Proving the Integral Question: m ≤ f(x) ≤ M on [a,b]

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In summary: But you need to show that \int_0^{\pi/2} sin(x^3)dx< \pi/2. You might note that with u= x^3, du= 3x^2dx, x^2= u^(2/3), and when x= 0, u= 0 and when x= (\pi/2)^{1/3}, u= \pi/8 so \int_0^{\pi/2} sin(x^3)dx= \int_0^{(\pi/2)^{1/3}} sin(u)(1/3)u^(-1/3)du and I doubt that you can
  • #1
(a) Show that [Broken]

(b) Show that [Broken] cannot be equal to [tex]\frac{\pi}{2}[/tex]

I 'm not quiet sure what the question means by saying "show", What do I have to show? I appreciate your guidance and hints...

My attempt at (a);

If [tex]m \leq f(x) \leq M [/tex] on [a,b] then [Broken]
We have
m = [tex](\sqrt{1 + 8}) = 3[/tex]
M = [tex](\sqrt{0 + 8}) = 2.8[/tex]

a = 1
b = 0



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  • #2
Well the good thing was that you had the right approach =] But you must solve M and m exactly! Sqrt 8 is not 2.8, its 2sqrt2! and the factors of (b-a) refer to the same b and a that are on the integral! It should be straight forward right after you sub them in =]

EDIT: Removed work on b), benorin's is much better =]
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  • #3
"Show" is like "prove" only a lot less formal/rigorous.

Your attempt at (a) is mostly correct, put instead [tex]m=\sqrt{8}=2\sqrt{2}[/tex] and [tex]M=3[/tex]. All that remains is to substitute n,M,a, and b into the inequality you cited.

(b) Think about why the inequality you cited is true... (this should include two rectangles with ares (b-a)m and (b-a)M): if the function being integrated is not a constant, could the integral have either (b-a)m or (b-a)M as its value?

BTW, (b) can be done another way: recall that for a < c < b, [tex]\int_a^b = \int_a^c+\int_c^b[/tex] and use the above inequality twice.
  • #4
Thank you guys,(a) now makes perfect sense!

But I still don't get (b)...

O.K... we use the property of the definite integral that;

[tex]\int_{a}^{b} f(x)dx = \int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx ;[/tex] [Broken]

a = 0
b = [tex]\frac{\pi}{2}[/tex]
c = ?

What is c? I don't know but since [tex] a < c < b[/tex], do you think we can use the mean value theorem for integrals? i.e. if [tex]f[/tex] is continuous on [a,b], there exists c in (a,b) such that;

[tex]\frac{1}{b-a} \int_{a}^{b} f(t)dt = f(c)[/tex]

That would be the average value of f on interval [a,b].

[tex]\frac{1}{\frac{\pi}{2} - 0} \int_{0}^{\frac{\pi}{2}} sin(x)^3 dx[/tex]

[tex]0.63 \times [cos(\frac{\pi}{2})^3 - cos(0)][/tex]

[tex]0.63 \times -1.74[/tex]

f(c) = -1.096

So now we substitute all in [tex]\int_a^b = \int_a^c+\int_c^b[/tex] :

[tex]\int_a^b = -1.74[/tex]
[tex]\int_a^c = -0.72[/tex]
[tex]\int_c^b = 1.01[/tex]

It doesn't work though.:frown:

Please help me... because I don't know what to do to prove this question. :confused:

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  • #5
You have the anti derivative wrong! In fact there is no easy anti derivative! There is another method: First, show the integrand is non-negative within the interval of integration. Also note that the sine function can not exceed 1. So all values within this interval must be between 0 and 1. If the value of the integral was pi/2, over an interval length pi/2, what MUST the value of this sine function be at all values over the interval? And is it that value?
  • #6
couldnt you just evaluate the integrals and wouldn't that show what the answer is equal (or not equal) to?
  • #7
That would be a good idea if that integral was easy to evaluate :(
  • #8
We have [tex]sin \theta \leq \frac{1}{2} for all \theta \in [0, \frac{\pi}{6}][/tex]

And [tex]sin \theta \leq 1[/tex] for all [tex]\theta[/tex] so that [tex]sin (x^3) \leq \frac{1}{2}[/tex] when [tex]x \in [0, \frac{\pi}{4}][/tex]

Thus by that theorem we have:

[tex]\int_{0}^{\frac{\pi}{2}} sin(x^3) dx = \int_{0}^{\frac{\pi}{4}}\frac{1}{2}dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2} 1 dx[/tex]

[tex]= \frac{3 \pi}{8} < \frac{\pi}{2}[/tex] yay!

How's that?
  • #9
Why would you need "[tex]sin \theta \leq \frac{1}{2} for all \theta \in [0, \frac{\pi}{6}][/tex][/quote]?
It is certainly true that [itex]sin(\theta)\le 1[/itex] for all [itex]\theta[/itex] and
[itex]\int_0^{\pi/2} 1 dx= \pi/2[/itex]

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