Area Under a Straight Line

The simplest curve to which the area idea of Section 1 applies is a straight line $y = mx + c$. Suppose the line stays at or above the $x$-axis for every $x$ in $[a,b]$ (with $b > a$). By the area principle, the area of the region bounded by the line, the $x$-axis, and the ordinates $x=a$ and $x=b$ is

Evaluating the integral. An antiderivative of $mx + c$ is $\dfrac{mx^2}{2} + cx$, so by the Fundamental Theorem of Calculus,

This can be simplified using the factorisation $b^2 - a^2 = (b-a)(b+a)$:

Now, $\dfrac{m(a+b)}{2} + c$ can be rewritten as $\dfrac{(ma+c) + (mb+c)}{2}$, which is exactly the average of the line's two end-heights, $y_1 = ma + c$ (the height at $x=a$) and $y_2 = mb + c$ (the height at $x=b$). So the formula becomes

A consistency check with elementary mensuration. This is precisely the familiar trapezium-area formula from coordinate geometry -- "half the sum of the parallel sides, times the distance between them," where the two parallel sides are the vertical segments of length $y_1$ and $y_2$ at $x=a$ and $x=b$, and the distance between them is $b-a$. When one of the ordinates shrinks to zero (say $a=0$ and the line passes through the origin, so $y_1=0$), the trapezium degenerates to a triangle, and the formula reduces to $\text{Area} = \frac12\, y_2\, b$ -- exactly $\frac12 \times \text{base} \times \text{height}$. This match is a valuable self-check: the integration method for a line must always agree with the mensuration formula you already know, and any disagreement signals an arithmetic slip in the limits or the antiderivative, not a flaw in the method itself. …