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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....

So now I am absolutely stuck!

How do I even start to solve this one?

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    But your function is $\tan(4x)$, so $m=\tan(4\cdot \frac{\pi}{16})$ and $M=\tan(4\cdot \frac{\pi}{12})$. – 2017-01-12
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    i thought that was implied when I wrote f(pi/12).... – 2017-01-12
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    do you know what I am doing wrong... even when I plug into ti-89 i get same results... – 2017-01-12
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    sorry, I oversaw that. But how do you know that your estimates are incorrect? – 2017-01-12
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    because the system tells me that the result is incorrect... – 2017-01-12
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    could the people who down voted this question tell me why they did so, that way i can improve the question. I do not know how to improve the question as I tried to make as "stackOverFlow" admissable as possible. I understand that everyone else in the world is a master at mathematics excluding me so feedback would be helpful... :/ – 2017-01-12

1 Answers 1

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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:

  • $7\frac{\pi}{48} \approx 0.458 $
  • $\int_{\tfrac{\pi}{16}}^{\tfrac{\pi}{12}} 7\tan(4x) \,\mbox{d}x = \tfrac{\ln 2}{8} \approx 0.606 $
  • $7\sqrt{3}\frac{\pi}{48} \approx 0.794 $
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    thank you for that. basically i just needed to keep going with the property as specified. – 2017-01-14