# How to Interpret and Calculate the Definite Integral

Integrals are mainly split into two categories: Definite and indefinite integrals. The definite integral is used for integrals that are limited by a function graph $f\left(x\right)$, the $x$-axis and two values on the $x$-axis (areas and volumes). The graph may lie both above and below the $x$-axis. It is important that you know which areas you are looking at, as it may affect the answer when you calculate.

Theory

### TheDefiniteIntegral

$\begin{array}{cc}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.17em}{0ex}}dx=F\left(b\right)-F\left(a\right)& \\ {F}^{\prime }\left(x\right)=f\left(x\right)& \end{array}$

 ${\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.17em}{0ex}}dx=F\left(b\right)-F\left(a\right),\phantom{\rule{2em}{0ex}}{F}^{\prime }\left(x\right)=f\left(x\right)$

Note! If the graph of $f\left(x\right)$ lies below the $x$-axis between $a$ and $b$, ${\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.17em}{0ex}}dx$ calculates those parts of the area as negative. Therefore, in the figure above, the definite integral will not calculate the shaded area. Instead, it calculates the area that lies above the $x$-axis minus the area that lies below the $x$-axis. To find the shaded area, you have to split the integral at the zeros and take the sum of the definite integrals where the function is positive minus the sum of the definite integrals where the function is negative.

Note! Each individual exercise determines whether you should take into account the geometrical interpretation or not. If the exercise asks you to find the definite integral, you should calculate directly and not think about how the graph lies relative to the $x$-axis. But, if the exercise asks you to find an area, you have to take the sign into account and split the integral into the positive and negative parts!

Example 1

Calculate the definite integral ${\int }_{0}^{\frac{\pi }{3}}\mathrm{sin}x\phantom{\rule{0.17em}{0ex}}dx$

In this case, you can solve the integral directly. It looks like this: $\begin{array}{llll}\hfill {\int }_{0}^{\frac{\pi }{3}}\mathrm{sin}x\phantom{\rule{0.17em}{0ex}}dx& =-\mathrm{cos}x|{}_{0}^{\frac{\pi }{3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\mathrm{cos}\frac{\pi }{3}-\left(-\mathrm{cos}0\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\frac{1}{2}-\left(-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Calculate the area between $x=0$, $x=\frac{3\pi }{2}$ and the graph of $\mathrm{cos}x$

In this case you have to find the area, so you must figure out where the function lies both over and under the $x$-axis. The first thing to do is to find the zeros of the function in the interval: $\begin{array}{cc}\mathrm{cos}x=0& \\ ⇓& \\ x=\frac{\pi }{2}+n2\pi \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}x=-\frac{\pi }{2}+n2\pi & \end{array}$

You see that $x=\frac{\pi }{2}$ lies in the interval $x\in \phantom{\rule{-0.17em}{0ex}}\left[0,b\frac{3\pi }{2}\right]$. The second zero in the interval is $-\frac{\pi }{2}+2\pi =\frac{3\pi }{2}$.

From the figure, you can see that the graph lies above the $x$-axis on the interval $\phantom{\rule{-0.17em}{0ex}}\left[0,\frac{\pi }{2}\right)$ (Area 1) and under the $x$-axis on the interval $\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ (Area 2). Therefore, you have to calculate the integral for each of them.

Area 1: $\begin{array}{llll}\hfill {A}_{1}& ={\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}x\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}x|{}_{0}^{\frac{\pi }{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\frac{\pi }{2}-\left(-\mathrm{sin}0\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1-0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Area 2: $\begin{array}{llll}\hfill {A}_{2}& ={\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\mathrm{cos}x\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}x|{}_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\frac{3\pi }{2}-\phantom{\rule{-0.17em}{0ex}}\left(\mathrm{sin}\frac{\pi }{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-1-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The area of the whole region is therefore:

Example 3

What happens to the area if you do not take into account how the graph lies relative to the $x$-axis, when it lies both above and under?

In the figure below, you can see that the area you are calculating consists of the areas $A$, $-A$ and $B$. That is, the area should be like in Example 2:

 $\phantom{\rule{-0.17em}{0ex}}|A|+\phantom{\rule{-0.17em}{0ex}}|-A|+\phantom{\rule{-0.17em}{0ex}}|B|=2A+B$

But, since parts of the graph lie above the $x$-axis ($A$) and other parts of the graph lie under the $x$-axis ($-A$ and $B$), $A$ and $-A$ cancels each other out. You are then left with the area $B$—and the integral gives you the answer:

 $A+\left(-A\right)+B=A-A+B=B$

It is obvious that $B$ is not the area of $2A+B$. Therefore, you have to pay attention to whether the task is asking for an area or a calculation of the definite integral!

Note!
• In cases where the whole graph is above the $x$-axis, the calculation of the area and the theoretical calculation for the definite integral are the same.

• In cases where the whole graph is under the $x$-axis, the calculation of the area and the theoretical calculation for the definite integral are very similar. They just have opposite signs.