25. Integration

Dated: 30-10-2024

The Area Problem

The Problem

Given a continous function¹ \(f(x)\) defined over an interval² \([a, b]\), find area between the x-axis and the graph of the function.³

Solution

Let's first define the area to be a function³ of \(x\) where \(x\) is the distance from \(a\) and \(a \le x \le b\).
To know what this area looks like, Newton and Leibniz suggested to find

\[A^{\prime}(x) = \lim_{h \to 0} \frac{A(x + h) - A(x)}{h}\]

Here \(x\) is an arbitrary number with \(h\) being an arbitrary addition to it.
The difference \(A(x + h) - A(x)\) can be thought of as:

The area of this slice can be approximated by computing the area of a rectangle whose base is \(h\) and whose height is \(f(c)\) where \(c\) is the midpoint between \(x\) and \((x + h)\).

From the visualization, as the \(h \to 0\), \(c \to x\).

\[A^{\prime}(x) = \lim_{h \to 0} \frac{A(x + h) - A(x)}{h} \approx \lim_{h \to 0} \frac{f(c) \cdot h}{h} = \lim_{h \to 0} f(c) = f(x)\]

Integrations (Anti-Derivatives)

A function³ \(F(x)\) is called anti-derivative of \(f(x)\) on a given interval² if \(F^{\prime}(x) = f(x)\) for all values of \(x\) within that interval.²

Example

Imagine few functions³ such as:

\[f(x) = x^2\]

\[g(x) = x^2 - 3\]

\[h(x) = x^2 \pm c; c \in \mathbb{R}\]

Then

\[f^{\prime}(x) = g^{\prime}(x) = h^{\prime}(x) = 2x\]

Hence, the more generalized anti-derivative is \(F(x) \pm C\) or more simply, just \(F(x) + C\) where \(C \in \mathbb{R}\).

Indefinite Integral

If we have \(\frac{d}{dx} \left(F(x) + C\right) = f(x)\) then,

\[d \left(F(x) + C\right) = f(x) \cdot dx\]

\[F(x) + C = \int f(x) \cdot dx\]

\(\int\) and \(d\) annihilate each other.

The \(\int\) is called the integral symbol and the \(f(x)\) in this case is called the integrand.

Properties of Indefinite Integral

\[\int c f(x) \, dx = c \int f(x) \, dx\]
\[\int \left(f(x) + g(x)\right) dx = \int f(x) dx + \int g(x) dx\]
\[\int \left(f(x) - g(x)\right) dx = \int f(x) dx - \int g(x) dx\]