Page:Calculus Made Easy.pdf/230

and this will be the whole area from $0$ up to any value of $x$ that we may assign.

Therefore, the larger area up to the superior limit $x_{2}$ will be

$bx_{2}+{\frac {a}{3}}x_{2}^{3}+C$;

and the smaller area up to the inferior limit $x_{1}$ will be

$bx_{1}+{\frac {a}{3}}x_{1}^{3}+C$.

Now, subtract the smaller from the larger, and we get for the area $S$ the value,

${\text{area S}}=b(x_{2}-x_{1})+{\frac {a}{3}}(x_{2}^{3}-x_{1}^{3})$.

This is the answer we wanted. Let us give some numerical values. Suppose $b=10$, $a=0.06$, and $x_{2}=6$ and $x_{1}=6$. Then the area $S$ is equal to

$10(8-6)+{\frac {0.06}{3}}(8^{3}-6^{3})$

${\begin{aligned}&=20+0.02(512-216)\\&=20+0.02\times 296\\&=20+5.92\\&=25.92.\end{aligned}}$

Let us here put down a symbolic way of stating what we have ascertained about limits:

$\int _{x=x_{1}}^{x=x_{2}}y\,dx=y_{2}-y_{1}$,

where $y_{2}$ is the integrated value of $y\,dx$ corresponding to $x_{2}$, and $y_{1}$ that corresponding to $x_{1}$.