How do you find a definite integral by computing an area?

Hint: We first define the definition and the mathematical expression for area under the curve. For continuous function $y=f\left( x \right)$ in the interval of $a\le x\le b$, the integral takes form of $I=\int\limits_{a}^{b}{f\left( x \right)dx}$. Then we use an example to understand the concept better.

Complete step-by-step answer:
Let $y=f\left( x \right)$ be a single-valued, continuous and monotonic increasing function defined in the interval $a\le x\le b$.
The whole area under the curve gets divided into $n$ equal parts each of length $h$.
Therefore, the intervals are defined as $nh=b-a$.
We take the small length of $h$ as $dx$ in the integral form, which defines a small interval.
The integral form becomes $I=\int\limits_{a}^{b}{f\left( x \right)dx}$.
Now we take an example.
We try to find the area of the function $\int{\dfrac{{{x}^{2}}-2}{x+1}dx}$ from $\left[ 0,2 \right]$. Let $I=\int\limits_{0}^{2}{\dfrac{{{x}^{2}}-2}{x+1}dx}$.
Now we express the numerator of the faction ${{x}^{2}}-2$ as ${{x}^{2}}-2={{x}^{2}}-1-1$.
Now we break the total function into two sub functions.
We get $\dfrac{{{x}^{2}}-2}{x+1}=\dfrac{{{x}^{2}}-1-1}{x+1}=\dfrac{{{x}^{2}}-1}{x+1}-\dfrac{1}{x+1}$.
We apply the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to get ${{x}^{2}}-1=\left( x+1 \right)\left( x-1 \right)$.
From the fraction we get $\dfrac{{{x}^{2}}-1}{x+1}=\dfrac{\left( x+1 \right)\left( x-1 \right)}{\left( x+1 \right)}=\left( x-1 \right)$.
So, $I=\int\limits_{0}^{2}{\left( x-1 \right)dx}-\int\limits_{0}^{2}{\dfrac{1}{x+1}dx}$.
We know $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$ and $\int{\dfrac{1}{x}dx}=\log \left| x \right|+c$.
We can also form the integration $\int\limits_{0}^{2}{\dfrac{1}{x+1}dx}$ as $\int\limits_{0}^{2}{\dfrac{1}{x+1}d\left( x+1 \right)}$.
This is possible because of $d\left( x+1 \right)=dx$.
Therefore,
$\begin{align}
  & I=\int\limits_{0}^{2}{\left( x-1 \right)dx}-\int\limits_{0}^{2}{\dfrac{1}{x+1}dx} \\
 & =\left[ \dfrac{{{x}^{2}}}{2}-x \right]_{0}^{2}-\left[ \log \left| x+1 \right| \right]_{0}^{2} \\
\end{align}$
We put the values in the equations to get
$I=\left[ \dfrac{{{x}^{2}}}{2}-x \right]_{0}^{2}-\left[ \log \left| x+1 \right| \right]_{0}^{2}=\left[ 2-2 \right]-\left[ \log 3-\log 1 \right]=-\log 3$
Therefore, the definite integral of $\int{\dfrac{{{x}^{2}}-2}{x+1}dx}$ from $\left[ 0,2 \right]$ is equal to $-\log 3$.

Note: We need to understand that the area can also be determined with respect to the other axis. In that case we take the small interval as $dy$ and the interval would have to be on the Y-axis as $c\le y\le d$. The function remains as it is. The integral form becomes ${{I}^{'}}=\int\limits_{c}^{d}{ydy}$.