Page:Calculus Made Easy.pdf/232

value of the general integral in square brackets with the limits marked above and below, we get

${\begin{aligned}{\text{area}}\;&=\left[{\frac {1}{3}}\cdot {\frac {1}{2}}x^{2}\right]_{x=0}^{x=12}+C\\&=\left[{\frac {x^{2}}{6}}\right]_{x=0}^{x=12}+C\\&=\left[{\frac {12^{2}}{6}}\right]-\left[{\frac {0^{2}}{6}}\right]\\&={\frac {144}{6}}=24.\quad Ans.\end{aligned}}$

Let us satisfy ourselves about this rather surprising dodge of calculation, by testing it on a simple example. Get some squared paper, preferably some

Fig. 54.

that is ruled in little squares of one-eighth inch or one-tenth inch each way. On this squared paper plot out the graph of the equation,

$y={\frac {x}{3}}$.

The values to be plotted will be:

$x$	$0$	$3$	$6$	$9$	$12$
$y$	$0$	$1$	$2$	$3$	$4$

The plot is given in Fig. 54.