Question

\[2+\cfrac{1}{2+\cfrac{1}{4+\cfrac{1}{2+1\cfrac{4+...}}}}\]

Answer 1

that dint work out so well

Answer 2

not really, can you attach a picture?

Answer 3

\[2+\cfrac{1}{2+\cfrac{1}{4+\cfrac{1}{2+\cfrac{1}{4+\frac{1}{...}}}}}\]

Answer 4

the continued fraction expansion of sqrt(6) :)

Answer 5

\[\Large\begin{array}l
\int_{0}^{b}x\ cos(x)-x^{10}dx\\\\
=cos(x)-x\ sin(x)-10x^{9}; [0,b]\\\\
=cos(b)-b\ sin(b)-10b^{9}-cos(0)\\\\
=cos(b)-b\ sin(b)-10b^{9}-1\\\\
\end{array}
\]

Answer 6

\[\Large\begin{array}l
\int_{0}^{b}x\ cos(x)-x^{10}dx\\\\
=x\ sin(x)+cos(x)-\frac{x^{11}}{11}; [0,b]\\\\
=b\ sin(b)+cos(b)-\frac{b^{11}}{11} -cos(0)\\\\
=b\ sin(b)+cos(b)-\frac{b^{11}}{11} -1\\\\

\end{array}\]

Answer 7

\[\Large\begin{array}l
\int_{0}^{\sqrt{3-\sqrt{6}}}x^5-6x^3+4x-x\ dx\\\\
+\int_{\sqrt{3-\sqrt{6}}}^{\sqrt{3+\sqrt{6}}}x^5-6x^3+4x-x\ dx\\\\
A=\frac{x^6}{6}-\frac{3x^4}{2}+\frac{3x^2}{2}\\\\
=\frac{(\sqrt{3-\sqrt{6}})^6}{6}-\frac{3(\sqrt{3-\sqrt{6}})^4}{2}+\frac{3(\sqrt{3-\sqrt{6}})^2}{2}\\\\
=\frac{(81-33\sqrt{6})}{6}-\frac{3(15-6\sqrt{6})}{2}+\frac{3(3-\sqrt{6})}{2}\\\\
=\frac{(81-33\sqrt{6})-9(15-6\sqrt{6})+9(3-\sqrt{6})}{6}\\\\
=\frac{81-33\sqrt{6}-135+54\sqrt{6}+27-9\sqrt{6}}{6}\\\\
=\frac{-27+12\sqrt{6}}{6}\\\\
=\frac{-9+4\sqrt{6}}{2}\\\\


+\int_{\sqrt{3-\sqrt{6}}}^{\sqrt{3+\sqrt{6}}}-(x^5-6x^3+4x)+x\ dx\\\\
B=-\frac{x^6}{6}+\frac{3x^4}{2}-\frac{3x^2}{2}\\\\
=-\frac{x^6}{6}+\frac{3x^4}{2}-\frac{3x^2}{2}+A\\\\
=-\frac{(\sqrt{3+\sqrt{6}})^6}{6}+\frac{3(\sqrt{3+\sqrt{6}})^4}{2}-\frac{3(\sqrt{3+\sqrt{6}})^2}{2}\\\\+A\\\\
=\frac{-(\sqrt{3+\sqrt{6}})^6+9(\sqrt{3+\sqrt{6}})^4-9(\sqrt{3+\sqrt{6}})^2}{6}\\\\+A\\\\
=\frac{ -(81+33\sqrt{6})+9(15+6\sqrt{6})-9(3+\sqrt{6})}{6}\\\\+A\\\\
=\frac{ -81-33\sqrt{6}+135+54\sqrt{6}-27-9\sqrt{6}}{6}+A\\\\
=\frac{ 9+4\sqrt{6}}{2}+A\\\\
=\frac{ 9+4\sqrt{6}-9+4\sqrt{6}}{2}\\\\
=\frac{8\sqrt{6}}{2}\\\\
A+B=\frac{-9+4\sqrt{6}}{2}+\frac{8\sqrt{6}}{2}\\\\
=\frac{-9+12\sqrt{6}}{2}\\\\
12\sqrt{6}-9


\end{array}\]