Finding the area of known geometric figures is easy when you make use of their formulas. But as geometric figures become more complex and cannot be split into known geometric figures, you need a new method to figure out their areas.

A Riemann sum is a way to approximate the area between a graph, the $x$-axis and two given values $a$ and $b$ on the $x$-axis.

One way to approximate a cumbersome area is to split it into rectangles that have a width of $\mathrm{\Delta}x$, and a height equal to the value of the function where the left or right side of the rectangle meets the graph of the function. This way, you get a bunch of rectangles that either almost covers the whole actual area or covers a bit more than the actual area—and it’s easy to measure the area of rectangles. You can also have some rectangles that are larger and some that are smaller if that seems to fit better.

If you let the width $\mathrm{\Delta}x$ of the rectangles tend towards being infinitely small, the approximation will approach the actual area of the region. This sum of infinitely small rectangles is called a Riemann sum. The Riemann sum is equal to the integral of the function.

Mathematically, you say it like this:

Theory

Say you choose $a$ and $b$ on the $x$-axis and that the interval between those two points is divided into smaller parts like this:

$$a<{x}_{1}<{x}_{2}<\cdots <{x}_{n-1}<b$$ |

The distance between $a$ and ${x}_{1}$, ${x}_{1}$ and ${x}_{2}$, and so on, is equal to $\mathrm{\Delta}x$. Here, $\mathrm{\Delta}x$ is equal to the very narrow width of the rectangles.

Let ${c}_{1}$ be an arbitrary point in the interval $\phantom{\rule{-0.17em}{0ex}}\left[a,{x}_{1}\right)$, such that ${c}_{i}$ is an arbitrary point in the interval $\phantom{\rule{-0.17em}{0ex}}\left[{x}_{i-1},{x}_{i}\right)$. Then, the Riemann sum of any function $f(x)$ will look like this:

$$\sum _{i=1}^{n}f({c}_{i})\cdot \mathrm{\Delta}x$$ |

If the limit of the Riemann sum exists when $\mathrm{\Delta}x$ tends to $0$, you say that this limit is the Riemann integral of $f(x)$ in the interval $\phantom{\rule{-0.17em}{0ex}}\left[a,b\right]$:

$$\begin{array}{llll}\hfill {\int}_{a}^{b}f(x)\phantom{\rule{0.17em}{0ex}}dx& =\underset{\begin{array}{c}n\to \infty \\ \mathrm{\Delta}x\to 0\end{array}}{\mathrm{lim}}\sum _{i=1}^{n}f({x}_{i})\cdot \mathrm{\Delta}x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =F(b)-F(a),{F}^{\prime}(x)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =f(x)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

$$\begin{array}{llll}\hfill {\int}_{a}^{b}f(x)\phantom{\rule{0.17em}{0ex}}dx& =\underset{\begin{array}{c}n\to \infty \\ \mathrm{\Delta}x\to 0\end{array}}{\mathrm{lim}}\sum _{i=1}^{n}f({x}_{i})\cdot \mathrm{\Delta}x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =F(b)-F(a),\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{F}^{\prime}(x)=f(x)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

**Note!** Calculations of areas and volumes are an important application of integrals. Using integrals is also the main way to solve differential equations. Therefore, it’s absolutely vital you master integrals so that you can also master differential equations!

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What Are the Important Integration Rules?