Matematika - LS 24/25

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Nalezněte obecné řešení rovnice

\cos^2 y-y'x^{13}=0

y=\mathrm{tg} (\frac{x^{-12}}{-12}+C)

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y=\mathrm{arctg} (\frac{x^{14}}{14}+C)

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y=\mathrm{arctg} (\frac{x^{-12}}{-12}+C)

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y=\mathrm{arctg} (\frac{x^{-12}}{-12})+C

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Žádná z předchozích odpovědí není správně.

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Určete obsah rovinného obrazce ohraničeného křivkami y=2x^{2}-4x-1 a 2x+y=-1.

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Spočtěte \int_{3\pi}^{7\pi}\limits \sin \frac{x+9\pi}{6}\ \mathrm{d}x

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Spočtěte neurčitý integrál

\displaystyle\int\frac{x\cdot 2^x+1}{8x}dx.

\displaystyle\frac{x^2\cdot 2^x\cdot \ln2-1}{8x^2}+C

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\displaystyle\frac{x\cdot 2^x+2\ln2}{8x\cdot \ln 2}+C

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\displaystyle\frac{2^x+\ln2\cdot \ln |x|}{8\ln 2}+C

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\displaystyle\frac{(x-1)2^x+x}{4x^2}+C

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\displaystyle\frac{2^x+x\cdot 2^x\cdot \ln2}{8}+C

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\displaystyle\frac{2^x+2(x-1)}{8(x-1)}+C

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Určete Taylorův polynom druhého stupně funkce

f(x)=\sin(3x+\frac{\pi}{2})+5. se středem v bodě a=-\frac{\pi}{4}.

5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{4})+\frac{9\sqrt 2}{4}(x+\frac{\pi}{4})^2

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5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{4})+\frac{9\sqrt 2}{2}(x+\frac{\pi}{4})^2

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5+\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{4})+\frac{9\sqrt 2}{4}(x+\frac{\pi}{4})^2

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5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x-\frac{\pi}{4})+\frac{9\sqrt 2}{2}(x-\frac{\pi}{4})^2

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5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{4})-\frac{9\sqrt 2}{4}(x+\frac{\pi}{4})^2

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5-\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}(x+\frac{\pi}{4})+\frac{\sqrt 2}{4}(x+\frac{\pi}{4})^2

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