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Matematika - LS 24/25

Matematika - LS 24/25

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Spočtěte neurčitý integrál

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

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

2x^5-12x^4+24x^3-8x^2+C

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Spočtěte $\displaystyle\int_{0}^{1}\left(\frac{3}{2}\right)^x\ \mathrm{d}x$ \displaystyle\int_{0}^{1}\left(\frac{3}{2}\right)^x\ \mathrm{d}x (zaokrouhlete na dvě desetinná místa).

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Určete Taylorův polynom druhého stupně funkce

$f(x)=\cos(3x-\frac{\pi}{3})+6.$ f(x)=\cos(3x-\frac{\pi}{3})+6. se středem v bodě $a=\frac{\pi}{6}$ a=\frac{\pi}{6}.

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

$6+\frac{\sqrt 3}{2}-\frac{3}{2}(x+\frac{\pi}{6})+\frac{9\sqrt 3}{2}(x+\frac{\pi}{6})^2$ 6+\frac{\sqrt 3}{2}-\frac{3}{2}(x+\frac{\pi}{6})+\frac{9\sqrt 3}{2}(x+\frac{\pi}{6})^2

$6+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 3}{2}(x-\frac{\pi}{6})^2$ 6+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 3}{2}(x-\frac{\pi}{6})^2

$6+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})-\frac{9\sqrt 3}{4}(x-\frac{\pi}{6})^2$ 6+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})-\frac{9\sqrt 3}{4}(x-\frac{\pi}{6})^2

$6+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 3}{4}(x-\frac{\pi}{6})^2$ 6+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 3}{4}(x-\frac{\pi}{6})^2

$6+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})-\frac{9\sqrt 3}{2}(x-\frac{\pi}{6})^2$ 6+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})-\frac{9\sqrt 3}{2}(x-\frac{\pi}{6})^2

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Spočtete obsah rovinného obrazce ohraničeného křivkami o rovnicích x=-4, x=2, y=-5 a y=1+(1·x).

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Vypočtěte hodnotu $f(\frac{\pi}{2})$ f(\frac{\pi}{2}) , je-li funkce f partikulární řešení rovnice $y'- \sin x= 0$ y'- \sin x= 0 vyhovující počáteční podmínce y(0) = 8.