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

[roam]

(a) Show that http://img187.imageshack.us/img187/8605/indexd1af5.gif

(b) Show that http://img516.imageshack.us/img516/351/indexd2aw6.gif cannot be equal to \frac{\pi}{2}





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 m \leq f(x) \leq M on [a,b] then http://img340.imageshack.us/img340/9103/indexd3wi0.gif
We have
m = (\sqrt{1 + 8}) = 3
M = (\sqrt{0 + 8}) = 2.8

a = 1
b = 0



Thanks.

 

[Gib Z]

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 =]

 

[benorin]

"Show" is like "prove" only a lot less formal/rigorous.

Your attempt at (a) is mostly correct, put instead m=\sqrt{8}=2\sqrt{2} and M=3. 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, \int_a^b = \int_a^c+\int_c^b and use the above inequality twice.

 

[roam]

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;

\int_{a}^{b} f(x)dx = \int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx ;

http://img516.imageshack.us/img516/351/indexd2aw6.gif

a = 0
b = \frac{\pi}{2}
c = ?

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

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

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

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

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

0.63 \times -1.74

f(c) = -1.096

So now we substitute all in \int_a^b = \int_a^c+\int_c^b :

\int_a^b = -1.74
\int_a^c = -0.72
\int_c^b = 1.01

It doesn't work though.

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

Thanks,

 

[Gib Z]

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?

 

[tron_2.0]

couldnt you just evaluate the integrals and wouldn't that show what the answer is equal (or not equal) to?

 

[Gib Z]

That would be a good idea if that integral was easy to evaluate :(

 

[roam]

We have sin \theta \leq \frac{1}{2} for all \theta \in [0, \frac{\pi}{6}]

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

Thus by that theorem we have:

\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

= \frac{3 \pi}{8} &lt; \frac{\pi}{2} yay!

How's that?

 

[HallsofIvy]

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