„Benutzer:Hourssales/HA6" – Versionsunterschied

${\displaystyle 0=x^{3}-k^{2}x}${\displaystyle 0=x^{3}-k^{2}x}
${\displaystyle a=x_{0}=0}${\displaystyle a=x_{0}=0}
${\displaystyle b=x_{1};\ b>0}${\displaystyle b=x_{1};\ b>0}
${\displaystyle 0=x\left(x^{2}-k^{2}\right)}${\displaystyle 0=x\left(x^{2}-k^{2}\right)}
${\displaystyle 0=x^{2}-k^{2}}${\displaystyle 0=x^{2}-k^{2}}
${\displaystyle {x_{0}}^{2}=k^{2}}${\displaystyle {x_{0}}^{2}=k^{2}}

${\displaystyle \left|b\right|=\left|x_{1}\right|=\left|k\right|}${\displaystyle \left|b\right|=\left|x_{1}\right|=\left|k\right|}

${\displaystyle {\begin{aligned}{\frac {8}{2}}&=\int \limits _{0}^{k}\mathrm {f} (x),円\mathrm {d} x\4円&=\left|\int \limits _{0}^{k}(x^{3}-k^{2}x),円\mathrm {d} x\right|\\&=\left|\left[{\frac {1}{4}}x^{4}-{\frac {k^{2}}{2}}x^{2}\right]_{0}^{k}\right|\\&=\left|{\frac {1}{4}}k^{4}-{\frac {k^{2}}{2}}k^{2}\right|\\&={\frac {1}{4}}k^{4}\\k&={\sqrt[{4}]{16}}={\sqrt {\sqrt {16}}}=2\end{aligned}}}${\displaystyle {\begin{aligned}{\frac {8}{2}}&=\int \limits _{0}^{k}\mathrm {f} (x),円\mathrm {d} x\4円&=\left|\int \limits _{0}^{k}(x^{3}-k^{2}x),円\mathrm {d} x\right|\\&=\left|\left[{\frac {1}{4}}x^{4}-{\frac {k^{2}}{2}}x^{2}\right]_{0}^{k}\right|\\&=\left|{\frac {1}{4}}k^{4}-{\frac {k^{2}}{2}}k^{2}\right|\\&={\frac {1}{4}}k^{4}\\k&={\sqrt[{4}]{16}}={\sqrt {\sqrt {16}}}=2\end{aligned}}}

${\displaystyle {\begin{aligned}\mathrm {f} (x)&=-x^{2}+4\\\mathrm {g} (x)&=-x+2\\A&=\int \limits _{-1}^{2}\mathrm {f} (x),円\mathrm {d} x-\int \limits _{-1}^{2}\mathrm {g} (x),円\mathrm {d} x\\&=\int \limits _{-1}^{2}(-x^{2}+4)\mathrm {d} x-\int \limits _{-1}^{2}(-x+2)\mathrm {d} x\\&=\left[-{\frac {1}{3}}x^{3}+4x\right]_{-1}^{2}-\left[-{\frac {1}{2}}x^{2}+2x\right]_{-1}^{2}\\&={\Bigl (}{\bigl (}-{\frac {1}{3}}(2)^{3}-4(2){\bigr )}-{\bigl (}{\frac {1}{3}}(-1)^{3}+4(-1){\bigr )}{\Bigr )}-{\Bigl (}{\bigl (}-{\frac {1}{2}}(2)^{2}+2(2){\bigr )}-{\bigl (}-{\frac {1}{2}}(-1)^{2}+2(-1){\bigr )}{\Bigr )}\\A&={\frac {9}{2}}\end{aligned}}}${\displaystyle {\begin{aligned}\mathrm {f} (x)&=-x^{2}+4\\\mathrm {g} (x)&=-x+2\\A&=\int \limits _{-1}^{2}\mathrm {f} (x),円\mathrm {d} x-\int \limits _{-1}^{2}\mathrm {g} (x),円\mathrm {d} x\\&=\int \limits _{-1}^{2}(-x^{2}+4)\mathrm {d} x-\int \limits _{-1}^{2}(-x+2)\mathrm {d} x\\&=\left[-{\frac {1}{3}}x^{3}+4x\right]_{-1}^{2}-\left[-{\frac {1}{2}}x^{2}+2x\right]_{-1}^{2}\\&={\Bigl (}{\bigl (}-{\frac {1}{3}}(2)^{3}-4(2){\bigr )}-{\bigl (}{\frac {1}{3}}(-1)^{3}+4(-1){\bigr )}{\Bigr )}-{\Bigl (}{\bigl (}-{\frac {1}{2}}(2)^{2}+2(2){\bigr )}-{\bigl (}-{\frac {1}{2}}(-1)^{2}+2(-1){\bigr )}{\Bigr )}\\A&={\frac {9}{2}}\end{aligned}}}

${\displaystyle {\begin{aligned}\mathrm {f} (x)&=sqrtx\\\mathrm {g} (x)&=-{\frac {3}{4}}x+3\\A&=\int \limits _{0}^{s}\mathrm {f} (x),円\mathrm {d} x+\int \limits _{s}^{4}\mathrm {g} (x),円\mathrm {d} x\\\mathrm {f} (x)&=\mathrm {g} (x)\\\mathrm {Y} 1&=-x^{2}+4\\\mathrm {Y} 2&=-{\frac {3}{4}}x+3\\&\to \mathrm {GTR} \to [\mathrm {GRAPH} ]\to [\mathrm {ISCT} ]\\s&=2{,}0789\\A&=\int \limits _{0}^{s}sqrtx,円\mathrm {d} x+\int \limits _{s}^{4}(-{\frac {3}{4}}x+3),円\mathrm {d} x\\&=\left[{\frac {2}{3}}x{^{\frac {3}{2}}}\right]_{0}^{s}+\left[-{\frac {3}{8}}^{2}+3x\right]_{s}^{4}\\&=\left({\frac {2}{3}}s^{\frac {3}{2}}\right)+\left(-{\frac {3}{8}}4^{2}+3(4)\right)-\left(-{\frac {3}{8}}s^{2}+3s\right)\\&=3{,}3823\end{aligned}}}${\displaystyle {\begin{aligned}\mathrm {f} (x)&=sqrtx\\\mathrm {g} (x)&=-{\frac {3}{4}}x+3\\A&=\int \limits _{0}^{s}\mathrm {f} (x),円\mathrm {d} x+\int \limits _{s}^{4}\mathrm {g} (x),円\mathrm {d} x\\\mathrm {f} (x)&=\mathrm {g} (x)\\\mathrm {Y} 1&=-x^{2}+4\\\mathrm {Y} 2&=-{\frac {3}{4}}x+3\\&\to \mathrm {GTR} \to [\mathrm {GRAPH} ]\to [\mathrm {ISCT} ]\\s&=2{,}0789\\A&=\int \limits _{0}^{s}sqrtx,円\mathrm {d} x+\int \limits _{s}^{4}(-{\frac {3}{4}}x+3),円\mathrm {d} x\\&=\left[{\frac {2}{3}}x{^{\frac {3}{2}}}\right]_{0}^{s}+\left[-{\frac {3}{8}}^{2}+3x\right]_{s}^{4}\\&=\left({\frac {2}{3}}s^{\frac {3}{2}}\right)+\left(-{\frac {3}{8}}4^{2}+3(4)\right)-\left(-{\frac {3}{8}}s^{2}+3s\right)\\&=3{,}3823\end{aligned}}}

${\displaystyle {\begin{aligned}\Delta E&=\int \limits _{0}^{2}4(-5e^{-0{,}1t}),円\mathrm {d} t\\&=-45{,}46,円\mathrm {Wh} \end{aligned}}}${\displaystyle {\begin{aligned}\Delta E&=\int \limits _{0}^{2}4(-5e^{-0{,}1t}),円\mathrm {d} t\\&=-45{,}46,円\mathrm {Wh} \end{aligned}}}

${\displaystyle {\begin{aligned}\mathrm {f} (x)&=e^{0{,}1x}+6\\\mathrm {g} (x)&=e^{0{,}1x}+6{,}5\\&\mathrm {Fassungsverm{\ddot {o}}gen} \\V_{Fass}&=\pi \int \limits _{0}^{25}\mathrm {f} (x)^{2},円\mathrm {d} x\\V_{Fass}&=\pi \int \limits _{0}^{25}(e^{0{,}1x}+6)^{2},円\mathrm {d} x\\V_{Fass}&=9358{,}7,円\mathrm {cm} ^{3}=9{,}4,円\mathrm {l} \\V_{G}&=\pi \int \limits _{0}^{25{,}5}(e^{0,1x}+6{,}5,円\mathrm {d} x\\V_{G}&=10767{,}5,円\mathrm {cm} ^{3}=11,円\mathrm {l} \\V_{K}&=V_{G}-V_{F}\\V_{K}&=1408{,}8,円\mathrm {cm} ^{3}\end{aligned}}}${\displaystyle {\begin{aligned}\mathrm {f} (x)&=e^{0{,}1x}+6\\\mathrm {g} (x)&=e^{0{,}1x}+6{,}5\\&\mathrm {Fassungsverm{\ddot {o}}gen} \\V_{Fass}&=\pi \int \limits _{0}^{25}\mathrm {f} (x)^{2},円\mathrm {d} x\\V_{Fass}&=\pi \int \limits _{0}^{25}(e^{0{,}1x}+6)^{2},円\mathrm {d} x\\V_{Fass}&=9358{,}7,円\mathrm {cm} ^{3}=9{,}4,円\mathrm {l} \\V_{G}&=\pi \int \limits _{0}^{25{,}5}(e^{0,1x}+6{,}5,円\mathrm {d} x\\V_{G}&=10767{,}5,円\mathrm {cm} ^{3}=11,円\mathrm {l} \\V_{K}&=V_{G}-V_{F}\\V_{K}&=1408{,}8,円\mathrm {cm} ^{3}\end{aligned}}}

${\displaystyle {\begin{aligned}\mathrm {P} _{1}&(1{,}0\vert 0{,}1)\\\mathrm {P} _{2}&(1{,}5\vert 0{,}2)\\\mathrm {P} _{3}&(2{,}5\vert 0{,}5)\\\mathrm {P} _{4}&(4{,}0\vert 0{,}9)\\\mathrm {P} _{5}&(5{,}0\vert 1{,}0)\\&[\mathrm {GTR} ]\to [\mathrm {STAT} ]\to [\mathrm {REG} {\mathopen {:}},円x^{3}]\\\mathrm {f} (x)&=-0{,}015x^{3}+0{,}109x^{2}+0{,}052x-0{,}071\end{aligned}}}${\displaystyle {\begin{aligned}\mathrm {P} _{1}&(1{,}0\vert 0{,}1)\\\mathrm {P} _{2}&(1{,}5\vert 0{,}2)\\\mathrm {P} _{3}&(2{,}5\vert 0{,}5)\\\mathrm {P} _{4}&(4{,}0\vert 0{,}9)\\\mathrm {P} _{5}&(5{,}0\vert 1{,}0)\\&[\mathrm {GTR} ]\to [\mathrm {STAT} ]\to [\mathrm {REG} {\mathopen {:}},円x^{3}]\\\mathrm {f} (x)&=-0{,}015x^{3}+0{,}109x^{2}+0{,}052x-0{,}071\end{aligned}}}

${\displaystyle {\begin{aligned}s_{2}&=\int \limits _{0}^{2}\mathrm {f} (x),円\mathrm {d} x&=0{,}1905\mathrm {km} \\s_{5}&=\int \limits _{0}^{5}\mathrm {f} (x),円\mathrm {d} x&=2{,}4405\mathrm {km} \\s_{6}&=s_{5}+1&=3{,}4405\mathrm {km} \end{aligned}}}${\displaystyle {\begin{aligned}s_{2}&=\int \limits _{0}^{2}\mathrm {f} (x),円\mathrm {d} x&=0{,}1905\mathrm {km} \\s_{5}&=\int \limits _{0}^{5}\mathrm {f} (x),円\mathrm {d} x&=2{,}4405\mathrm {km} \\s_{6}&=s_{5}+1&=3{,}4405\mathrm {km} \end{aligned}}}

${\displaystyle {\begin{aligned}&\mathrm {Der\ Rotationsk{\ddot {o}}rper\ hat\ eine\ r{\ddot {a}}umliche\ Parabelform,wenn\ die\ untere\ Intervall-} \\&\mathrm {grenze\ gr{\ddot {o}}sser\ oder\ gleich\ 1\ ist.\ Im\ Intervall\ 0\ \leq \ x\ <\ 1\ ist\ die\ Funktion\ jedoch} \\&\mathrm {nicht\ definiert,im\ gegebenen\ Intervall\ existiert\ also\ {\color {red}kein\ K{\ddot {o}}rper}.} \\V&=\pi \int \limits _{0}^{2}sqrt{x-1}^{2},円\mathrm {d} x\\&=\pi \int \limits _{0}^{2}(x-1),円\mathrm {d} x\\&=\pi \left[{\frac {1}{2}}x^{2}-x\right]_{0}^{2}\\&=\pi \left({\frac {4}{2}}-2\right)\\V&=0\\&\mathrm {F{\ddot {u}}r\ das\ Intervall\ I=[1;2]:} \\V&=\pi \int \limits _{1}^{2}(x-1),円\mathrm {d} x\\&=\pi \left[{\frac {1}{2}}x^{2}-x\right]_{1}^{2},円\mathrm {d} x\\&=\pi {\bigl (}\left({\frac {4}{2}}-2\right)-\left({\frac {1}{2}}-1\right){\bigr )}\\V&={\frac {\pi }{2}}\end{aligned}}}${\displaystyle {\begin{aligned}&\mathrm {Der\ Rotationsk{\ddot {o}}rper\ hat\ eine\ r{\ddot {a}}umliche\ Parabelform,wenn\ die\ untere\ Intervall-} \\&\mathrm {grenze\ gr{\ddot {o}}sser\ oder\ gleich\ 1\ ist.\ Im\ Intervall\ 0\ \leq \ x\ <\ 1\ ist\ die\ Funktion\ jedoch} \\&\mathrm {nicht\ definiert,im\ gegebenen\ Intervall\ existiert\ also\ {\color {red}kein\ K{\ddot {o}}rper}.} \\V&=\pi \int \limits _{0}^{2}sqrt{x-1}^{2},円\mathrm {d} x\\&=\pi \int \limits _{0}^{2}(x-1),円\mathrm {d} x\\&=\pi \left[{\frac {1}{2}}x^{2}-x\right]_{0}^{2}\\&=\pi \left({\frac {4}{2}}-2\right)\\V&=0\\&\mathrm {F{\ddot {u}}r\ das\ Intervall\ I=[1;2]:} \\V&=\pi \int \limits _{1}^{2}(x-1),円\mathrm {d} x\\&=\pi \left[{\frac {1}{2}}x^{2}-x\right]_{1}^{2},円\mathrm {d} x\\&=\pi {\bigl (}\left({\frac {4}{2}}-2\right)-\left({\frac {1}{2}}-1\right){\bigr )}\\V&={\frac {\pi }{2}}\end{aligned}}}