Q. 18 - Calculus

Question

The definite integral of a function $f$ on an interval $\left[a,b\right]$ is defined as a limit of Riemann sums. How can it be that the sum of the areas of infinitely many rectangles that are each “infinitely thin” is a finite number? On the one hand, shouldn’t it be infinite, since we are adding up infinitely many rectangles? On the other hand, shouldn’t it always be zero, since the width of each of the rectangles is approaching zero as $n\to \infty$?

Step-by-Step Solution

Answer

The sums of the areas of infinitely many rectangles which are infinitely thin is a finite number is proved.

Step 1 . Given information

The definite integral of a function $f$ on an interval $\left[a,b\right]$ is defined as a limit of Riemann sums.

Step 2 . The above proof can be shown with an example.

Let the definite integral be,

${\int }_2^5(5-x)dx$.

The right sum defined for $n$ rectangles on $\left[a,b\right]$ is, $\sum _{k=1}^{n}f\left(x_k\right)\Delta x$.

where, $\Delta x=\frac{b-a}{n},x_k=a+k\Delta x.$

The interval is, $\left[2,5\right]$.

Now,

$$\begin{gathered} ∆x=\frac{5-2}{n} \\ =\frac{3}{n} \end{gathered}$$

And,

$$\begin{gathered} x_k=2+k\left(\frac{3}{n}\right) \\ =2+\frac{3k}{n} \end{gathered}$$

Step 3 . The right sum is,

$$\begin{gathered} \sum _{k=1}^n\left(5-\left(2-\frac{3k}{n}\right)\right)\left(\frac{3}{n}\right)=\left(\frac{3}{n}\right)\sum _{k=1}^n\left(3-\frac{3k}{n}\right) \\ =\frac{3}{n}\sum _{k=1}^{n}3-\frac{3}{n}\sum _{k=1}^n\frac{3k}{n} \\ =\frac{3}{n}\left(3n\right)-\left(\frac{3}{n}\right)\left(\frac{3}{n}\right)\left(\frac{n(n+1)}{2}\right) \\ =\frac{3}{n}\left(3n\right)-\frac{9}{n^2}\left(\frac{n(n+1)}{2}\right) \end{gathered}$$

Therefore, the right sum is, $\frac{3}{n}\left(3n\right)-\frac{9}{n^2}\left(\frac{n(n+1)}{2}\right)$.

Step 4 . Now applying limit on infinite rectangles for infinitely thin rectangles,

$$\begin{gathered} {\int }_2^5(5-x)dx=\underset{n\to \infty }{\lim }\left(\frac{3}{n}\left(3n\right)-\frac{9}{n^2}\left(\frac{n(n+1)}{2}\right)\right) \\ =\underset{n\to \infty }{\lim }\left(9-\frac{9}{n^2}\left(\frac{n^2+n}{2}\right)\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{18-9n^2-9n}{2n^2}\right) \\ =\frac{9}{2} \end{gathered}$$

The exact value is finite.

Therefore, the sums of the areas of infinitely many rectangles which are infinitely thin is a finite number.