Respuesta :
The correct question is:
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) ≤ (integral from a to b)f(x)dx ≤ M(b − a).
Use this property to estimate the value of the (integral from π/16 to π/12) 5 tan(4x)dx.
Answer:
The estimated value of the integral is 0.447030768
Step-by-step explanation:
Given f(x) = 5tan(4x)
f(x) is an increasing function on the interval (-infinity, infinity)
This implies that x must be in the interval (-π/8, π/8)............................(1)
But we are given the interval
[π/16, π/12] ....................................(2)
Since this interval is contained in (1), we have π/16 to be the minimum, and π/12 to be the maximum.
Now, the minimum value is
f(π/16) = 5tan(4π/16)
= 5tan(π/4)
= 5 × 1
m = 5
Maximum value is
f(π/12) = 5tan(4π/12)
= 5tan(π/3)
= 5 × √3
M = 5√3
Now, because
m(b - a) ≤ (integral from a to b) f(x)dx ≤ M(b - a)
We have
5(π/12 - π/16) ≤ (Integral from (π/16 to π/12) 5tan(4x)dx ≤ 5√3(π/12 - π/16)
5π/48 ≤ (Integral from (π/16 to π/12) 5tan(4x)dx ≤ 5√3π/48
Let's take the midpoint of this interval to be the approximate value of the integral.
(Integral from (π/16 to π/12) 5tan(4x)dx is approximately
(5π/48 + 5√3π/48)/2
= 0.447030768
The answer for Smaller Value =[tex]\frac{21\pi}{144}[/tex] and Larger Value =[tex]\frac{21\sqrt 3\pi}{144}[/tex], and calculation can be defined as follows:
Estimation value:
If
[tex]m\leq f(x)\leq M(b-a); a\leq x\leq b[/tex] where m is the absolute minimum and M is the absolute maximum on the interval [tex][a,b][/tex], then
[tex]\to m(b-a)\leq\int_{a}^{b} f(x)dx\leq M(b-a).........................(i)[/tex]
function
[tex]\to f(x)=7 \tan (4x) \\\\ \to a=\frac{\pi}{16} \\\\ \to b=\frac{\pi}{12}[/tex]
We are to determine m and M here where m and M are the absolute minima and maxima of the function f(x).
The function [tex]f(x) = 7 \tan( 4x)[/tex] is increasing in the interval [tex][\frac{\pi}{6},\frac{\pi}{12}].[/tex]
Therefore the minimum value of
[tex]\to f(x) = f( \frac{\pi}{16})[/tex]
[tex]= 7 \tan (4 \times \frac{\pi}{16})\\\\= 7 \tan(\frac{\pi}{4})\\\\= 7 \times 1\\\\=7[/tex]
And maximum value:
[tex]\to f(x) = f( \frac{\pi}{12})[/tex]
[tex]= 7 \tan (4 \times \frac{\pi}{12})\\\\= 7 \tan(\frac{\pi}{3})\\\\=7\times \sqrt{3}\\\\= 7\sqrt{3}[/tex]
Therefore
[tex]\to m=7 \\\\ \to M=7\sqrt{3}[/tex]
Using
[tex]\to f(x)= 7 \tan(4x)\\\\ \to a= \frac{\pi}{16} \\\\ \to b=\frac{\pi}{12}\\\\ \to m= 7 \\\\ \to M=7\sqrt{3}[/tex].......... (i)
[tex]=7(\frac{\pi}{12}-\frac{\pi}{16})\leq \int_{\frac{\pi}{16}}^{\frac{\pi}{12}}7 \tan (4x)dx \leq 7\sqrt 3(\frac{\pi}{12}-\frac{\pi}{16})\\\\= 7(\frac{12\pi}{144}-\frac{9\pi}{144})\leq \int_{\frac{\pi}{16}}^{\frac{\pi}{12}}7 \tan (4x)dx \leq 7\sqrt 3(\frac{12\pi}{144}-\frac{9\pi}{144})\\\\= 7(\frac{3\pi}{144})\leq \int_{\frac{\pi}{16}}^{\frac{\pi}{12}}7 \tan (4x)dx \leq 7\sqrt 3(\frac{3\pi}{144})\\\\= \frac{21\pi}{144}\leq \int_{\frac{\pi}{16}}^{\frac{\pi}{12}}7 \tan (4x)dx \leq \frac{21\sqrt 3\pi}{144}\\[/tex]
[tex]\text{Smaller Value} =\frac{21\pi}{144}\\\\\text{Larger Value} =\frac{21\sqrt 3\pi}{144}\\\\[/tex]
Note:
Please find the complete question in the attached file.
Find out more information about the estimated value here:
brainly.com/question/3639595
