contestada

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

If m fx 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 mb a b a fx dx Mb a Use this property to estimat class=

Respuesta :

LammettHash

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

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

and

[tex]m = \min \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/16} = 7[/tex]

Then the integral is bounded by

[tex]\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)[/tex]

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

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