# Integrating Polynomials

Essentially, we can consider integration to be the reverse process of differentiation. Recall that once we differentiate an expression we obtain its derivative. Similarly, when we integrate an expression we obtain what is known as its **integral**. We sometimes refer to an integral as an **antiderivative**, especially when discussing the Fundamental Theorem of Calculus.

## The Fundamental Theorem of Calculus - Part 1

The **Fundamental Theorem of Calculus **is split into two parts. The formal definitions are too advanced for this page but we will explore what the theorem states loosely. The first part of the theorem, often termed the **First Fundamental Theorem of Calculus**, relates to **indefinite integration**.

### Indefinite Integration

Consider differentiating $y=x^2$ to get $\frac{dy}{dx}=2x$. We obtain this derivative when differentiating $y=x^2+2$, $y=x^2-5$ or $y=x^2+c$ where $c$ is ANY constant. This is because when differentiating, constants disappear. Considering integration as the reverse of differentiation, it is not possible to retrieve the original constant without further information. Hence, when integrating we add on a generic constant known as the **constant of integration**. This is known as **indefinite integration**. See Integrating Polynomials below for more information.

The **First Fundamental Theorem of Calculus **loosely states that, for continuous functions f(x), the indefinite integral $F(x)$ (or antiderivative) of $f(x)$ exists. We find it through integration i.e. $f(x)=F'(x)$. See Definite Integrals for the second part of the theorem. Taking any expression, say $y$, we write its **indefinite integral **as

$\int y\,dx$.

We read it as ‘the integral of $y$ with respect to $x$’.

## Integrating Polynomials

In order to integrate a polynomial, first recall how to differentiate a polynomial.

Differentiating a polynomial term requires first multiplying down by the power then reducing the power by one. If integration is the reverse process, then we can integrate a polynomial term by **increasing the power by one then dividing by the new power**. For example, consider $y=x^3$, differentiating gives $\frac{dy}{dx}=3x^2$. Adding one to the power of this and then dividing by the new power gets us back to the original expression: $y=\frac{3x^3}{3}+c=x^3+c$, where we include the constant of integration as mentioned above. We can use integral notation to show this for a general term:

$\int x^n\, dx=\frac{1}{n+1}x^{n+1}+c$

for $n\ne -1$. We can find the constant we have additional information such as the coordinates of a point. See Example 2. In some exam questions, you may be required to write the expression as a polynomial first. You can also see this in Example 2.

## Questions by Topic

## Examples of Integration

- Find $y$ if $\frac{dy}{dx}=6x^2-3x-4$
- If $f'(x)=4x^{\frac{1}{2}}-\frac{1}{2}x^{\frac{3}{2}}$, find $f(x)$.
- Find $\int y\,dx$ if $y=(2x-3)^2$.

1.

$\begin{array}{l}\int\left(6x^2-3x-4\right)dx\\=\frac{6}{3}x^3-\frac{3}{2}x^2-4x+c\\=2x^3-\frac{3}{2}x^2-4x+c\end{array}$

2.

$\begin{array}{l}\int\left(4x^{\frac{1}{2}}-\frac{1}{2}x^{\frac{3}{2}}\right)dx\\=\frac{4}{\frac{3}{2}}x^{\frac{3}{2}}-\frac{\frac{1}{2}}{\frac{5}{2}}x^{\frac{5}{2}}+c\\=\frac{8}{3}x^{\frac{3}{2}}-\frac{1}{5}x^{\frac{5}{2}}+c\end{array}$

3.

$\begin{array}{l}\int\left(2x-3\right)^2dx\\=\int\left(4x^2-12x+9\right)dx\\=\frac{4}{3}x^3-\frac{12}{2}x^{2}+9x+c\\=\frac{4}{3}x^{3}-6x^2+9x+c\end{array}$.

The curve $C$ with equation $y=f(x)$ passes through the point $(1,2)$. Given that

$f'(x)=\frac{x\times\sqrt[3]{x}-x^5}{x^2}$

find the equation of $C$.

Before we can integrate $f'(x)$ we must write it as a polynomial:

$f'(x)=\frac{x\times x^{\frac{1}{3}}-x^5}{x^2}=\frac{x^{\frac{4}{3}}-x^5}{x^2}=x^{-\frac{2}{3}}-x^3$

It follows that

$f(x)=\int\left(x^{-\frac{2}{3}}-x^3\right)dx=3x^{\frac{1}{3}}-\frac{1}{4}x^4+c$

We can use the fact that the point $(1,2)$ lies on $f(x)$ to find $c$:

$2=3-\frac{1}{4}+c\hspace{10pt}\Longrightarrow\hspace{10pt}c=-\frac{3}{4}$.

Hence,

$f(x)=3x^{\frac{1}{3}}-\frac{1}{4}x^4-\frac{3}{4}$