Calculus

# Introduction to Integration Calculus: Definitions, Formulas, & Examples

## Rachel McLean

Subject Matter Expert

In this article, we’ll discuss the definition of integral calculus, indefinite vs definite integration, key integration rules, and three essential integration methods.

## What is Integral Calculus?

You are probably already familiar with differentiation, which is the process used to calculate the instantaneous rate of change of a function. What is the difference between integration and differentiation? Well, you can think about integration as the reverse operation of differentiation. Together, differentiation and integration make up the essential operations of calculus and are related by the fundamental theorems of calculus.

Dr. Hannah Fry discusses the fundamental theorem of calculus:

When you integrate some function $f(x)$, you find its antiderivative function, which is often denoted $F(x)$. This function can ‌calculate the area underneath the curve of $f(x)$.

The notation for integrating $f(x)$ looks like this:

$\int f(x)\,dx = F(x) + C$

Here’s a guide for interpreting this integral notation:

• What is $\int$?

The symbol $\int$ is called the integral sign. This symbol indicates that we’re calculating the antiderivative function of $f(x)$.

• What is $f(x)$?

The function $f(x)$ is called the integrand, and it’s the function we’re taking the integral of.

• What is $dx$ in calculus?

These letters represent the differential $dx$. The differential $dx$ indicates that we’re integrating $f(x)$ with respect to the variable $x$.

• What is $F(x)$?

$F(x)$ is the antiderivative function that gives back $f(x)$ when differentiated.

• What is $C$?

The capital letter $C$ represents a constant value called the constant of integration. We’ll talk more about what the constant of integration means later.

• Why is $F(x)$ called the antiderivative function?

When you take the derivative of $F(x)$, you get back $f(x)$ again. To help think about the relationship between a function $f$ and its antiderivative, you can ask the question, “What function $F(x)$ has the derivative $f(x)$?” Their relationship can be thought of like this:

If $\int f(x)\,dx = F(x) + C$, then $F’(x) = f(x)$

Or, it might be easier to think about it like this:

$\int f’(x)\,dx = f(x) + C$

For example,

This relationship is guaranteed by the First Fundamental Theorem of Calculus:

$F'(x) = \frac{d}{dx} \int_a^x f(t)\,dt = f(x)$

Because of this relationship, you might also hear integration referred to as anti-differentiation.

Dr. Tim Chartier discusses why we need antiderivatives to begin with:

## Standard Integration Rules and Theorems

Assuming that $f$ and $g$ are continuous functions, here is a list of the most important integration rules and properties that you should know:

### Sum Rule

$\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx$

### Difference Rule

$\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx$

### Constant Multiplier Rule

$\int kf(x)\,dx = k\int f(x)\,dx$ for some constant $k$

### Power Rule

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for some real number $n$

### Constant Rule

$\int a\,dx = ax + C$ for some constant $a$

### Reciprocal Rules

$\int \frac{1}{x}\,dx = \int x^{-1}\,dx = \ln{|x|} + C$

$\int\frac{1}{ax+b}\,dx = \frac{1}{a} \ln{(ax+b)} + C$

### Exponential and Logarithmic Function Rules

$\int e^x\,dx = e^x + C$

$\int a^x\,dx = \frac{a^x}{\ln{(a)}} + C$, for any positive real number $a$

$\int \ln{(x)}\,dx = x\ln{(x)}-x + C$

### Trigonometric Function Rules

For these rules, assume that x is in radians.

$\int \sin{x}\,dx = -\cos{x} + C$

$\int \cos{x}\,dx = \sin{x} + C$

$\int \sec ^2 x\,dx = \tan{x} + C$

$\int \csc ^2 x\,dx = -\cot{x} + C$

$\int \sec{x}\tan{x}\,dx = \sec{x} + C$

$\int \csc{x}\cot{x}\,dx = -\csc{x} + C$

$\int \frac{dx}{\sqrt{1-x^2}}=\sin ^{-1}x + C$

$\int \frac{-dx}{\sqrt{1-x^2}}=\cos ^{-1}x + C$

$\int \frac{dx}{1+x^2}=\tan ^{-1}x + C$

$\int \sin{(ax)}\,dx = \frac{-\cos{(ax)}}{a}+C$ for some real number $a$

$\int \cos{(ax)}\,dx = \frac{\sin{(ax)}}{a}+C$ for some real number $a$

### Absolute Value Rule

$\int |x|\,dx = \frac{x |x|}{2} + C$

Here are a few short examples to practice these integration rules.

• Practice the power rule:

$\int x^4\,dx = \frac{x^{5}}{5} + C$

• Practice the constant multiplier rule:

$\int 7x^6\,dx = 7\int x^6\,dx = 7 \cdot \frac{x^7}{7} = x^7 + C$

• Practice the constant rule:

$\int 32\,dx = 32x + C$

• Practice the reciprocal rules:

$\int\frac{1}{2x+17}\,dx = \frac{1}{2} \ln{(2x+17)} + C$

• Practice the sum/difference rules:

$\int (x + sin{(x)} - 3^x)\,dx = \int x \,dx + \int \sin{(x)}\,dx - \int 3^x \,dx= \frac{x^2}{2} -\cos{x} - \frac{3^x}{\ln{(3)}} + C$

## Indefinite vs Definite Integrals

There are two types of integrals: indefinite and definite.

Dr. Hannah Fry talks more about indefinite and definite integrals:

The indefinite integral finds the general antiderivative function of $f(x)$, while the definite integral finds the area under the curve of $f(x)$ on a specific interval.

These types of integrals have different outputs. The definite integral outputs a unique number that represents the area enclosed by a function’s curve and the x-axis over some interval $[a, b]$. The indefinite integral outputs a function’s antiderivative function, accompanied by the constant of integration $C$.

So, what exactly is the constant of integration $C$? Let’s figure this out by looking at an example. Consider the function $f(x) = 5x^4$. Using the power rule, observe that $\int (5x^4) \,dx = x^5 + C = F(x)$.

Let’s try plugging some different $C$ values into the antiderivative function $F(x) = x^5 + C$. When we take the derivative of the antiderivative function $F(x)$, we should get our original function $f(x)$ back again.

• If $C$= 4, then $F(x) = x^5+4$, and its derivative is $F’(x) = 5x^4$

• If $C$= 0, then $F(x) = x^5$, and its derivative is $F’(x) = 5x^4$

• If $C$= -150, then $F(x) = x^5-150$, and its derivative is $F’(x) = 5x^4$

Even though their last constant term is different, all of the above $F(x)$ functions have the same derivative $F’(x)$, and $F’(x)$ is always equal to our original integrand function $f(x) = 5x^4$.

This is because the derivative of any constant is zero. Remember that for an integrand $f(x)$, its antiderivative function answers the question, “What function $F(x)$ has the derivative $f(x)$?” Any of the above $F(x)$ functions will satisfy this question.

Since there are infinite constant values we can plug into $C$, the constant of integration $C$ and the antiderivative function $F(x)$ together represent an infinite family of functions. That’s why it’s called “indefinite” integration since there is not one unique antiderivative function.

By contrast, definite integrals give us a unique answer. We also don’t need to worry about the constant of integration anymore. To evaluate the definite integral of a function $f(x)$ on the interval $[a, b]$, we use this notation:

$\int_{a}^{b} f(x)\,dx = A$

The letters $a$ and $b$ are called integral bounds or limits. The letter $a$ represents the lower bound, while $b$ represents the upper bound. We can think of this notation as the area bounded by $f(x)$, the x-axis, and the lines $x=a$ and $x=b$.

To find the definite integral of a function on $[a, b]$, we take the difference between the indefinite integral of the function evaluated at $a$ and the indefinite integral of the function evaluated at $b$. This is called the Second Fundamental Theorem of Calculus.

This theorem clarifies the relationship between the integral and the antiderivative function:

$\int_{a}^{b} f(x)\,dx = F(x)\Big|_a^b = F(b) - F(a)$

Here are the four steps for evaluating a definite integral:

• Step 1 - Find the indefinite integral $F(x)$ using the integral rules.

• Step 2 - Find $F(b)$ by plugging $b$ into $F(x)$.

• Step 3 - Find $F(a)$ by plugging $a$ into $F(x)$.

• Step 4 - Take the difference $F(b) - F(a)$. Since we subtract these values, the constant of integration $C$ cancels out, which is why we can ignore it.

For example, let’s evaluate $f(x) = x + 1$ on $[2, 4]$. The notation for this looks like $\int_{2}^{4} (x + 1)\,dx$.

• Step 1 - Using the power and constant rules, $\int( x + 1)\,dx = \frac{x^2}{2} + x$.

• Step 2 - $F(4) = \frac{4^2}{2} + 4= 12$

• Step 3 - $F(2) = \frac{2^2}{2} + 2 = 4$

• Step 4 - $F(4) - F(2) = 12 - 4 = 8$

This value represents the area under the curve of $f(x)$ on $[2, 4]$.

## 3 Ways to Calculate Integrals

Below, we’ll discuss three primary techniques for evaluating more complex integrals.

### 1. U-Substitution

U-Substitution reverses the chain rule for derivatives and is used to integrate composite functions. We need to rewrite our integral in terms of $u$ and $du$, so that it looks like this:

$\int f(g(x))g’(x)\,dx = \int f(u)\,du$

Here are the four steps to integrating with u-substitution:

1. Pick $u$, the "inside" part of the chain rule.

2. Differentiate $u$ to find $du$. If necessary, rearrange the problem algebraically so that $du$ perfectly matches what’s left inside the integral.

3. Substitute $u$ and $du$ into the integrand and integrate.

4. Substitute the original values back into the resulting function and add the constant of integration to your final answer.

#### Example

Let’s evaluate $\int \frac{10x}{(5x^2+1)^2}\,dx$. Let $u = 5x^2 + 1$. Then, $du = 10x\,dx$. Now, we’ll substitute $u$ and $du$ into the integral.

$\int f(g(x))g’(x)\,dx = \int f(u)\,du$
$\int \frac{10x}{(5x^2+1)^2}\,dx = \int \frac{du}{u^2}$
$= \int u^{-2}\,du$
$= \frac{u^{-1}}{-1}$
$= \frac{-1}{5x^2+1} + C$

### 2. Integration by Parts

Integration by parts uses this formula to integrate a product of functions :

$\int udv = uv - \int v\,du$

We must choose one function in the integrand to represent $u$ and the other to represent $dv$.

Here are four steps to integrate with integration by parts:

1. Separate the integrand into a product of functions by choosing $u$ and $dv$.

2. Differentiate $u$ to find $du$ and integrate $dv$ to find $v$.

3. Plug $u$, $v$, and $du$ into the integration by parts formula.

4. Solve and simplify.

### Example

Let’s evaluate $\int x \sin(x)\,dx$. We’ll let $dv = \sin{(x)}\,dx$ since the integral of this function is easy to find . Then, $u = x$ since that’s what is left over. Now, we need to differentiate $u$ to find $du$ and integrate $dv$ to find $v$.

Using the power rule on $u = x$ and solving for $du$, we find that $du = 1dx$. Integrating $dv$ using the trigonometry rules, we find that $v = \int \sin{(x)} = -\cos{(x)}$. Now we can plug these values into our formula.

$\int u\,dv = uv - \int v\,du$
$\int x \sin(x)\,dx = x(-\cos{(x)} - \int -\cos{(x)}\,dx$
$= -x\cos{(x)} + \int \cos{(x)}\,dx$
$= -x\cos{(x)} + \sin{(x)}$

### 3. Integration by Partial Fractions

Integration by partial fractions is used to integrate rational functions. This method is hard to understand without an example, so be sure to try the example exercise.

Here are nine steps to integrating with this method:

1. Factor the denominator of the function.

2. Decompose the function into a sum of its parts by assigning an unknown variable to each term of the denominator.

3. Combine all terms into one by finding the common denominator, making sure you multiply each numerator appropriately.

4. Multiply out the numerator.

5. Set up an equation that equates the $x$ terms of the original function’s numerator with the $x$ terms in your new equation’s numerator.

6. Set up a second equation that equates the constant terms of the original function’s numerator with the constant terms in your new numerator.

7. Solve for the unknown variables.

8. Plug your solved variables into Step 2.

9. Integrate using the reciprocal rules.

#### Example

Let’s evaluate $\int \frac{x+4}{x^2+x-12}dx$. First, we need to factor the denominator.

$\int \frac{x+4}{x^2+x-12}\,dx = \int \frac{x+4}{(x+4)(x-3)}\,dx$

Now we can do steps 2 - 4.

$\frac{x+4}{(x+4)(x-3)} = \frac{A}{x+4} + \frac{B}{x-3}$
$=\frac{A(x-3) + B(x+4)}{(x+4)(x-3)}$
$=\frac{Ax-3A+Bx+4B}{(x+4)(x-3)}$

Now, we can solve for A and B with steps 5 - 7.

Following those steps, our first equation is $Ax + Bx = x$, which simplifies to $A+B=1$. Our second equation is $-3A+4B=4$. Solving this system of equations, we find that $A = 0$ and $B = 1$.

We can now finish with steps 8 - 9:

$\int \frac{x+4}{(x+4)(x-3)}\,dx = \int \frac{0}{x+4}dx + \frac{1}{x-3}\,dx$
$= \ln{(x-3)}$

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