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:
$$ m(b − a) ≤ \int f(x)dx \ a ≤ M(b − a).$$
Use this property to estimate the value of the integral.
$$ \int_\frac{π}{16}^\frac{π}{12} 7*tan (4x) dx $$
first things first, I know that f(x) is an increasing function from:
$$ \frac{π}{16} \ to \frac{π}{12} $$
Therefore: I should use the Comparison properties of the Integral to Solve this which states that:
If:
$$ m \le f(x) \le M \ for a \le x \le b,$$
then:
$$ m(b-a) \le \int_a^bf(x)dx \le M(b-a) $$
When graph this problem it looks like the graph is increasing and approach m = 7 and M = ~10
However, I just dont know how to calculate these Estimates.
I tried using that property to solve by using:
m = f(pi/16) = 7
and
M = f(pi/12) = 12.12436
but that was incorrect....
How do I start to solve this one?
It looks like you did the right thing, or at least intended to approach this correctly.
You are right that $\tan(4x)$ is increasing on $[\tfrac{\pi}{16},\tfrac{\pi}{12}]$ so: $$\tan(4\tfrac{\pi}{16}) \le \tan(4x) \le \tan(4\tfrac{\pi}{12}) \quad \mbox{for} \quad x \in [\tfrac{\pi}{16},\tfrac{\pi}{12}]$$ which means that using your notation, you have for $f(x)=7\tan(4x)$: $$m = 7\tan(4\tfrac{\pi}{16}) = 7\tan(\tfrac{\pi}{4}) = 7\quad \mbox{and} \quad M = 7\tan(4\tfrac{\pi}{12}) = 7\tan(\tfrac{\pi}{3}) = 7\sqrt{3}$$ From the theorem you mention, you then have with $b-a = \tfrac{\pi}{12}-\tfrac{\pi}{16}=\tfrac{\pi}{48}$: $$m(b-a) \le \int_{a}^{b} f(x) \,\mbox{d}x \le M(b-a) \to 7\frac{\pi}{48} \le \int_{\tfrac{\pi}{16}}^{\tfrac{\pi}{12}} 7\tan(4x) \,\mbox{d}x \le 7\sqrt{3}\frac{\pi}{48}$$
To have an idea, the values are approximately: