The Integral Test

If f is positive, continuous, and decreasing for x\geq 1 and a_{n} = f(n) , then \displaystyle\sum_{n=1}^\infty{a_n} and \displaystyle\int_{1}^{\infty}f(x) dx either both converge or both diverge.

To prove this theorem we first partition the interval [1, n] into n-1 unit intervals. The total areas of the inscribed rectangles and the circumscribed triangles are as follows:

\displaystyle\sum_{i=2}^{n}f(i) = f(2) + f(3) + .... +f(n) (inscribed area)

\displaystyle\sum_{i=1}^{n-1}f(i) = f(1) + f(2) + ... + f(n-1) (circumscribed area)

The precise area under the graph of f from x=1 to x=n lies between the inscribed and circumscribed areas which implies \displaystyle\sum_{i=2}^{n}f(i) \leq \displaystyle\int_{1}^{n}f(x) dx \leq \displaystyle\sum_{i=1}^{n-1} f(i) . Using the nth partial sum, S_n = f(1) + f(2) +... + f(n) , we can write this inequality as S_n -f(1) \leq \displaystyle\int_1^n f(x) dx \leq S_{n-1} . Now assuming that \int_1^\infty f(x) dx converges to L, it follows that for n \geq 1 , S_{n} - f(1) \leq L\Rightarrow S_{n} \leq L + f(1)  . Consequently, {S_n} is bounded and monotonic, and thereby by the Bounded Monotonic Sequence Theorem it converges. So, \sum a_n converges.

Arc Length

In this post I want to explain how a definite integral can be used to find the arc length of a curve because I think that it is fascinating that integrals have functions other than just finding the area under curves. Let us start with a definition:

A RECTIFIABLE curve is one that has a finite arc length. A sufficient condition for the graph of a function f to be rectifiable between (a, f(a)) and (b, f(b)) is that  f' be continuous on [a, b] . Such a function is continuously differentiable on [a, b] , and its graph on the interval [a, b] is a smooth curve.

Now let’s consider a function y = f(x) that is continuously differentiable on the interval [a, b] . We can approximate the graph of f by line segments whose endpoints are determined by the partition:

a = x_{0} < x_{1} < x_{2} <....< x_{n} = b

By letting \Delta x_{i} = x_{i} - x_{i-1} and \Delta y_{i} = y_{i} - y_{i-1} , we can approximate the length of the graph by the following:

\displaystyle S \approx \sum_{i=1}^{n} \sqrt{(x_{i} - x_{i-1})^2 +(y_{i} - y_{i-1})^2}

= \displaystyle \sum_{i=1}^{n} \sqrt{(\Delta x_{i})^2 + (\Delta y_{i})^2}

= \displaystyle \sum_{i=1}^{n} \sqrt{(\Delta x_{i})^2 +(\frac{\Delta y_{i}}{\Delta x_{i}})^2(\Delta x_{i})^2}

= \displaystyle \sum_{i=1}^{n} \sqrt{1 + (\frac{\Delta y_{i}}{\Delta x_{i}})^2}(\Delta x_{i})

This approximation becomes more and more precise as ||\Delta|| \to 0 (n \to \infty) . Therefore the length of the graph is S = \displaystyle \lim_{||\Delta|| \to 0} \displaystyle \sum_{i=1}^{n} \sqrt{1 + (\frac{\Delta y_{i}}{\Delta x_{i}})^2}(\Delta x_{i})

Because f'(x) exists for each x in (x_{i-1}, x_i) , the Mean Value Theorem guarantees the existence of c_i in (x_{i-1}, x_i) such that f(x_i) - f(x_{i-1}) = f'(c_i)(x_i - x_{i-1})

\frac{\Delta y_i}{\Delta x_i} = f'(c_i)

Because f' is continuous on [a, b] , it follows that \sqrt{1 + [f'(x)]^2} is also continuous and therefore integrable on [a, b] , which implies that S = \displaystyle \lim_{||\Delta|| \to 0} \displaystyle \sum_{i=1}^{n} \sqrt{1 + [f'(c_i)]^2}(\Delta x_i)                                                                               = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx