28. Area as Limits

Dated: 30-10-2024

Definition of Area

Let us define area under a curve, to be a sum of areas of rectangles which are created as a result of dividing the interval¹ \([a, b]\) into \(n\) slices.

We are taking \([a, b]\) into consideration. The width of this slice is \(b - a\).
Then we divide this width into \(n\) slices. The width of each slice will be \(\frac{b - a}{n}\)

Then we take the sum of areas of these rectangles.
If \(R_n\) is the nth rectangle and \(R\) is the whole region which comes out as a result of summation then

\[A = area(R) = \lim_{n \to \infty} area(R_n)\]

The width of each rectangle would be \(\Delta x = \frac{b - a}{n}\).
The height of each rectangle would be \(f(x_i)\) where \(i\) is the index, pointing to each rectangle at a time.
Hence, area of ith rectangle is \(f(x_i) \cdot \Delta x\)

\[area(R_n) = f(x_1) \cdot \Delta x + f(x_2) \cdot \Delta x + \ldots + f(x_n) \cdot \Delta x\]

\[= \Delta x \left(f(x_1) + f(x_2) + \ldots + f(x_n)\right)\]

\[= \Delta x \sum_{i = 1}^{n} f(x_i)\]

\[\because A = area(R) = \lim_{n \to \infty} area(R_n)\]

\[A = \lim_{n \to \infty} \sum_{i = 1}^{n} f(x_i) \cdot \Delta x\]

Example

Problem

Find area under the line² \(y = x\) over the interval¹ \([a, b]\).

Solution

Approach 1

As we know, the graph creates a triangle, we can divide it into pieces

Then, to find the desired area, we perform the following

The bigger triangle has base \(2 - 0 = 2\) and height being \(2\).
Therefore, \(A_1 = \frac 1 2 (2) (2) = 2\)
The smaller triangle has base \(1 - 0 = 0\) and height being \(1\) as well.
Therefore, \(A_2 = \frac 1 2 (1)(1) = \frac 1 2\)
The area of our desired section is \(A_3 = A_1 - A_2\).

\[A_3 = 2 - \frac 1 2 = \frac 3 2 = 1.5\]

Approach 2

We will use our previous conclusion

\[A = \lim_{n \to \infty} \sum_{i = 1}^{n} f(x_i) \cdot \Delta x\]

Here \(\Delta x = \frac{2 - 1}{n} = \frac 1 n\)
Let's discuss \(f(x_i)\). Since our equation is \(f(x) = y = x\), therefore, \(f(x_i) = x_i\).
We know that the width is \(\Delta x = \frac 1 n\) and each \(x_i\) is apart by \(\Delta x\) distance.
Hence, we can select each slice using \(i \cdot \frac 1 n\).
We will also add \(1\) which acts as our offset because the interval¹ starts from \(1\).

\[\therefore f(x_i) = 1 + \frac i n\]

Plug these values in the equation above,

\[A = \lim_{n \to \infty} \sum_{i = 1}^{n} \left(1 + \frac i n\right) \frac 1 n\]

Using the properties of sigma³

\[= \lim_{n \to \infty} \frac 1 n \sum_{i = 1}^{n} \left(1 + \frac i n\right)\]

\[= \lim_{n \to \infty} \frac 1 n \left(\sum_{i = 1}^{n} 1 + \sum_{i = 1}^{n} \frac i n\right)\]

\[= \lim_{n \to \infty} \frac 1 n \left(n + \frac 1 n \sum_{i = 1}^n i\right)\]

\[= \lim_{n \to \infty} \frac 1 n \left(n + \frac 1 n \cdot \frac{n (n + 1)}{2} \right)\]

\[= \lim_{n \to \infty} \frac 1 n \left(n + \frac{(n + 1)}{2} \right)\]

\[= \lim_{n \to \infty} \frac 1 n \left(\frac{2n + (n + 1)}{2} \right)\]

\[= \lim_{n \to \infty} \frac 1 n \left(\frac{3n + 1}{2} \right)\]

\[= \lim_{n \to \infty} \left(\frac{3n + 1}{2n} \right)\]

\[= \lim_{n \to \infty} \left(\frac{3}{2} + \frac{1}{2n} \right)\]

\[\because \lim_{n \to \infty} \frac{1}{2n} = 0\]

\[= \lim_{n \to \infty} \left(\frac{3}{2} + \frac{1}{2n} \right) = \frac 3 2 = 1.5\]