Integration Notation

As a reminder, integration notation is important to understand. There are two main integrals you will encounter:

 

integration_notation
A B C
Indefinite Integral without bounds Function Variable to be integrated
Definite Integral with upper bound m and lower bound n Function Variable to be integrated

 

In the table above, you can see that the only difference between indefinite and definite variables is that definite variables specify the interval for which the area needs to be calculated.

integration_example_graph

 

A B C
\int \limits_2^5 \int \limits_-2^0 \int \limits_7^10

 

To get the area between an interval, you simply need to take the area of the upper limit minus the area of the lower limit.

integrate_step_by_step

 

Step 1 Find the integral of \int \limits_-2^1 (2x + 3) dx \int \limits_-2^1 (2x^{2} + 3x)
Step 2 Get the area of the upper limit (2(1)^{2} + 3(1)) = 5
Step 3 Get the area of the lower limit (2(-2)^{2} + 3(-2)) = 2
Step 4 Get the area between the limits 5-2 = 3

 

The best Maths tutors available
1st lesson free!
Intasar
4.9
4.9 (26 reviews)
Intasar
£36
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Paolo
4.9
4.9 (13 reviews)
Paolo
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (7 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Ayush
5
5 (28 reviews)
Ayush
£60
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Farooq
5
5 (14 reviews)
Farooq
£40
/h
1st lesson free!
Tom
5
5 (9 reviews)
Tom
£22
/h
1st lesson free!
Intasar
4.9
4.9 (26 reviews)
Intasar
£36
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Paolo
4.9
4.9 (13 reviews)
Paolo
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (7 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Ayush
5
5 (28 reviews)
Ayush
£60
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Farooq
5
5 (14 reviews)
Farooq
£40
/h
1st lesson free!
Tom
5
5 (9 reviews)
Tom
£22
/h
First Lesson Free>

Integration Rules

You can think of integration as the opposite as taking the derivative. Let’s compare each by seeing the rule for taking the integral and derivative or a power.

integrate_derivative

 

Start with a power function f = 2x^{3}
Take the derivative f’ = 2*3x^{3-1} = 6x^{2}
Take the integral of the derivative \int f'(x) = \int 6x^{2} dx = \frac{6x^{2+1}}{2+1} = \frac{6x^{3}}{3} = 2x^{3}

 

In the table above, we can see that by taking the integral of the derivative, we have come back to the original function. Take a look at more integration rules and examples below.

 

Integral Function Result Example
Constant \int a dx ax \int 2 dx = 2x + c
Power \int x^{n} dx \frac{x^{n+1}}{n+1} \int x dx = \frac{x^{1+1}}{1+1} = \frac{x^{2}}{2}
Fraction of x \int \frac{1}{x} dx ln|x| \int \frac{1}{3} dx = ln|3|
Natural Log \int e^{x} dx e^{x} \int e^{2} dx = e^{2}
Exponential \int a^{x} dx \frac{a^{x}}{ln(a)} \int 2^{x} dx = \frac{2^{x}}{ln(2)}
Log \int ln(x) dx x*ln(x) - x \int ln(2x) dx =2x(ln(2x)) - 2x
Sum of two functions \int f(x) + g(x) dx \int f(x) dx + \int g(x) dx \int (x+2) + (x^3) dx = \int (x+2) dx + \int (x^{3}) dx
Two functions subtracted \int f(x) - g(x) dx \int f(x) dx - \int g(x) dx \int (x+2) - (x^3) dx = \int (x+2) dx - \int (x^{3}) dx

 

Integration by Parts

When you have two functions multiplied by each other, you can use integration by parts. The rule is as follows:

integrate_by_parts

 

Element Description
u(x), u The first function
v(x), v The second function
u’(x), u’ The derivative of the first function

 

Say you have the following function to integrate:

integrate_eulers

We can use the rule in order to find the integral.

 

Element Description Result
u(x), u The first function x
v(x), v The second function e^{x}
u’(x), u’ The derivative of the first function 1

 

Plugging this into the rule, we get:

integrate_eulers_by_parts

Now we solve for each term:

 

A x * e^{x}
B e^{x}
C \int 1*e^{x} dx = e^{x}

 

The final result is:

 

    \[ x*e^{x} - e^{x} \]

 

Integration by Substitution

Also called u-substitution, Integration by substitution can be used if you have two functions, one of which can be written as the derivative of the first function.

integrate_parts_example

The goal is to simplify the integration process. So, we can pick the inside function as our ‘u.’

integrate_by_parts_example

Now that we selected a u and have attained the derivative, what we need to do now is get the dx term by itself.

u_substitution

Now, we simply replace the function that we’ve chosen as our u by the u term and replace the dx term with what we just solved.

u_substitution_example

Now, we simplify.

integrate_u_substitution_example

Recall that a function multiplied by a constant in an integral is just that constant multiplied by the integral.

integrateion_by_substitution

Now, instead of the complicated formula we had in the beginning, we can solve the integral easily.

integration_rule

The final step is to simply substitute the ‘u’ back into the equation.

integration_substitution

 

Problem 1

Integrate the following function using any of the rules above:

 

    \[ \int 3(x+1)^{2} dx \]

 

Problem 2

Integrate the following function using any of the rules above:

 

    \[ \int \limits_2^5 (4x + x)^{2} dx \]

 

Problem 3

Integrate the following functions using integration by parts:

 

    \[ \int 4x(x+2)^{3} dx \]

 

Problem 4

Integrate the following function using u-substitution.

 

    \[ \int x^{4}e^{x^{5}} dx \]

 

Problem 5

Integrate the following functions using any of the rules above:

 

    \[ \int 2x(x+1)^{3} - 12x(\sqrt{10 - 2x^{2}})dx \]

 

Solution Problem 1

Let’s integrate this function using u-substitution:

 

    \[ \int 3(x+1)^{2} dx \]

 

    \[ u = x+1 \]

    \[ du = 1 dx \]

 

    \[ \int 3u^{2} du = 3(\frac{u^{2+1}}{2+1}) \]

 

    \[ \int 3u^{2} du = 3(\frac{(x+1)^{3}}{3}) \]

 

Solution Problem 2

Let’s integrate the function:

 

    \[ \int \limits_2^5 (4x + x)^{2} dx = \int \limits_2^5 (5x)^{2} dx = \int \limits_2^5 25x^{2} dx \]

 

    \[ 25 * \int \limits_2^5 \frac{x^{3}}{3}} \]

 

    \[ 25*(\frac{5^{3}}{3}} - \frac{2^{3}}{3}}) = 975 \]

 

Solution Problem 3

For this function we use integration by parts:

 

    \[ \int 4x(x+2)^{3} dx \]

 

The integration of (x+2)^{3} is exactly the same as in problem 1.

    \[ 4x * (\frac{(x+1)^{4}}{4}) - \int (4 * ((x+1)^{4}*\frac{1}{4})) dx \]

 

Let’s focus on the second term.

    \[ 4*\frac{1}{4} \int ((x+1)^{4}) dx \]

The integration of (x+1)^{4} is the same as problem 1.

 

    \[ 4x * (\frac{(x+1)^{4}}{4}) - (\frac{(x+1)^{5}}{5}) \]

 

Solution Problem 4

First, you need to pick which function will be your u term.

u_substitution_rule

Next, take the derivative and solve for dx.

u_substitution_example_integration

Plug everything back into the equation and simplify.

integration_by_substitution

Lastly, plug the u term back in.

eulers_example

 

Solution Problem 5

In this integral, you first need to use the difference rule.

integration_by_parts

Use integration by parts for the first integral.

integration_parts

Use u-substitution for the second integral.

integration_parts_substitution

Finally, put it all together.

integration_parts_example

 

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 5.00/5 - 1 vote(s)
Loading...

Danica

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