# 8

All that is needed to answer this problem is a graph of the integrand :

> plot( sec(x)^3, x=0..Pi/6 );

In particular, since the integrand is increasing on the interval , the Riemann sum formed using the right endpoints must be an overestimate of the actual value of the integral. In fact, the actual value of the integral is found to be

> Int( sec(x)^3, x=0..Pi/6 );

> value( % );

> evalf( % );

which is smaller than the value given in the problem. You can use Maple yourself, by the way, to compute the left- right- or middle-Riemann sums as follows.

> with(student):

> rightbox( sec(x)^3, x=0..Pi/6, 3);

> rightsum( sec(x)^3, x=0..Pi/6, 3); evalf(%);

> rightsum( sec(x)^3, x=0 .. Pi/6, 500); evalf(%);

>