What is Integration?

 

Integration is an operation which helps you determine the area under the graph of a function.

integration_rules

 

A B C
Integral Integral sign Function Tells us which variable we’re integrating
Example \int 2b + 4 db

 

Notice that we don’t always have to use ‘dx,’ as this part of an integration formula will depend on which variable we’re integrating.

 

When we plot the example from the table above, the integration of this function can give us the area underneath it.

integration_definite

 

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Indefinite Integral

An indefinite integral is an integral whose bounds are limited to the graph of the function.

limit_integration

 

A B C
Integral Integral sign which has no upper or lower bound Function Tells us which variable we’re integrating

 

As you can see, an indefinite integral has no upper or lower bound. This means that this integrated formula can give us the area under any point on the graph:

integration_indefinite

 

Definite Integral

A definite integral has bounds that are defined in the integration equation.

definite_integration_limits

 

A B C
Integral Integral sign which has upper bound of m and lower bound of n Function Tells us which variable we’re integrating

 

As you can see, the main difference between an indefinite and a definite integral is that a definite integral is defined by bounds of the interval [n,m].

integration_limits

In order to get the area of the interval between n and m, we simply do the following:

integration_definite_example

 

Step 1 Integrate the function \int \limits_2^5 (2x + 1) dx = \int \limits_2^5 (x^2 + x)
Step 2 Plug in m to the integrated equation (5)^2 + 5 = 30
Step 3 Plug n to the integrated equation (2)^2 + 2 = 6
Step 4 Get m - n 30 - 6 = 24

 

Integration Rules

Now that you know the logic behind what integration is and why we use it, take a look at integration rules.

Constant

 

Integral Function Rule
Constant \int a dx ax

 

Here are some examples:

 

Integral Function Result
1 \int 4 dx 4x + c
2 \int 1 dx x + c
3 \int 2 dx 2x + c

 

Integration can be thought of doing the opposite of finding the derivative. Take a look at the example below.

 

Function Derivative Integral
4x + 2 4*1(x^{1-1}) + 2*0(x^{0-1}) \int f' dx = \int 4 dx
f = 4x + 2 f’ =4 \int f' = 4s + c

 

We can never get the exact constant of the original formula, but we can put ‘c’ there to stand in for the constant that was there.

 

Fraction

There are two ways you can integrate a fraction. First, convert the fraction into a regular number.

integration_power

Instead of integrating the fraction, we can use common exponent rules to turn it into two numbers multiplied by each other.

integration_power_rule

Another way you can integrate a fraction is to see if it follows the rule for integrating reciprocals.

integration_natural_log

You can combine any of the methods we discussed, as well as use them with integration by parts and u-substitution.

 

Sum

The sum rule of integration is a simple rule. It says that if you have to integrate the sum of two functions, the result is equal to the integration of both functions added together.

integration_sum_rule

 

What Example
f(x) First function x+3
g(x) Second function 5x

 

Let’s take a look at an example.

integration_power_sum

First, we simplify the expression.

integration_sum_rule_example

Next, because it is the addition of two functions, we can integrate them separately.

sum_rule_example

Difference

The difference rule of integration is similar to the sum rule. It says that if you have to integrate the difference of two functions, the result is equal to the integration of one function minus the other.

difference_rule_example

 

What Example
f(x) First function 2x
g(x) Second function (that we subtract from the first) 4x+3

 

Let’s take a look at an example.

integration_sum_example

First, we simplify the expression.

integration_sum_rule_difference

Next, because it is the difference of two functions, we can integrate them separately.

example_integration

 

Power

Powers are a very important part of integration. The rule is as follows.

power_rule_integration
x The variable we’re integrating
n The exponent, or power

 

What about when you don’t have any power? Take a look at this example.

power_integration

Well, in this case, the power is actually there. It is equal to one, because a number to the power of 1 is just itself. So, it would be the following.

power_integration_example

 

Integration by Parts

Integration by parts is exactly as it sounds. When you have two functions multiplied together, you can use integration by parts to integrate them.

integration_parts_rule

 

What Example
u The first function 3x
v The second function (x+2)^{2}
u’ The derivative of the first function 3

 

Keep in mind, however, that you should only integrate by parts if you have two functions that will not produce more functions that need to be integrated by parts. To understand this, let’s take an example that we actually can use this method for.

sin_integration

Now, we follow the rule.

integration_parts_sin

Simplifying only the highlighted term, we get the following.

cos_integration_parts

Now, imagine if we didn’t have this term, but instead had this.

sin_cos_integration

When we simplify this term, we end up with another two functions multiplied by one another. This means that we would need to keep integrating by parts infinitely. If you end up with this, it is better to integrate by u-substitution.

 

Integration by Substitution

Integration by substitution, also called u-substitution, is an easy way to integrate. Let’s take a look at the notation and a simple example.

u_substitution_integration

 

u The expression we want to replace 4x
du The derivative of the expression 4 dx
dx Solve for dx \frac{du}{4}

 

Continuing with our example from above, let’s take a look at the steps in u-substitution.

 

1 Replace the u and dx values in the formula \int \sqrt{u} \frac{du}{4} =

\int  \sqrt{u} du \frac{1}{4}

2 Simplify the expression \frac{1}{4} \int  \sqrt{u} du =

\frac{1}{4} \int  u^{\frac{1}{2}} du

3 Integrate like normal \frac{1}{4} (\frac{ u^{\frac{1}{2} + \frac{2}{2}}}{\frac{1}{2} + \frac{2}{2}} du
4 Simplify \frac{1}{4} (\frac{u^{\frac{3}{2}}}{\frac{3}{2}})
5 Plug the original value of u back in to get the final result \frac{1}{4}*\frac{2}{3} (u^{\frac{3}{2}})

= \frac{1}{6}(4x)^{\frac{3}{2}}

 

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Danica

Located in Prague and studying to become a Statistician, I enjoy reading, writing, and exploring new places.