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December 13th, 2017, 10:31 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 98 Thanks: 2  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. 
December 13th, 2017, 10:48 AM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 475 Thanks: 262 Math Focus: Dynamical systems, analytic function theory, numerics 
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.

December 13th, 2017, 11:34 AM  #3 
Member Joined: Jan 2016 From: Blackpool Posts: 98 Thanks: 2 
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?

December 13th, 2017, 11:40 AM  #4 
Member Joined: Jan 2016 From: Blackpool Posts: 98 Thanks: 2 
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. 
December 13th, 2017, 02:32 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,718 Thanks: 1806 
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 $(0,\, \pi/2)$. 
December 13th, 2017, 06:53 PM  #6 
Senior Member Joined: Sep 2016 From: USA Posts: 475 Thanks: 262 Math Focus: Dynamical systems, analytic function theory, numerics 
A more general approach which doesn't require exploiting any properties of $\cos x$ is as follows. 1. $\cos x$ is nonnegative on $[0,\frac{\pi}{2}]$ and $\cos x$ is nonnegative 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. 
December 13th, 2017, 08:14 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond  
December 14th, 2017, 05:56 PM  #8 
Senior Member Joined: Sep 2016 From: USA Posts: 475 Thanks: 262 Math Focus: Dynamical systems, analytic function theory, numerics  
December 14th, 2017, 06:27 PM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond 
"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 nonnegative on [0,π/2] and −cosx is nonnegative on [π/2,π]." Nitpicking perhaps, but there it is. 

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