Consider the equation \(y = 4\). If we plot this on two axes, x and y, it has no slope; it's just a horizontal line. We could also write the equation as \(y = 0x + 4\) because the value of y is unaffected by the value of x. We could also write it as \(y = f(x)\).

The integral of a function can be used to find the area between the function and the x-axis and is written \(I = \int_a^b f(x) dx\). It's essentially the sum of tiny changes in x (\(dx\)) multiplied by their corresponding y values (\(f(x)\)). The equation reminds us that we need to restrict ourselves to a lower and an upper value of x (otherwise the area would be infinite).

Let's choose a = x = 0 (the y-axis) and b = x = 3 as those two bounds.

\(I = \int_0^3 4 dx\)

\(I = \left[4 x\right]_0^3\)

\(I = \left(4\times3\right) - \left(4\times0\right)\)

\(I = 12\)

We're not restricted to 0 as the lower bound, but it exposes an interesting property: the area under a-to-b is equal to the area under 0-to-b less the area under 0-to-a. Let's try the same integral as above, but from 0-to-1.

\(I = \int_0^1 4 dx\)

\(I = \left[4 x\right]_0^1\)

\(I = \left(4\times1\right) - \left(4\times0\right)\)

\(I = 4\)

We might then subtract 4 from 12 to arrive at 8, but why not do this in one go?

\(I = \int_1^3 4 dx\)

\(I = \left[4 x\right]_1^3\)

\(I = \left(4\times3\right) - \left(4\times1\right)\)

\(I = 8\)