# Calculating the Area Under a Curve: An Intuitive Introduction

##### Introduction

One of the most fundamental problems that gave birth to integral calculus was the calculation of areas bounded by curves. The Greeks used method of exhaustion to remedy this problem, particularly on finding the area of a circle. After Descartes’ and Fermat’s invention of Coordinate Geometry, algebraic solutions were utilized. One of the strategies used to find the area under the function $f$ between $a$ and $b$ is to divide it into sub-intervals and form rectangles as shown in the first figure. Clearly, as the number of rectangles increases, the sum of all the areas of the rectangles gets closer and closer to the area of under the curve.

##### A Specific Case

To explain further, let us take a specific case of the function $f(x) = x^2$ and find the area under the curve bounded by $x=1$, $x = 2$, and the x-axis. As we have mentioned above, we will divide the intervals, and for convenience, we use equal intervals, and construct rectangles as shown in the second figure. In this case, we see to it that the upper left corner of all the rectangles touch the graph; the right graph or the center of the graph may also be used.

If we let $A_1, A_2, A_3, A_4$ and $A_5$ be the areas of the rectangles above (ordered from left to right), then the total area of the rectangles are $A_1 + A_2 + A_3 + A_4 + A_5$

which equals $\displaystyle \sum_{i=1}^5 A_i$

using the summation notation.

If we let $x_1, x_2, x_3, x_4$ and $x_5$ be the coordinates of the leftmost part of the rectangles, then the upper left corner of the first rectangle that touches the graph have coordinates $(x_1, f(x_1))$ where $f(x_1)$ is  its height  (Can you see why?).  The base of the rectangle is $x_2 - x_1$. If we let $\Delta x_1 = x_2 - x_1$, the base of the rectangle, then the area of the first rectangle is $f(x_1) \Delta x_1$. Using this notation, the sum of the areas of the five rectangles is $f(x_1) \Delta x_1 + f(x_2) \Delta x_2 + f(x_3) \Delta x_3 + f(x_4) \Delta x_4 + f(x_5) \Delta x_5$

which is equal to $\displaystyle\sum_{i = 1}^5 f(x_i) \Delta x_i$.

Now, we note two things. First, the sum of the areas of all the rectangles between the $x = 1$ and $x = 2$ always less than the area under the curve. Second, as we have mentioned above, as we increase the number of rectangles, the better our approximation.

So, to approximate the area under the curve, we create as many rectangles as we can.  If we create $n$ rectangles where $n$ for a very large $n$, the sum of the areas of all $n$ rectangles  is $\displaystyle\sum_{i = 1}^n f(x_i) \Delta x_i$.

Now, as $n$ gets larger and larger, or as $n$ approaches infinity, the sum of the areas of rectangles approaches the area under the curve. In fact, the limit  of the sum of the areas of the rectangles as $n$ approaches infinity is equal to the area under the curve. In calculus, the area under the curve $f(x)$ between $x = a$ and $x = b$ is denoted by $\int_{a}^{b} f(x) dx$

Using both notations, we have $\displaystyle \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x_i= \int_{a}^b f(x) d x$.

So, in our specific case above, $\displaystyle\lim_{n \to \infty} \sum_{i=1}^n x_i^2 \Delta x = \int_{1}^2 x^2 dx$.

The notation $\int_{a}^{b} f(x) dx$ means the area under the curve $f(x)$ from $a$ to $b$.