Area Problem

Section 5.1 Area Problem

For a given function defined on a closed interval, let’s try to find the area that its graph traps above the axis minus the area it traps below the axis. For example, if we plot \(f(x)=x\) on \(-1 \le x \le 2\) we get

Notice that this graph traps some area above the axis (in a triangle with legs of length 2 and 2 over \(0 \le x \le 2\)) and some area below the axis (in a triangle with legs of length 1 and 1 over \(-1 \le x \le 0\)). Do you see them? Here, let’s highlight them.

If we subtract the area below from the area above we get \(2 - 0.5 = 1.5\text{.}\) Of course, that is a case where we have well established formulas for calculating areas (triangles in this case). What if the graph produces shapes for which we do *not* have familiar geometric formulas. Consider, for example, \(f(x)=x^2-1\) on the interval \(0 \le x \le 2\text{.}\)

Now what? That’s the area problem. Namely, find a way to compute area trapped above the axis minus the area trapped below the axis for a given function. This area trapped above minus area trapped below needs some notation and language. So....

Definition 5.1.1. Definition of Integral.

For a function \(f\) definied on a closed interval \([a,b]\) that produces familiar geometric shapes whose areas we know, we denote the area trapped above the axis minus the area trapped below the axis by

\begin{equation*} \int_{a}^{b} f(x)dx \end{equation*}

and we refer to this as *the integral of \(f\) on \([a,b]\)*.

A little thought suggests that, for familiar geometric shapes, integrals have the following *linearity* properties.

\begin{equation*} \int_{a}^{b} (rf(x)+sg(x))dx = r\int_{a}^{b} f(x)dx + s\int_{a}^{b} g(x)dx \end{equation*}

And when \(a \lt c \lt b\text{,}\)

\begin{equation*} \int_{a}^{b} f(x)dx = \int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx \end{equation*}

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