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

• roam
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

#### roam

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

(b) Show that http://img516.imageshack.us/img516/351/indexd2aw6.gif [Broken] 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 [Broken]
We have
m = $$(\sqrt{1 + 8}) = 3$$
M = $$(\sqrt{0 + 8}) = 2.8$$

a = 1
b = 0

Thanks.

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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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"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.

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 [Broken]

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

What is c? I don't know but since $$a < c < b$$, 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.

Thanks,

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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?

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

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

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:

$$\int_{0}^{\frac{\pi}{2}} sin(x^3) dx = \int_{0}^{\frac{\pi}{4}}\frac{1}{2}dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2} 1 dx$$

$$= \frac{3 \pi}{8} < \frac{\pi}{2}$$ yay!

How's that?

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
$\int_0^{\pi/2} 1 dx= \pi/2$

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