# Integration

In this section, we will explore the world of integration. We will do so by learning the examples on integration, integration rule, and other integral concepts. You can check out the interactive simulations to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page.

Let's recap a bit before diving into integration. You already have an intuitive idea of derivatives.

If we differentiate a function \(f\) in an interval \(I\), then we get a family of functions in \(I\).

What if the values of functions are known in \(I\)?

Can you determine the function \(f\)?

Yes, we do it using the method of integration.

Basically, integration is a way of uniting the part to find a whole.

A pizza is integration and each slice of the pizza is the derivative.

Here, the heart of the matter is the integration of a function in its domain of definition.

Let's move further and learn about integration and some of its powerful techniques.

**Lesson Plan**

**What Is Meant By Integration?**

Given the derivative \(f’\) of the function \(f\), the natural question arises is can we determine the function\(f\)?

The functions that give the derivative, that is, the function \(f\) is called antiderivative or integral of \(f’\).

The process of finding the antiderivative is called integration.

For example, the derivative of \(f(x)=x^{3}\) is \(f’(x)=3x^{2}\) and the antiderivative of \(g(x)=3x^{2}\) is \(f(x)=x^{3}\).

**How Integration Is an Inverse Process of Differentiation?**

By definition of integration, we know that integration is an inverse process of differentiation.

If we are given the derivative of a function, the process of finding the original function is called integration.

Consider a function \(f(x)=\dfrac{x^{3}}{3}\).

We know that, \(\dfrac{d}{dx}\left(\dfrac{x^{3}}{3}\right)=x^{2}\;\;\;\;\;\; \cdots (1)\)

Here, we say \(\dfrac{x^{3}}{3}\) is the integral of \(x^{2}\).

In fact, there exists infinite integrals of this function because the derivative of any real constant \(C\) is zero and we can write **(1)** as \(\dfrac{d}{dx}\left(\dfrac{x^{3}}{3}+C\right)=x^{2}\)

So, it is an integration rule to add an arbitrary constant \(C\) from the set of real numbers.

**Notation**

If \(\dfrac{dy}{dx}=f(x)\), we write \(y=\int f(x) dx\).

Here, \(\int f(x) dx\) represents the whole class of integral.

This is read as "Integral of \(f\) with respect to \(x\)".

**Formulas of Integration**

We already know formulas of derivatives of some important functions.

Here are the corresponding integrals of these functions.

**What Are the Different Methods of Integration?**

Sometimes, the inspection is not enough to find the integral of some functions.

There are additional methods to reduce the function in the standard form to find its integral.

Prominent methods are given below.

**Method 1: Substitution**

Suppose, we have to find \(y=\int f(x) dx\).

Let \(x=g(t)\). Then, \(\dfrac{dx}{dt}=g'(t)\).

So, \(y=\int f(x) dx\) can be written as \(y=\int f(g(t)) g'(t)dt\).

For example, let's find the integral of \(f(x)=\sin{mx}\) using substitution.

Let \(mx=t\). Then, \(m\dfrac{dx}{dt}=1\).

So,

\[\begin{align}y&=\int \sin{mx}dx\\&=\dfrac{1}{m}\int \sin{t}dt\\&=-\dfrac{1}{m} \cos{t}+C\\&=-\dfrac{1}{m} \cos{mx}+C\end{align}\]

\(y=\int \sin{mx}dx\) can be written as \(y=\int f(g(t)) g'(t)dt\).

Here is a simulation for you to practice integration by substitution.

**Method 2: Partial Fractions**

Suppose we have to find \(y=\int \dfrac{P(x)}{Q(x)} dx\) where \(\dfrac{P(x)}{Q(x)}\) is an improper rational function.

We reduce it in such a way that \(\dfrac{P(x)}{Q(x)}=T(x)+\dfrac{P_{1}(x)}{Q(x)}\).

Here, \(T(x)\) is polynomial in \(x\) and \(\dfrac{P_{1}(x)}{Q(x)}\) is proper rational function.

The following table shows some rational function and their corresponding rational functions.

For example, let's find the integral of \(f(x)=\dfrac{1}{(x+1)(x+2)}\) using partial fraction.

By using partial fraction we have \(\dfrac{1}{(x+1)(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{x+2} \cdots (1)\).

We will determine the values of \(A\) and \(B\).

On comparing in equation (1), we get \(1=A(x+2)+B(x+1)\).

From this, we have a set of two linear equations.

\[\begin{align}A+B&=0\\2A+B&=1\end{align}\]

On solving these equations we get, \(A=1\) and \(B=-1\).

So, equation (1) can be written as \(\dfrac{1}{(x+1)(x+2)}=\dfrac{1}{x+1}-\dfrac{1}{x+2}\).

Now, solving the integral

\[\begin{align}\int \left(\dfrac{1}{(x+1)(x+2)}\right)dx\\=\int \left(\dfrac{1}{x+1}-\dfrac{1}{x+2}\right)dx\\=\log{|x+1|}-\log{|x+2|}+C\\=\log{\left|\dfrac{x+1}{x+2}\right|}+C\end{align}\]

Experiment with the simulation below to practice this method.

**Method 3: Integration by Parts**

This Integration rule is used to find the integral of two functions.

By product rule of derivatives, we have \(\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\;\;\;\;\;\;\; \cdots (1)\)

Integration on both sides of equation** **(1), we get \(\int u\dfrac{dv}{dx}dx=uv-\int v\dfrac{du}{dx} dx\;\;\;\;\;\;\; \cdots (2)\)

Equation (2) can be written as \(uv=\int u\dfrac{dv}{dx}dx+\int v\dfrac{du}{dx} dx\)

Let \(u=f(x)\) and \(\dfrac{dv}{dx}=g(x)\).

Then, we have \(\dfrac{du}{dx}=f'(x)\) and \(v=\int g(x)dx\).

So, Equation (2) becomes \[\begin{align}\int f(x) g(x)dx\\=f(x) \int g(x)dx-\int [f'(x) \int g(x)dx]dx\end{align}\]

For example, let's find the integral of \(xe^{x}\) using integration by parts.

\[\begin{align}\int xe^{x}dx&=x \int e^{x}dx-\int \left[\dfrac{dx}{dx}\int e^{x}dx\right]dx\\&=x e^{x}-\int [e^{x}]dx\\&=x e^{x}-e^{x}+C\end{align}\]

Try this method in the simulation below.

- Integration is an inverse process of differentiation.
- Always add the constant of integration after determining the integral of the function.
- If two functions, say \(f(x)\) and \(g(x)\) have same derivatives, then \(|f(x)-g(x)|=C\), where \(C\) is some constant.

**Solved Examples**

Example 1 |

Susane wants to integrate the function \(f(x)=2x \sin{(x^2+1)}\) with respect to \(x\).

Can you help her?

**Solution**

Observe that the derivative of \(x^{2}+1\) is \(2x\).

So, we will proceed with integration by substitution.

Let \(x^{2}+1=z\).

Then, \(2x dx=dz\).

\[\begin{align}\int f(x) dx&=\int 2x \sin{(x^2+1)}dx\\&=\int \sin{z}dz\\&=-\cos{z}+C\\&=-\cos{(x^{2}+1)}+C\end{align}\]

\(\therefore\) \(\int 2x \sin{(x^2+1)}dx=-\cos{(x^{2}+1)}+C\) |

Example 2 |

Find the integral \(\int \dfrac{x^2+1}{x^2-5x+6}dx\).

**Solution**

Observe that the function \(\dfrac{x^2+1}{x^2-5x+6}\) is an improper rational function.

So, by long division, this can be written as \(\dfrac{x^2+1}{x^2-5x+6}=1+\dfrac{5x-5}{x^2-5x+6}dx\).

By factorization, we have \(x^2-5x+6=(x-2)(x-3)\).

So, \(1+\dfrac{5x-5}{x^2-5x+6}=1+\dfrac{5x-5}{(x-2)(x-3)}\)

Let's proceed with partial fraction on \(\dfrac{5x-5}{(x-2)(x-3)}\).

\[\begin{align}\dfrac{5x-5}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}\end{align}\].

On comparing, we get, \(5x-5=A(x-3)+B(x-2)\).

From this, we have a set of two linear equations.

\[\begin{align}A+B&=5\\3A+2B&=5\end{align}\]

On solving these equations we get, \(A=-5\) and \(B=10\).

So, \(1+\dfrac{5x-5}{x^2-5x+6}=1-\dfrac{5}{x-2}+\dfrac{10}{x-3}\).

Integrate on both sides of equation,

\[\begin{align}I&=\int \left(1+\dfrac{5x-5}{x^2-5x+6}\right)dx\\&=\int \left(1-\dfrac{5}{x-2}+\dfrac{10}{x-3}\right)dx\\&=\int dx-5\int \dfrac{dx}{x-2}+10\int \dfrac{dx}{x-3}\\&=x-5\log{|x-2|}+10\log{|x-3|}+C\end{align}\]

\(\therefore\) \(x-5\log{|x-2|}+10\log{|x-3|}+C\) |

Example 3 |

Find \(\int e^{x}\sin{x}dx\).

**Solution**

Let \(I_{1}=\int e^{x}\sin{x}dx\).

Since, it is the product of two functions, we will use integration by parts to find the integral.

\[\begin{align}I_{1}&=\int e^{x}\sin{x}dx\\I_{1}&=e^{x} \int \sin{x}dx-\\&\int \dfrac{d}{dx}(e^x)\left(\int \sin{x}dx\right)dx\\I_{1}&=-e^{x}\cos{x}-\int \left(-e^{x}\cos{x}\right)dx\\I_{1}&=-e^{x}\cos{x}+e^{x} \int \cos{x}dx-\\&\int \left[\dfrac{d}{dx}(e^x)\int \cos{x}dx\right]dx\\I_{1}&=-e^{x}\cos{x}+e^{x} \sin{x}-\int e^x \sin{x}dx\\I_{1}&=-e^{x}\cos{x}+e^{x} \sin{x}-I_{1}+C\\2I_{1}&=-e^{x}\cos{x}+e^{x} \sin{x}\\I_{1}&=\dfrac{-e^{x}\cos{x}+e^{x} \sin{x}+C}{2}\\I_{1}&=\dfrac{e^x}{2}\left(\sin{x}-\cos{x}\right)+C\end{align}\]

\(\therefore\) \(\int e^{x}\sin{x}dx=\dfrac{e^x}{2}\left(\sin{x}-\cos{x}\right)+C\) |

Example 4 |

Hailey is busy doing her maths assignment.

She is stuck on one question.

The question says "suppose \(\dfrac{d}{dx}f(x)=4x^3-\dfrac{3}{x^4}\) and \(f(2)=0\), find the function \(f(x)\)."

Can you help her solve this?

**Solution**

Integrate on both sides of the equation \(\dfrac{d}{dx}f(x)=4x^3-\dfrac{3}{x^4}\).

\[\begin{align}\int \dfrac{d}{dx}f(x) dx&=\int \left(4x^3-\dfrac{3}{x^4}\right)dx\\f(x)&=4\left(\dfrac{x^4}{4}\right)-\dfrac{3x^{-3}}{-3}+C\\f(x)&=x^4+\dfrac{1}{x^{3}}+C\end{align}\]

Use the condition \(f(2)=0\) to find the value of \(C\).

\[\begin{align}f(2)&=0\\2^4+\dfrac{1}{2^{3}}+C&=0\\16+\dfrac{1}{8}+C&=0\\C&=-\dfrac{129}{8}\end{align}\]

\(\therefore\) The function is \(f(x)=x^4+\dfrac{1}{x^{3}}-\dfrac{129}{8}\). |

Example 5 |

Calculate \(\int \cos^{2}{x}dx\).

**Solution**

Using the trigonometric identity, we have \(\int \cos^{2}{x}dx=\int\left(\dfrac{1+\cos{(2x)}}{2}\right)\).

So, \(\begin{align}\int \cos^{2}{x}dx&=\int\left(\dfrac{1+\cos{(2x)}}{2}\right)dx\\&=\dfrac{1}{2}\int dx+\dfrac{1}{2}\int \cos{(2x)}dx\end{align}\)

Let \(2x=z\).

Then, \(2dx=dz\).

Using this substitution we get,

\[\begin{align}\int \cos^{2}{x}dx&=\dfrac{1}{2}\int dx+\dfrac{1}{2}\int \dfrac{1}{2}\cos{z}dz\\&=\dfrac{x}{2}+\dfrac{1}{4}\sin{z}+C\\&=\dfrac{x}{2}+\dfrac{1}{4}\sin{(2x)}+C\end{align}\]

\(\therefore\) \(\int \cos^{2}{x}dx=\dfrac{x}{2}+\dfrac{1}{4}\sin{(2x)}+C\) |

- Are all functions integrable? Give reasons for your answer.
- When a polynomial function is integrated, how is the degree of the function gets changed?
- Find the indicated integral.

\[\begin{align}\int \dfrac{dx}{3x^2+13x-10}\end{align}\]

**Interactive Questions**

**Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.**

**Let's Summarize**

We hope you enjoyed learning about Integration** **with the examples and practice questions. Now, you will be able to easily solve problems on integration, examples on integration, integration rule, and integrals.

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**Frequently Asked Questions (FAQs)**

## 1. What is the integration of 1?

The integration of 1 is \(x+C\)

## 2. How integration is used in real life?

The applications of integration in real-life is mentioned below.

- In electrical engineering, we need a cable to connect two substations that are miles apart from each other. Integration helps us to find the exact length of the cable.
- In physics, integration is used to find the center of mass, the center of gravity, the velocity of an object, etc.
- In epidemiology, integral calculus is used to study the spread of an infectious disease.

## 3. Why is the integration important?

Calculus is centered on the concepts of derivatives and integration.

In mathematics, we use integration to find the areas, volumes, displacement, etc.

In fact, the concept of integration in calculus gave birth to integral calculus.