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#1 2006-12-06 12:36:49

mikau
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Registered: 2005-08-22
Posts: 1,504

properties of integrals

use the properties of integrals to veryfy that the definite integral of x^4 cos x from 0 to 1 is between 0 and 0.2.

The only theorem we have for this sort of thing (that I know of) is that the definite integral of f(x) from a to b is less then or equal to (b -a)* max (of f(x)) and greater than or equal to (b-a)* min (of f(x))

but in this case, the max value of the function is in fact at x = 1, and f(1) is about 0.5. Thus we can deduce the integral should be less than or equal to (about) 0.5. But that doesn't solve the problem.

Btw, graphing this function reveals it is increasing on the interval from 0 to 1.


We could try to split the integral up, and integrate from zero to b, and from b to 1 where b is some value between 0 and 1. Then find the maximum possible values for these two integrals, and see if the sum of the max values is less then or equal to 0.2, but we know from the getgo that on the interval [b to 1] the max value of the integral is f(1) (since f(x) is increasing on 0 to 1) which is 0.5. So the sum is already too large. The first half cannot be negative because cos (x) and x^4 are both positive on [0, 1].

You could note that x^4 cos x <= cos (x) on the interval [0, 1] but that function has a max of 1 and a total area of about 0.84 on that interval. So that doesn't help us either.

Suggestions? Ideas?


A logarithm is just a misspelled algorithm.

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#2 2006-12-06 13:52:26

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: properties of integrals

I'm guessing that you're not allowed to just integrate it?

By using integration by parts a lot of times, I got that:
∫x^4cosxdx = x^4sinx + 4x^3cosx - 12x^2sinx - 24xcosx + 24sinx.
Evaluating that between 0 and 1 gives the integral as ~0.133.

So it's definitely true, but I'd imagine that there's an easier way.

Both x^4 and cosx are always positive between 0 and 1, so we can trivially see that the integral will be more than 1. Showing that it's less than 0.2 is harder though.

I think you're on the right track with your idea of splitting the integral up, but perhaps you need to split it into more than 2 pieces.


Why did the vector cross the road?
It wanted to be normal.

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#3 2006-12-07 12:42:30

mikau
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Registered: 2005-08-22
Posts: 1,504

Re: properties of integrals

aye, no integration allowed.

But as far as splitting into parts, that won't help us (if my thinking is correct) no matter how many pieces we break it into. We have to prove that the absolute maximum possible value of the integral is less then 0.2. The theorem tells us that it must be less then or equal to the max value of the function on that interval times the width of the interval (and similarly for minimums).  The minumum value does restrict how small it could possibly be, but all we're interested is how large it could possible be. But since f(1) has a value of about 0.5 then the max of the last interval is already greater then 0.2 so that idea fails.

But my book didn't present any other inequality properties of integrals, other then:
the integral of f(x) dx is greater then zero if f(x) > 0 for all x and
the integral of f(x) dx is less then zero if f(x) < 0 for all x.

But those don't help us here either.


A logarithm is just a misspelled algorithm.

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#4 2006-12-07 14:16:59

mathsyperson
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Registered: 2005-06-22
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Re: properties of integrals

But if you split it up into lots of pieces, then the interval of each piece would be less and so it wouldn't matter that some of the intervals would have maxes that were more than 0.2.

After all, integration is just what happens when you split the graph up into so many pieces that the interval becomes infinitissimal.


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It wanted to be normal.

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#5 2006-12-07 14:21:40

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: properties of integrals

aye, but how are we going to find the size of these split portions without actually integrating? Or even the limits of their size?


A logarithm is just a misspelled algorithm.

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#6 2006-12-07 14:55:57

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: properties of integrals

It might be worth proving that x^4cosx is always increasing between 0 and 1.
That way, we can say that the maximum of the function in [a,b] (0<a,b<1) is just f(b).

So to do that, we need to show that there are no values of x in (0,1) where f'(x) = 0. That shows that there are no turning points and so the function must always be increasing (or decreasing, but we can discount that by showing that f(1) > f(0)).

By the product rule, f'(x) = 4x^3cosx - x^4sinx.
So, to find points where f'(x) = 0, we solve 4x^3cosx = x^4sinx.

4cosx = xsinx
tanx = 4/x

But in [0,1], max(tanx) = tan 1 ≈1.6 and min(4/x) = 4/1 = 4, which means that tanx can never equal 4/x within that range. Which in turn means that within that range, f(x) is always increasing.

So now we know that the maximum within [a,b] is f(b), which would make working out the maximum of each piece you split the integral up into easier.


Why did the vector cross the road?
It wanted to be normal.

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#7 2006-12-07 15:49:36

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: properties of integrals

Oops. I forgot that when we split the integral into pieces, the width of the interval shrinks. (duh..))

But your proof that its always increasing is good and I think vital to the problem. thanks, mathsy!


A logarithm is just a misspelled algorithm.

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