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 The Basic Formula of 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 \(f(x)=x^{3}\).

**What are the Three Integration Methods? **

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

These 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 \(y=\int f(g(t)) g'(t)dt\).

**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.

Hence we have:

**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.

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

**What Are The Different Types of Integration?**

In general, there are two types of integrals.

**Definite Integration**

These are the integrations that have a pre-existing value of limits; thus making the final value of integral definite.

**Indefinite Integration**

These are the integrations that do not have a pre-existing value of limits; thus making the final value of integral indefinite.

Here, \(c\) is the integration constant.

**What Are the Different Integration Formulas?**

We already know the formulas of derivatives of some important functions.

Here are the corresponding integrals of these functions.

- 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.
- In general, there are two types of integrations:

Definite Integrations (the value of the integral is definite)

Indefinite Integrations (the value of the integral is indefinite)

**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 |

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 3 |

Rachel wants to understand the difference between integral and integration.

Can you help her?

**Solution**

The process of finding the antiderivative is called integration.

On the other hand, the value of the function found by the process of integration is called an integral.

\(\therefore\) Integration and differentiation are different. |

- 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.

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

**Let's Summarize**

The mini-lesson targeted the fascinating concept of Integration Formulas. The math journey around integration formulas started with what a student already knew and went on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever.

**About Cuemath**

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

**Frequently Asked ****Questions ****(FAQs)**

**1. What is the formula of integral UV? **

Ans: The formula for integral UV is used to integrate the product of two functions.

**2. What is the derivative formula?**

Ans: The derivative formula is the instantaneous rate of change of a function.

**3. How do you integrate LN? **

Ans: Use the following steps to integrate LN:

1) Take U= LN and V= 1.

2) Apply integration by parts rule

**4. How do you integrate?**

Ans: We can use the below steps to integrate:

1) Firstly define a small part of an object in certain dimensions which on adding repetitively makes the whole object.

2) Use integration formulae over that small part along the varying dimensions.

**5. What is the use of Integration?**

Ans: The integration is used to find the area of any objects. Real-life examples are to find the center of mass of an object.

**6. What are Integration techniques? **

Ans: Substitution, Integration by parts, reverse chain rule, and partial fraction expansion are some integration techniques.