Vyřešit pi({left(95/2right)}^2 | Microsoft Math Solver

\pi ( { \left( \frac{ 95 }{ 2 } \right) }^{ 2 }

Vyhodnotit

\frac{9025\pi }{4}\approx 7088,218424662

Postup řešení

\pi ( { \left( \frac{ 95 }{ 2 } \right) }^{ 2 }

Výpočtem \frac{95}{2} na 2 získáte \frac{9025}{4}.

\pi \times \frac{9025}{4}

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\pi ( { \left( \frac{ 95 }{ 2 } \right) }^{ 2 }

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How i could prove this:F(x)=\int_{0}^{\pi/2} \cos^{x}\! \theta \cos(x\theta){\mathrm{d}}\theta= \frac{\pi}{2^{a+1}}, \quad

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https://math.stackexchange.com/q/868337

Let's write (u,v) = \Big( \sin(x-y) \:\: , \:\: \frac{1}{2} \cos(x-y) - \frac{1}{2} \cos(x+y) \Big) As Daniel Fischer noted, x-y=\arcsin u\tag1 because -\pi/2\le x-y\le \pi/2. Furthermore, ...

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https://math.stackexchange.com/questions/325096/find-iint-xyx2-y21-2-mathrm-da

Alternatively, you can consider the region as pairs (x,y) with x\in[0,1] and y\in [\sqrt{1-x^2},1]: \int_{0}^1 \int_{\sqrt{1-x^2}}^1 xy\sqrt{x^2+y^2} dy dx For any particular x, it is easy ...

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\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} ...

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Zkopírováno do schránky

\pi \times \frac{9025}{4}

Výpočtem \frac{95}{2} na 2 získáte \frac{9025}{4}.

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