Understanding the Sign of Integrals: Explaining x cos x without Evaluation

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I have been asked to explain without evaluating the integrals why the integral of x cos x from 0 to pi/2 is positive and the integral of x cos x from pi/2 to pi is negative. Also would I expect x cos x from 0 to pi to be positive, zero or negative? And why ?

How can I do this without evaluting it? :frown: some help would be much appreciated, thanks!
 
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Do you know what the graphs look like? Keep in mind that a positive function times a negative function is negative, etc, and the integral of a positive function is positive, etc.
 
If f(x)\geq 0 for x in [a,b], then \int_{a}^{b}f(x)dx \geq 0.
 
Do not double-post, please.
Haven't you got enough information on this page (you posted it 3 days ago remember?)? If you have any further questions, why don't consider to post it there, instead of starting a brand new thread?
This proves that some never read the hints in the posts others have given, and think about it, instead, they are looking for a complete solution!
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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