Question

If

m ≤ f(x) ≤ M
for
a ≤ x ≤ b,
where m is the absolute minimum and M is the absolute maximum of f on the interval [a, b], then
m(b − a) ≤
b
a
f(x) dx ≤ M(b − a).
Use this property to estimate the value of the integral.
⁄12 7 tan(4x) dx

Answer 1

It's easy to show that
`7\tan(4x)` is strictly increasing on
`x\in\left[0,\frac\pi8\right]`. This means

`M = \max \left\{7\tan(4x) \mid \frac\pi{16} \le x \le \frac\pi{12}\right\} = 7\tan(4x) \bigg|_(x=\pi/12) = 7\sqrt3`

and

`m = \min \left\{7\tan(4x) \mid \frac\pi{16} \le x \le \frac\pi{12}\right\} = 7\tan(4x) \bigg|_(x=\pi/16) = 7`

Then the integral is bounded by

`\displaystyle 7\left(\frac\pi{12} - \frac\pi{16}\right) \le \int_(\pi/16)^(\pi/12) 7\tan(4x) \, dx \le 7\sqrt3 \left(\frac\pi{12} - \frac\pi{16}\right)`

`\implies \displaystyle \boxed{(7\pi)/(48)} \le \int_(\pi/16)^(\pi/12) 7\tan(4x) \, dx \le \boxed{(7\sqrt3\,\pi)/(48)}`