Matematika - LS 24/25

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Určete Taylorův polynom druhého stupně funkce

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

3+\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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3+\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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3+\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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3+\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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3+\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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3+\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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Spočtěte \int_{-\pi}^{2\pi}\limits \cos \frac{x+10\pi}{6}\ \mathrm{d}x

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Určete obsah rovinného obrazce ohraničeného křivkami x-y=1 a y=-\frac{x^{2}}{2}-2x-5.

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Spočtěte neurčitý integrál

\int 12x\left(x-4\right)^2dx.

3x^4-96x^2+C

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2x^5-24x^4+96x^3+C

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3x^4-32x^3+96x^2+C

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2x^5-24x^4+96x^3-64x^2+C

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\frac{12}{5}x^5-2x^4+\frac{16}{3}x^3+C

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4x^4-48x^3+192x^2+C

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Určete Taylorův polynom druhého stupně funkce

f(x)=\cos(5x+\pi)+5. se středem v bodě a=-\frac{\pi}{6}.

5+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{6})+\frac{25\sqrt 3}{2}(x-\frac{\pi}{6})^2

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5+\frac{\sqrt 3}{2}-\frac{5}{2}(x+\frac{\pi}{6})+\frac{25\sqrt 3}{2}(x+\frac{\pi}{6})^2

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5+\frac{\sqrt 3}{2}-\frac{1}{2}(x+\frac{\pi}{6})+\frac{\sqrt 3}{4}(x+\frac{\pi}{6})^2

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5+\frac{\sqrt 3}{2}-\frac{5}{2}(x+\frac{\pi}{6})-\frac{25\sqrt 3}{4}(x+\frac{\pi}{6})^2

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5+\frac{\sqrt 3}{2}-\frac{5}{2}(x+\frac{\pi}{6})+\frac{25\sqrt 3}{4}(x+\frac{\pi}{6})^2

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5+\frac{\sqrt 3}{2}-\frac{5}{2}(x+\frac{\pi}{6})-\frac{25\sqrt 3}{2}(x+\frac{\pi}{6})^2

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Spočtěte \int_{4\pi}^{5\pi}\limits \cos \frac{4x-7\pi}{6}\ \mathrm{d}x

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