Basic Integration Rules

Basic Integration Rules


In many cases in calculus, it is known that the rate of change of one variable with respect to another, but we do not have a formula that relates the variables. In other words, it is known that $\dfrac{dy}{dx}$, but we need to know $y$ in terms of $x$.
The process of finding $y$ from $\dfrac{dy}{dx}$, or $f(x)$ from $f'(x)$, is the reverse process of differentiation. We call it antidifferentiation or integration.
$$ y\xrightarrow{\text{differentiation}} \dfrac{dy}{dx} \\
y\xleftarrow[\text{or integration}]{\text{antidifferentiation}} \dfrac{dy}{dx} \\
f(x)\xrightarrow{\text{differentiation}} f'(x) \\
f(x)\xleftarrow[\text{or integration}]{\text{antidifferentiation}} f'(x)$$
Consider $\dfrac{dy}{dx}=x^3$.
From our work on differentiation, we know that when we differentiate power functions, the index reduces by $1$. We hence know that $y$ must involve $x^4$.
If $y=x^4$ then $\dfrac{dy}{dx}=4x^3$, so if we start with $y=\dfrac{1}{4}x^4$ then $\dfrac{dy}{dx}=x^4$.
However, in all of the cases $y=\dfrac{1}{4}x^4+1$, $y=\dfrac{1}{4}x^4+-2$, $y=\dfrac{1}{4}x^4+100$ and $y=\dfrac{1}{4}x^4+12$ we find $\dfrac{dy}{dx}=x^3$.
There are infinitely many functions of the form $y=\dfrac{1}{4}x^4+c$ where $c$ is an arbitrary constant that will give $\dfrac{dy}{dx}=x^3$.
Ignoring the arbitrary constant, we say that $\dfrac{1}{4}x^4$ is the antiderivative of $x^3$. It is the simplest function that, when differentiated, gives $x^2$.

Example 1

Find the antiderivative of $x^5$.

\( \begin{align} \displaystyle
\dfrac{d}{dx}x^6 &= 6x^5 \\
\dfrac{d}{dx}\dfrac{1}{6}x^6 &= x^5
\end{align} \)
Therefore the antiderivative of $x^5$ is $\dfrac{1}{6}x^6$.

Example 2

Find the antiderivative of $\sqrt{x}$.

\( \begin{align} \displaystyle
\sqrt{x} &= x^{\frac{1}{2}} \\
\dfrac{d}{dx}x^{\frac{3}{2}} &= \dfrac{3}{2}x^{\frac{1}{2}} \\
\dfrac{d}{dx}\dfrac{2}{3}x^{\frac{3}{2}} &= x^{\frac{1}{2}} \\
\dfrac{d}{dx}\dfrac{2}{3}\sqrt{x^3} &= \sqrt{x}
\end{align} \)
Therefore the antiderivative of $\sqrt{x}$ is $\dfrac{2}{3}\sqrt{x^3}$.

Example 3

Find the antiderivative of $(2x+1)^3$.

\( \begin{align} \displaystyle
\dfrac{d}{dx}(2x+1)^4 &= 4(2x+1)^3 \times \dfrac{d}{dx}(2x+1) \\
&= 4(2x+1)^3 \times 2 \\
&= 8(2x+1)^3 \\
\dfrac{d}{dx}\dfrac{1}{8}(2x+1)^4 &= (2x+1)^3
\end{align} \)
Therefore the antiderivative of $(2x+1)^3$ is $\dfrac{1}{8}(2x+1)^4$.


\( \begin{align} \displaystyle
x^3 &= \dfrac{d}{dx}\Big(\dfrac{1}{4}x^4\Big) \\
x^3 &= \dfrac{d}{dx}\Big(\dfrac{1}{4}x^4 +1\Big) \\
x^3 &= \dfrac{d}{dx}\Big(\dfrac{1}{4}x^4 +10\Big) \\
x^3 &= \dfrac{d}{dx}\Big(\dfrac{1}{4}x^4 -7\Big) \\
\therefore \int{x^3}dx &= \dfrac{1}{4}x^4+c
\end{align} \)

Example 4

If $y=x^4+3x^3$, find $\dfrac{dy}{dx}$. Hence find $\displaystyle \int{(4x^3+9x^2)}dx$.

\( \begin{align} \displaystyle
\dfrac{d}{dx}(x^4+3x^3) &= 4x^3+9x^2 \\
\therefore \int{(4x^3+9x^2)}dx &= x^4+3x^3+c
\end{align} \)

Rules for Integration

$$ \large \displaystyle \int{x^n}dx = \dfrac{1}{n+1}x^{n+1}+c$$

Example 5

Find $\displaystyle \int{x^6}dx$.

\( \begin{align} \displaystyle
\int{x^6}dx &= \dfrac{1}{6+1}x^{6+1} +c \\
&= \dfrac{1}{7}x^{7} +c
\end{align} \)

Example 6

Find $\displaystyle \int{\sqrt{x}}dx$.

\( \begin{align} \displaystyle
\int{\sqrt{x}}dx &= \int{x^{\frac{1}{2}}}dx \\
&= \dfrac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1} + c \\
&= \dfrac{1}{\frac{3}{2}}x^{\frac{3}{2}} + c \\
&= \dfrac{2}{3}\sqrt{x^3} + c
\end{align} \)

Example 7

Find $\displaystyle \int{5}dx$.

\( \begin{align} \displaystyle
\int{\sqrt{5}}dx &= \int{5 \times 1}dx \\
&= \int{5 \times x^0}dx \\
&= 5 \times \dfrac{1}{0+1}x^{0+1} + c \\
&= 5 \times x + c \\
&= 5x + c
\end{align} \)

Example 8

Find $\displaystyle \int{(3x^3 + 4x^4)}dx$.

\( \begin{align} \displaystyle
\int{(3x^3 + 4x^4)}dx &= \dfrac{3}{3+1}x^{3+1} + \dfrac{4}{4+1}x^{4+1} +c \\
&= \dfrac{3}{4}x^{4} + \dfrac{4}{5}x^{5} +c
\end{align} \)

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