Basic Integration Rules

Antiderivatives

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 $\dfrac{1}{8}(2x+1)^4$.

Integration

$ \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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