Page:Calculus Made Easy.pdf/238

of the base $ON$. But before we can find the area we must ascertain the length of the base, so as to know up to what limit we are to integrate. At $N$ the ordinate $y$ has zero value; therefore, we must look at the equation and see what value of $x$ will make $y=0$. Now, clearly, if $x$ is $0$, $y$ will also be $0$, the curve passing through the origin $O$; but also, if $x=1$, $y=0$; so that $x=1$ gives us the position of the point $N$.

Then the area wanted is

${\begin{aligned}&=\int _{x=0}^{x=1}(x-x^{2})\,dx\\&=\left[{\tfrac {1}{2}}x^{2}-{\tfrac {1}{3}}x^{3}\right]_{0}^{1}\\&=\left[{\tfrac {1}{2}}-{\tfrac {1}{3}}\right]-[0-0]\\&={\tfrac {1}{6}}.\end{aligned}}$

But the base length is $1$.

Therefore, the average ordinate of the curve $={\tfrac {1}{6}}$.

[N.B.–It will be a pretty and simple exercise in maxima and minima to find by differentiation what is the height of the maximum ordinate. It must be greater than the average.]

The mean ordinate of any curve, over a range from $x=0$ to $x=x_{1}$, is given by the expression,

${\text{mean y}}={\frac {1}{x_{1}}}\int _{x=0}^{x=x_{1}}y\cdot dx$.

One can also find in the same way the surface area of a solid of revolution.