# Américo Tavares

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bio website problemasteoremas.wordpress.c… location Queluz, Portugal age 63 member for 3 years, 8 months seen 2 mins ago profile views 6,613

I am a retired electrical engineer, Portuguese. My math weblog is Problemas e Teoremas.

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 1m comment Uncomfortable using Leibniz notation for the chain rule. You are welcome. I'm a retired engineer. 40m revised Uncomfortable using Leibniz notation for the chain rule. edited tags 46m answered Uncomfortable using Leibniz notation for the chain rule. 15h revised If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then what values of $(a,b)$ would work? deleted 1 character in body; edited title 15h revised How do the steps of this definite integral work? edited tags 15h revised Solve $\log_2 (1+\frac{1}{x-1})<1$ edited tags 17h comment Evaluate $\iiint xyz$ @Pedro Tamaroff I know it's not a very compelling reason. But I didn't like that downvote in part because there was no explanation. It might be because I did not evaluate correctly the integral or I detailed the computation too much or any other reason. Anyhow I left my computation as a comment. 18h comment Evaluate $\iiint xyz$ @Juris Thanks for your comment but note that we are evaluating $$\iiint_{E}xyz\,dV=\int_{z=0}^{9}\int_{y=0}^{z}\int_{x=0}^{y}xyz\,dx\,dy \,dz$$ and not $$\iiint_{E}\,dV=\int_{z=0}^{9}\int_{y=0}^{z}\int_{x=0}^{y}\,dx\,dy \,dz$$ 18h comment Evaluate $\iiint xyz$ This was part of my now deleted answer that got a downvote\begin{eqnarray*} I &=&\iiint_{E}xyz\,dV=\int_{z=0}^{9}\int_{y=0}^{z}\int_{x=0}^{y}xyz\,dx\,dy \,dz \\ &=&\int_{0}^{9}\,dz\int_{0}^{z}dy\int_{0}^{y}xyz\,dx \\ &=&\int_{0}^{9}\left( \int_{0}^{z}\left( \int_{0}^{y}xyz\,dx\right) \,dy\right) \,dz \\ &=&\dots\\ &=&\left. \frac{1}{48}z^{6}\right\vert _{0}^{9} \\ &=&\frac{531\,441}{48}=\frac{177\,147}{16}. \end{eqnarray*} 18h revised Evaluate $\iiint xyz$ edited tags; edited title 20h revised What is the value of that integral? corrected TeX code for \sin^2 (x) and added (improper-integrals) tag 1d comment How do I show that $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$ This is a particular case of the algebraic identity ($x,y>0$) $$\sqrt{x+\sqrt{y}}=\sqrt{\frac{x+\sqrt{x^2-y}}{2}}+\sqrt{\frac{x-\sqrt{x^2-y}}{‌​2}}$$ $$\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+\sqrt{5^2-24}}{2}}+\sqrt{\frac{5-\sqrt{5^2-24‌​}}{2}}=\sqrt{3}+\sqrt{2}$$ 1d revised Evaluate $\int x \sqrt{1 - x^4} \,\mathrm{d}x$ edited tags 1d comment Evaluate $\int x \sqrt{1 - x^4} \,\mathrm{d}x$ +1 for the addendum. 1d comment Evaluate $\int x \sqrt{1 - x^4} \,\mathrm{d}x$ You coud try first the substitution $u=x^2$. Your integral becomes $\frac{1}{2}\int\sqrt{1-u^2}du$. Then you could use e.g. $u=\cos\theta.$ Apr15 revised What is the easiest way to integrate $y=\frac {x+4}{\sqrt{-x^2-2x+3}}$? Added a few details Apr15 revised How do I solve this definite integral? Fixed TeX Apr15 revised Vector Calculus Surface Integral (Limits of Integration) Improved TeX Apr15 revised What is the easiest way to integrate $y=\frac {x+4}{\sqrt{-x^2-2x+3}}$? Added note Apr15 revised What is the easiest way to integrate $y=\frac {x+4}{\sqrt{-x^2-2x+3}}$? added 84 characters in body