (2.)\:\: \displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} \\[5ex] x = a\sin\theta \\[3ex] a^2 - x^2 \\[3ex] = a^2 - (a\sin\theta)^2 \\[3ex] = a^2 - a^2\sin^2\theta \\[3ex] = a^2(1 - \sin^2\theta) \\[3ex] \sin^2\theta + \cos^2\theta = 1 ... Pythagorean\;\;Identity \\[3ex] \cos^2\theta = 1 - \sin^2\theta \\[3ex] = a^2\cos^2\theta \\[3ex] \sqrt{a^2 - x^2} \\[3ex] = \sqrt{a^2\cos^2\theta} \\[3ex] = a\cos\theta \\[5ex] x = a\sin\theta \\[3ex] \dfrac{dx}{d\theta} = a\cos\theta \\[5ex] dx = a\cos\theta d\theta \\[5ex] x = a\sin\theta \\[3ex] a\sin\theta = x \\[3ex] \sin\theta = \dfrac{x}{a} \\[5ex] \theta = \sin^{-1}\left(\dfrac{x}{a}\right) \\[5ex] \implies \\[3ex] \displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} \\[5ex] = \displaystyle\int \dfrac{a\cos\theta d\theta}{a\cos\theta} \\[5ex] = \displaystyle\int d\theta \\[3ex] = \theta + C \\[3ex] = \sin^{-1}\left(\dfrac{x}{a}\right) + C \\[5ex] \therefore \color{red}{\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right) + C} \\[7ex] $ $ (3.)\:\: \displaystyle\int \sqrt{a^2 - x^2} dx \\[3ex] x = a\sin\theta \\[3ex] a^2 - x^2 \\[3ex] = a^2 - (a\sin\theta)^2 \\[3ex] = a^2 - a^2\sin^2\theta \\[3ex] = a^2(1 - \sin^2\theta) \\[3ex] \sin^2\theta + \cos^2\theta = 1 ... Pythagorean\;\;Identity \\[3ex] \cos^2\theta = 1 - \sin^2\theta \\[3ex] = a^2\cos^2\theta \\[3ex] \sqrt{a^2 - x^2} \\[3ex] = \sqrt{a^2\cos^2\theta} \\[3ex] = a\cos\theta \\[5ex] x = a\sin\theta \\[3ex] \dfrac{dx}{d\theta} = a\cos\theta \\[5ex] dx = a\cos\theta d\theta \\[5ex] x = a\sin\theta \\[3ex] a\sin\theta = x \\[3ex] \sin\theta = \dfrac{x}{a}...eqn.(1) \\[5ex] \theta = \sin^{-1}\left(\dfrac{x}{a}\right)...eqn.(2) \\[5ex] \sin^2\theta = \left(\dfrac{x}{a}\right)^2 = \dfrac{x^2}{a^2} \\[5ex] \cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{x^2}{a^2} = \dfrac{a^2 - x^2}{a^2} \\[5ex] \cos\theta = \sqrt{\dfrac{a^2 - x^2}{a^2}} = \dfrac{\sqrt{a^2 - x^2}}{a}...eqn.(3) \\[5ex] \implies \\[3ex] \displaystyle\int \sqrt{a^2 - x^2} \\[3ex] = \displaystyle\int (a \cos\theta)(a \cos\theta) d\theta \\[3ex] = \displaystyle\int a^2\cos^2\theta d\theta \\[3ex] = a^2\displaystyle\int \cos^2\theta d\theta \\[5ex] $ Keep $a^2$
Let's focus on integrating $\cos^2\theta$ with respect to $\theta$

$ \cos 2\theta = \cos\theta\cos\theta - \sin\theta\sin\theta ... Sum\;\;Formula \\[3ex] \cos 2\theta = \cos^2\theta - \sin^2\theta \\[3ex] \sin^2\theta + \cos^2\theta = 1 ... Pythagorean\;\;Identity \\[3ex] \sin^2\theta = 1 - \cos^2\theta \\[3ex] \implies \\[3ex] \cos 2\theta = \cos^2\theta - (1 - \cos^2\theta) \\[3ex] \cos 2\theta = \cos^2\theta - 1 + \cos^2\theta \\[3ex] \cos 2\theta = 2\cos^2\theta - 1 \\[3ex] 2\cos^2\theta - 1 = \cos 2\theta \\[3ex] 2\cos^2\theta = \cos 2\theta + 1 \\[3ex] \cos^2\theta = \dfrac{\cos 2\theta + 1}{2} \\[5ex] \implies \\[3ex] \displaystyle\int \cos^2\theta d\theta \\[5ex] = \displaystyle\int \left(\dfrac{\cos 2\theta + 1}{2}\right) d\theta \\[5ex] = \displaystyle\int \left(\dfrac{1 + \cos 2\theta}{2}\right) d\theta \\[5ex] = \displaystyle\int \dfrac{1}{2} d\theta + \displaystyle\int \dfrac{\cos 2\theta}{2} d\theta \\[5ex] = \dfrac{1}{2}\theta + \displaystyle\int \dfrac{\cos 2\theta}{2} d\theta \\[5ex] $ Keep $\dfrac{1}{2}\theta$

Let's focus on integrating $\dfrac{\cos 2\theta}{2}$ with respect to $\theta$

$ \displaystyle\int \dfrac{\cos 2\theta}{2} d\theta \\[5ex] = \dfrac{1}{2} \displaystyle\int \cos 2\theta d\theta \\[5ex] $ Keep $\dfrac{1}{2}$

Let's focus on integrating $\cos 2\theta$ with respect to $\theta$

$ \displaystyle\int \cos 2\theta d\theta \\[5ex] \underline{Integration\;\;by\;\;Algebraic\;\;Substitution} \\[3ex] let\;\; p = 2\theta \\[3ex] \dfrac{dp}{d\theta} = 2 \\[5ex] \dfrac{d\theta}{dp} = \dfrac{1}{2} \\[5ex] d\theta = \dfrac{dp}{2} \\[5ex] = \displaystyle\int \cos p \dfrac{dp}{2} \\[5ex] = \dfrac{1}{2}\displaystyle\int \cos p dp \\[5ex] = \dfrac{1}{2}\sin p \\[5ex] = \dfrac{1}{2}\sin 2\theta \\[5ex] = \dfrac{\sin 2\theta}{2} \\[5ex] \sin 2\theta = \sin (\theta + \theta) = \sin \theta \cos \theta + \cos \theta \sin \theta ... Sum\;\;Formula \\[3ex] \sin 2\theta = 2\sin\theta\cos\theta \\[3ex] \implies \\[3ex] = \dfrac{2\sin\theta\cos\theta}{2} \\[5ex] = \sin\theta\cos\theta \\[3ex] $ Reviewing everything we kept beginning from the last one
Get back the $\dfrac{1}{2}$ we kept...by multiplying

$ = \dfrac{1}{2}\sin\theta\cos\theta \\[5ex] $ Get back the $\dfrac{1}{2}\theta$ we kept...by adding

$ = \dfrac{1}{2}\theta + \dfrac{1}{2}\sin\theta\cos\theta \\[5ex] = \dfrac{1}{2}(\theta + \sin\theta\cos\theta) \\[5ex] $ Get back the $a^2$ we kept...by multiplying

$ = a^2 * \dfrac{1}{2}(\theta + \sin\theta\cos\theta) \\[5ex] = \dfrac{a^2}{2}(\theta + \sin\theta\cos\theta) \\[5ex] $ The question had only two variables: a and x; not $\theta$, not $\sin\theta$ and not $\cos\theta$
So, we need to substitute for the values of $\theta$, $\sin\theta$, and $\cos\theta$
Remember we expressed them in terms of a and x
In other words, please review eqns.(1), (2.), and (3.)

$ = \dfrac{a^2}{2}\left[\sin^{-1}\left(\dfrac{x}{a}\right) + \dfrac{x}{a} * \dfrac{\sqrt{a^2 - x^2}}{a}\right] \\[5ex] = \dfrac{a^2}{2}\left[\sin^{-1}\left(\dfrac{x}{a}\right) + \dfrac{x\sqrt{a^2 - x^2}}{a^2}\right] \\[5ex] \therefore \color{red}{\displaystyle\int \sqrt{a^2 - x^2}dx = \dfrac{a^2}{2}\left[\sin^{-1}\left(\dfrac{x}{a}\right) + \dfrac{x\sqrt{a^2 - x^2}}{a^2}\right] + C} \\[7ex] $ Please NOTE:
(1.) The coefficient of $x^2$ must be 1, just as it is 1 in all the three cases that we discussed.
If we are given a question where the coefficient of $x^2$ is not 1, we have to perform some operations to make sure it is 1.

(2.) Some functions may not look exactly like the ones discussed above, but may look similar.
In that case, we will need to do some operations before finding the integral of those functions using trigonometric substitution.

(3.) Depending on the function and if the question requires that we use trigonometric substitution without giving the substitution:
If the function involves:

$ (a.)\;\; \displaystyle\int \sqrt{a^2 + x^2} dx \;\;\;use\;\;the\;\;substitution:\;\;x = a\tan\theta \\[5ex] (b.)\;\; \displaystyle\int \sqrt{x^2 - a^2} dx \;\;\;use\;\;the\;\;substitution:\;\;x = a\sec\theta \\[5ex] $ Let us do some examples
Ask students to tell you the question numbers applicable to the concept you just taught them.

Integration by Parts

Integration by Parts also known as Integration of Products is the technique of integration used to integrate products (or quotients) of two terms.
A quotient or division can also be seen as a product because $2 \div 3 = \dfrac{2}{3}$ can also be written as $\dfrac{1}{3} * 2$

Recall:
From the Product Rule of Derivatives:

$ y = u * v \\[3ex] \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}...Product\;\;Rule \\[5ex] Integrate\;\;each\;\;term\;\;with\;\;respect\;\;to\;\;x \\[3ex] \displaystyle\int \dfrac{dy}{dx} \cdot dx = \displaystyle\int u\dfrac{dv}{dx} \cdot dx + \displaystyle\int v\dfrac{du}{dx} \cdot dx \\[5ex] \displaystyle\int dy = \displaystyle\int udv + \displaystyle\int vdu \\[5ex] y = \displaystyle\int udv + \displaystyle\int vdu \\[5ex] But\;\;y = uv \\[3ex] \implies \\[3ex] uv = \displaystyle\int udv + \displaystyle\int vdu \\[5ex] \displaystyle\int udv + \displaystyle\int vdu = uv \\[5ex] \therefore \displaystyle\int udv = uv - \displaystyle\int vdu...Integration\;\;by\;\;Parts\;\;Formula \\[5ex] $ Notable NOTES:
For the applicable question, we need to:
(1.) Express the function as a product of two terms
The first term shall be the u
This is the term to be differentiated so we can find du
The second term shall be the dv (not v)
This is the term to be integrated so we can find v

(2.) Because the second term has to be integrated, it is imporatant to choose a term that is easy to integrate. Because the first term has to be differentiated, it is important to choose a term that is easy to differentiate.
Here are some recommendations:
If these kind of functions are in the integral, use them as the first term: u in this order:
(a.) logarithmic functions
(b.) positive exponents of the independent variable
(c.) exponential functions

(3.) Some questions will require at least two operations of Integration of Parts
In other words, you may need to perform Integration by Parts more than one time for some questions.

No worries, guys. We'll get through this. 🙂

Let us do some examples
Ask students to tell you the question numbers applicable to the concept you just taught them.

Derivatives of Integrals

$ (1.)\;\; \dfrac{d}{dy} \displaystyle\int_{a(y)}^{b(y)} f(x) dx = f[b(y)] * b'(y) - f[a(y)] * a'(y) \\[7ex] (2.)\;\; \dfrac{d}{dx} \displaystyle\int_{a(x)}^{b(x)} f(x) dx = f[b(x)] * b'(x) - f[a(x)] * a'(x) \\[7ex] (3.)\;\; \dfrac{d}{dy} \displaystyle\int_{a(y)}^{b(y)} f(y) dx = f'(y)[b(y) - a(y)] + f(y)[b'(y) - a'(y)] \\[7ex] (4.)\;\; \dfrac{d}{dy} \displaystyle\int_{a(y)}^{b(y)} f(x) dy = f(x)[b'(y) - a'(y)] \\[7ex] (5.)\;\; \dfrac{d}{dy} \displaystyle\int_{a(y)}^{b(y)} f(y) dy = f[b(y)] * b'(y) - f[a(y)] * a'(y) \\[7ex] (6.)\;\; \dfrac{d}{dx} \displaystyle\int_{a(y)}^{b(y)} f(x) dy = f'(x)[b(y) - a(y)] \\[7ex] (7.)\;\; \dfrac{d}{dx} \displaystyle\int_{a(x)}^{b(x)} f(y) dx = f(y)[b'(x) - a'(x)] \\[7ex] (8.)\;\; \dfrac{d}{dx} \displaystyle\int_{a(y)}^{b(y)} f(x) dx = 0 \\[7ex] (9.)\;\; \dfrac{d}{dx} \displaystyle\int_{a(y)}^{b(y)} f(y) dx = 0 \\[7ex] (10.)\;\; \dfrac{d}{dx} \displaystyle\int_{a(y)}^{b(y)} f(y) dy = 0 \\[7ex] $

Integration by Hyperbolic Substitution