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December 13th, 2017, 10:31 AM   #1
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Determine the sign of the following integral without integrating it

Determine the sign of the following integral without integrating it:

\[\int_{0}^{\pi}x{\cos(x)}dx\]

Last edited by skipjack; December 13th, 2017 at 12:13 PM.
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December 13th, 2017, 10:48 AM   #2
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I'm not even sure how to give a hint here. If you know what an integral is, then this question should be completely trivial. Maybe try computing this integral (despite the instructions) so you know the answer, and then try to figure out why this answer should have been easy to see without computation.
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December 13th, 2017, 11:34 AM   #3
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My theory is that it will be negative because at x=pi we have pi*cos(pi)=-pi and at x=0 we have xcos(x)=0 since -pi is below the x axis the integral will be negative, is this correct?
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December 13th, 2017, 11:40 AM   #4
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I also looked at a graph and this makes it clear why it is negative. However, I can't put down in words the answer as it is still not clear to me from the definition of integration.

Last edited by skipjack; December 13th, 2017 at 12:12 PM.
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December 13th, 2017, 02:32 PM   #5
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As $\displaystyle \int_{\pi/2}^\pi x\cos(x)dx = -\int_{\pi/2}^0 (\pi - x)\cos(\pi - x)dx = \int_0^{\pi/2} (x - \pi)\cos(x)dx$,
$\displaystyle \begin{align*}\int_0^\pi\! x\cos(x)dx &= \int_0^{\pi/2} x\cos(x)dx + \int_{\pi/2}^\pi x\cos(x)dx \\
&= \int_0^{\pi/2} x\cos(x)dx + \int_0^{\pi/2} (x - \pi)\cos(x)dx \\
&= \int_0^{\pi/2} (2x - \pi)\cos(x)dx \end{align*}$

whose integrand is negative in the interval $(0,\, \pi/2)$.
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December 13th, 2017, 06:53 PM   #6
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A more general approach which doesn't require exploiting any properties of $\cos x$ is as follows.

1. $\cos x$ is non-negative on $[0,\frac{\pi}{2}]$ and $-\cos x$ is non-negative on $[\frac{\pi}{2},\pi]$. Thus, you have the decomposition into positive integrals
\[\int_0^\pi x \cos x \ ds = \int_0^{\frac{\pi}{2}} x \cos x \ dx - \int_\pi^{\frac{\pi}{2}} x \cos x \ dx \]
Thus, it suffices to show that the first integral is smaller than the second.

2. $\cos x \leq 1$ holds for all $x \in [0,\frac{\pi}{2}]$, so we have the bound
\[\int_0^{\frac{\pi}{2}} x \cos x \ dx \leq \int_0^{\frac{\pi}{2}} x \ dx = \frac{\pi^2}{8} \]

3. $x \leq \frac{\pi}{2}$ holds for all $x \in [\frac{\pi}{2}, \pi]$, so we have the bound on the second integral
\[-\int_\pi^{\frac{\pi}{2}} x \cos x \ dx \leq \frac{\pi}{2} \int_{\pi}^{\frac{\pi}{2}} \cos x \ dx = -\frac{\pi}{2} \]

4. Taking these bounds together, we have
\[ \int_0^\pi x \cos x \ dx \leq \frac{\pi^2}{8} - \frac{\pi}{2} = \frac{\pi}{2}(\frac{\pi}{4} - 1) < 0 \].

The main point of this approach is that it is completely general and illustrates a common "trick" in analysis which is to bound integrals of products by bounding their factors. Often, one breaks the domain into regions of monotonicity like this and the factor which must be bounded changes.
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December 13th, 2017, 08:14 PM   #7
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Quote:
Originally Posted by SDK View Post
A more general approach which doesn't require exploiting any properties of $\cos x$ is as follows.

1. $\cos x$ is non-negative on $[0,\frac{\pi}{2}]$ and $-\cos x$ is non-negative on $[\frac{\pi}{2},\pi]$.
Aren't those two statements contradictory?
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December 14th, 2017, 05:56 PM   #8
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Originally Posted by greg1313 View Post
Aren't those two statements contradictory?
Please elaborate? This is a basic fact about the $\cos$ function but the method of splitting into positive/negative integrals works in general. I don't know what contradiction you are referring to.
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December 14th, 2017, 06:27 PM   #9
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"A more general approach which doesn't require exploiting any properties of cos x is as follows."

Are the following not properties of the cosine function? "1. cos x is non-negative on [0,π/2] and −cosx is non-negative on [π/2,π]."

Nitpicking perhaps, but there it is.
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