30. The First Fundamental Theorem of Calculus

Dated: 30-10-2024

Theorem

If \(f(x)\) is a continuous function¹ over the interval² \([a, b]\) and \(F(x)\) is its anti-derivative³ then

\[\int_a^b f(x) dx = F(b) - F(a)\]

Proof

Let us divide the interval² into \(n\) subintervals, such that

\[a < x_1 < x_2 < \ldots < x_{n-1} < b\]

So our subintervals are

\[[a, x_1], [x_1, x_2], \ldots , [x_{n - 1}, b]\]

Let \(x_1^*\) be the midpoint⁴ of interval² \([a, x_1]\) and so on, then using the mean value theorem,⁵

\[F(x_1) - F(a) = F'(x_1^*)(x_1 - a) = f(x_1^*) \Delta x_1\]

\[F(x_2) - F(x_1) = F'(x_2^*)(x_2 - x_1) = f(x_2^*) \Delta x_2\]

And so on, eventually the last one looks like this:

\[F(b) - F(x_{n-1}) = F'(x_n^*)(b - x_{n-1}) = f(x_n^*)\Delta x_n\]

Adding all these equations together, we will get

\[F(b) - F(a) = \sum_{i=1}^{n} f(x_i^*)\Delta x_i\]

Now, we will increase the number of slices \(n\) in such a way that \(max (\Delta x_{i} \to 0)\)

\[F(b) - F(a) = \lim_{max(\Delta x_i \to 0)} \sum_{i=1}^{n} f(x_i^*)\Delta x_i = \int_a^b f(x)dx\]

We can also write it as

\[\int_{a}^{b} f(x) dx = F(x) \bigg]_{a}^{b}\]

Example

The Problem

Evaluate \(\int_0^6 f(x) dx\) if

\[ f (x) = \begin{cases} x^2 & x < 2 \\ 3x - 2 & x \ge 2 \end{cases} \]

Solution

First thing we notice is that it is a discontinous function¹ and the point of discontinuity is \(2\).
We can divide our interval² \([0, 6]\) into \([0, 2)\) and \([2, 6]\).
Then using the properties of integration,³

\[\int_0^6 f(x) dx = \int_0^2 f(x) dx + \int_2^6 f(x) dx\]

\[= \int_0^2 x^2 dx + \int_2^6 3x - 2 dx\]

\[= \frac{x^3}{3} \bigg]_0^2 + \bigg[\frac 3 2 x^2 - 2x \bigg]_2^6\]

\[= \left(\frac{2^3}{3} - \frac{0^3}{3}\right) + \left(\left(\frac 3 2 \cdot 6^2 - 2(6)\right) - \left(\frac 3 2 \cdot 2^2 - 2(2)\right)\right)\]

\[= \left(\frac 8 3 - 0\right) + \left( 42 - 2\right)\]

\[= \frac {128}{3}\]

Mean Value Theorem for Integrals

Here \(M\) is the largest value which \(f(x)\) can output and \(m\) is the smallest.
According to mean value theorem,⁵ there is some value \(c\) in \([a, b]\) where,

\[\int_a^b f(x) dx = f(c) \cdot (b - a)\]

From the diagram,

\[m(b - a) \le \int_a^b f(x) dx \le M(b - a)\]

The inequality⁶ suggests a ranking of areas in following descending order.

Area of rectangle with width \(b - a\) and height \(M\).
Area under the curve \(f(x)\).
Area of rectangle with width \(b - a\) and height \(m\).

We can re-arrange the inequality⁶ as follows:

\[m \le \frac {1}{b - a}\int_a^b f(x) dx \le M\]

\[\because \int_a^b f(x) dx = f(c) \cdot (b - a)\]

\[\therefore \frac {1}{b - a}\int_a^b f(x) dx = f(c)\]

Here \(f(c)\) is also called \(f_{avg}\), the average value of \(f(x)\) with respect to \(x\) over the interval² \([a, b]\).