<PRELOAD>
  miniaufgabe.js
</PRELOAD> 


==== 24. März 2025 bis 28. März 2025 ====
=== Montag 24. März 2025 ===
Leiten Sie von Hand und ohne Unterlagen ab:
<JS>miniAufgabe("#exoableiten","#solableiten",
[["a) $f(x)=\\frac{2}{3}x^{12}-\\frac{3}{8}x^{9}-\\frac{1}{7}x^{2}\\quad$ b) $f(x)=-\\frac{3}{8}\\ln(x)\\quad$ c) $f(x)=-\\frac{1}{3}\\cdot \\sqrt{x}\\quad$ d) $f(x)=-\\frac{1}{7}\\cdot 5^{x}\\quad$ ", "a) $f'(x)=8x^{11}-\\frac{27}{8}x^{8}-\\frac{2}{7}x\\quad$ b) $f'(x)=-\\frac{3}{8}\\cdot \\frac{1}{x}\\quad$ c) $f'(x)=-\\frac{1}{6}\\cdot \\frac{1}{\\sqrt{x}}\\quad$ d) $f'(x)=-\\frac{1}{7}\\cdot \\ln(5) \\cdot 5^x\\quad$ "], ["a) $f(x)=\\frac{1}{2}x^{11}-\\frac{3}{8}x^{5}-\\frac{1}{2}x^{4}\\quad$ b) $f(x)=\\frac{2}{7}\\ln(x)\\quad$ c) $f(x)=-\\frac{1}{7}\\cdot \\sqrt{x}\\quad$ d) $f(x)=\\frac{3}{7}\\cdot 2^{x}\\quad$ ", "a) $f'(x)=\\frac{11}{2}x^{10}-\\frac{15}{8}x^{4}-2x^{3}\\quad$ b) $f'(x)=\\frac{2}{7}\\cdot \\frac{1}{x}\\quad$ c) $f'(x)=-\\frac{1}{14}\\cdot \\frac{1}{\\sqrt{x}}\\quad$ d) $f'(x)=\\frac{3}{7}\\cdot \\ln(2) \\cdot 2^x\\quad$ "], ["a) $f(x)=-\\frac{3}{8}x^{12}-\\frac{1}{6}x^{11}+\\frac{2}{5}x^{9}\\quad$ b) $f(x)=\\frac{2}{3}\\ln(x)\\quad$ c) $f(x)=\\frac{1}{2}\\cdot \\sqrt{x}\\quad$ d) $f(x)=\\frac{1}{4}\\cdot 2^{x}\\quad$ ", "a) $f'(x)=-\\frac{9}{2}x^{11}-\\frac{11}{6}x^{10}+\\frac{18}{5}x^{8}\\quad$ b) $f'(x)=\\frac{2}{3}\\cdot \\frac{1}{x}\\quad$ c) $f'(x)=\\frac{1}{4}\\cdot \\frac{1}{\\sqrt{x}}\\quad$ d) $f'(x)=\\frac{1}{4}\\cdot \\ln(2) \\cdot 2^x\\quad$ "], ["a) $f(x)=\\frac{1}{2}x^{7}+\\frac{3}{5}x^{5}+\\frac{1}{7}x^{2}\\quad$ b) $f(x)=-\\frac{3}{2}\\ln(x)\\quad$ c) $f(x)=-\\frac{2}{5}\\cdot \\sqrt{x}\\quad$ d) $f(x)=-\\frac{3}{7}\\cdot 5^{x}\\quad$ ", "a) $f'(x)=\\frac{7}{2}x^{6}+3x^{4}+\\frac{2}{7}x\\quad$ b) $f'(x)=-\\frac{3}{2}\\cdot \\frac{1}{x}\\quad$ c) $f'(x)=-\\frac{1}{5}\\cdot \\frac{1}{\\sqrt{x}}\\quad$ d) $f'(x)=-\\frac{3}{7}\\cdot \\ln(5) \\cdot 5^x\\quad$ "], ["a) $f(x)=\\frac{2}{3}x^{11}-\\frac{2}{3}x^{6}-\\frac{3}{7}x^{2}\\quad$ b) $f(x)=\\frac{1}{2}\\ln(x)\\quad$ c) $f(x)=\\frac{4}{3}\\cdot \\sqrt{x}\\quad$ d) $f(x)=\\frac{1}{2}\\cdot 2^{x}\\quad$ ", "a) $f'(x)=\\frac{22}{3}x^{10}-4x^{5}-\\frac{6}{7}x\\quad$ b) $f'(x)=\\frac{1}{2}\\cdot \\frac{1}{x}\\quad$ c) $f'(x)=\\frac{2}{3}\\cdot \\frac{1}{\\sqrt{x}}\\quad$ d) $f'(x)=\\frac{1}{2}\\cdot \\ln(2) \\cdot 2^x\\quad$ "], ["a) $f(x)=\\frac{3}{7}x^{12}+\\frac{3}{8}x^{6}+\\frac{1}{4}x^{2}\\quad$ b) $f(x)=-\\frac{1}{4}\\ln(x)\\quad$ c) $f(x)=\\frac{1}{2}\\cdot \\sqrt{x}\\quad$ d) $f(x)=\\frac{1}{2}\\cdot 5^{x}\\quad$ ", "a) $f'(x)=\\frac{36}{7}x^{11}+\\frac{9}{4}x^{5}+\\frac{1}{2}x\\quad$ b) $f'(x)=-\\frac{1}{4}\\cdot \\frac{1}{x}\\quad$ c) $f'(x)=\\frac{1}{4}\\cdot \\frac{1}{\\sqrt{x}}\\quad$ d) $f'(x)=\\frac{1}{2}\\cdot \\ln(5) \\cdot 5^x\\quad$ "], ["a) $f(x)=-\\frac{1}{6}x^{7}+\\frac{1}{7}x^{5}+\\frac{3}{8}x^{2}\\quad$ b) $f(x)=\\frac{3}{7}\\ln(x)\\quad$ c) $f(x)=-\\frac{2}{3}\\cdot \\sqrt{x}\\quad$ d) $f(x)=-\\frac{1}{6}\\cdot 5^{x}\\quad$ ", "a) $f'(x)=-\\frac{7}{6}x^{6}+\\frac{5}{7}x^{4}+\\frac{3}{4}x\\quad$ b) $f'(x)=\\frac{3}{7}\\cdot \\frac{1}{x}\\quad$ c) $f'(x)=-\\frac{1}{3}\\cdot \\frac{1}{\\sqrt{x}}\\quad$ d) $f'(x)=-\\frac{1}{6}\\cdot \\ln(5) \\cdot 5^x\\quad$ "], ["a) $f(x)=\\frac{1}{2}x^{7}-\\frac{2}{9}x^{4}-\\frac{1}{3}x^{3}\\quad$ b) $f(x)=-\\frac{1}{3}\\ln(x)\\quad$ c) $f(x)=\\frac{4}{5}\\cdot \\sqrt{x}\\quad$ d) $f(x)=\\frac{1}{5}\\cdot 5^{x}\\quad$ ", "a) $f'(x)=\\frac{7}{2}x^{6}-\\frac{8}{9}x^{3}-x^{2}\\quad$ b) $f'(x)=-\\frac{1}{3}\\cdot \\frac{1}{x}\\quad$ c) $f'(x)=\\frac{2}{5}\\cdot \\frac{1}{\\sqrt{x}}\\quad$ d) $f'(x)=\\frac{1}{5}\\cdot \\ln(5) \\cdot 5^x\\quad$ "], ["a) $f(x)=-\\frac{1}{2}x^{12}-\\frac{2}{3}x^{10}-\\frac{1}{4}x^{3}\\quad$ b) $f(x)=-\\frac{1}{2}\\ln(x)\\quad$ c) $f(x)=-\\frac{1}{9}\\cdot \\sqrt{x}\\quad$ d) $f(x)=\\frac{1}{3}\\cdot 3^{x}\\quad$ ", "a) $f'(x)=-6x^{11}-\\frac{20}{3}x^{9}-\\frac{3}{4}x^{2}\\quad$ b) $f'(x)=-\\frac{1}{2}\\cdot \\frac{1}{x}\\quad$ c) $f'(x)=-\\frac{1}{18}\\cdot \\frac{1}{\\sqrt{x}}\\quad$ d) $f'(x)=\\frac{1}{3}\\cdot \\ln(3) \\cdot 3^x\\quad$ "], ["a) $f(x)=-\\frac{4}{9}x^{12}-\\frac{1}{2}x^{6}+\\frac{1}{2}x^{3}\\quad$ b) $f(x)=-\\frac{3}{8}\\ln(x)\\quad$ c) $f(x)=\\frac{1}{6}\\cdot \\sqrt{x}\\quad$ d) $f(x)=-\\frac{1}{2}\\cdot 4^{x}\\quad$ ", "a) $f'(x)=-\\frac{16}{3}x^{11}-3x^{5}+\\frac{3}{2}x^{2}\\quad$ b) $f'(x)=-\\frac{3}{8}\\cdot \\frac{1}{x}\\quad$ c) $f'(x)=\\frac{1}{12}\\cdot \\frac{1}{\\sqrt{x}}\\quad$ d) $f'(x)=-\\frac{1}{2}\\cdot \\ln(4) \\cdot 4^x\\quad$ "]],
" <br> ");
</JS>
<HTML>
<div id="exoableiten"></div>

</HTML>
<hidden Lösungen>

<HTML>
<div id="solableiten"></div>
<div style='font-size:12px;color:gray;'>ruby ableiten-von-hand.rb 1</div>
</HTML>

</hidden>


=== Donnerstag 27. März 2025 ===
Berechnen Sie:<JS>miniAufgabe("#exopolynome_integrieren","#solpolynome_integrieren",
[["$\\displaystyle \\int_{-3}^{2}\\left(-2x^{2}-2x-3\\right)\\mathrm{d}x$", "$\\displaystyle \\int_{-3}^{2}\\left(-2x^{2}-2x-3\\right)\\mathrm{d}x = -\\frac{2}{3}x^{3}-x^{2}-3x \\,\\,\\Bigr\\rvert_{-3}^{2} = $\\begin{multline*}-\\frac{2}{3} \\cdot \\left(2\\right)^{3}-1 \\cdot \\left(2\\right)^{2}-3 \\cdot 2-\\left(-\\frac{2}{3} \\cdot \\left(-3\\right)^{3}-1 \\cdot \\left(-3\\right)^{2}-3\\left(-3\\right)\\right) = \\\\\n-\\frac{2}{3} \\cdot 8-1 \\cdot 4-6+\\frac{2}{3} \\cdot \\left(-3\\right)^{3}+1 \\cdot \\left(-3\\right)^{2}+3\\left(-3\\right) = \\\\\n-\\frac{16}{3}-4-6+\\frac{2}{3}\\left(-27\\right)+1 \\cdot 9-9 = \\\\\n-\\frac{16}{3}-4-6-18+9-9 = \\\\\n-\\frac{100}{3}\\\\\n\n\\end{multline*}"], ["$\\displaystyle \\int_{-2}^{3}\\left(2x^{2}-2x-3\\right)\\mathrm{d}x$", "$\\displaystyle \\int_{-2}^{3}\\left(2x^{2}-2x-3\\right)\\mathrm{d}x = \\frac{2}{3}x^{3}-x^{2}-3x \\,\\,\\Bigr\\rvert_{-2}^{3} = $\\begin{multline*}\\frac{2}{3} \\cdot \\left(3\\right)^{3}-1 \\cdot \\left(3\\right)^{2}-3 \\cdot 3-\\left(\\frac{2}{3} \\cdot \\left(-2\\right)^{3}-1 \\cdot \\left(-2\\right)^{2}-3\\left(-2\\right)\\right) = \\\\\n\\frac{2}{3} \\cdot 27-1 \\cdot 9-9-\\frac{2}{3} \\cdot \\left(-2\\right)^{3}+1 \\cdot \\left(-2\\right)^{2}+3\\left(-2\\right) = \\\\\n18-9-9-\\frac{2}{3}\\left(-8\\right)+1 \\cdot 4-6 = \\\\\n18-9-9+\\frac{16}{3}+4-6 = \\\\\n\\frac{10}{3}\\\\\n\n\\end{multline*}"], ["$\\displaystyle \\int_{-3}^{-2}\\left(4x^{2}+4x-2\\right)\\mathrm{d}x$", "$\\displaystyle \\int_{-3}^{-2}\\left(4x^{2}+4x-2\\right)\\mathrm{d}x = \\frac{4}{3}x^{3}+2x^{2}-2x \\,\\,\\Bigr\\rvert_{-3}^{-2} = $\\begin{multline*}\\frac{4}{3} \\cdot \\left(-2\\right)^{3}+2 \\cdot \\left(-2\\right)^{2}-2\\left(-2\\right)-\\left(\\frac{4}{3} \\cdot \\left(-3\\right)^{3}+2 \\cdot \\left(-3\\right)^{2}-2\\left(-3\\right)\\right) = \\\\\n\\frac{4}{3}\\left(-8\\right)+2 \\cdot 4+4-\\frac{4}{3} \\cdot \\left(-3\\right)^{3}-2 \\cdot \\left(-3\\right)^{2}+2\\left(-3\\right) = \\\\\n-\\frac{32}{3}+8+4-\\frac{4}{3}\\left(-27\\right)-2 \\cdot 9-6 = \\\\\n-\\frac{32}{3}+8+4+36-18-6 = \\\\\n\\frac{40}{3}\\\\\n\n\\end{multline*}"], ["$\\displaystyle \\int_{-2}^{2}\\left(-4x^{2}+3x-2\\right)\\mathrm{d}x$", "$\\displaystyle \\int_{-2}^{2}\\left(-4x^{2}+3x-2\\right)\\mathrm{d}x = -\\frac{4}{3}x^{3}+\\frac{3}{2}x^{2}-2x \\,\\,\\Bigr\\rvert_{-2}^{2} = $\\begin{multline*}-\\frac{4}{3} \\cdot \\left(2\\right)^{3}+\\frac{3}{2} \\cdot \\left(2\\right)^{2}-2 \\cdot 2-\\left(-\\frac{4}{3} \\cdot \\left(-2\\right)^{3}+\\frac{3}{2} \\cdot \\left(-2\\right)^{2}-2\\left(-2\\right)\\right) = \\\\\n-\\frac{4}{3} \\cdot 8+\\frac{3}{2} \\cdot 4-4+\\frac{4}{3} \\cdot \\left(-2\\right)^{3}-\\frac{3}{2} \\cdot \\left(-2\\right)^{2}+2\\left(-2\\right) = \\\\\n-\\frac{32}{3}+6-4+\\frac{4}{3}\\left(-8\\right)-\\frac{3}{2} \\cdot 4-4 = \\\\\n-\\frac{32}{3}+6-4-\\frac{32}{3}-6-4 = \\\\\n-\\frac{88}{3}\\\\\n\n\\end{multline*}"], ["$\\displaystyle \\int_{-2}^{2}\\left(-4x^{2}+2x-3\\right)\\mathrm{d}x$", "$\\displaystyle \\int_{-2}^{2}\\left(-4x^{2}+2x-3\\right)\\mathrm{d}x = -\\frac{4}{3}x^{3}+x^{2}-3x \\,\\,\\Bigr\\rvert_{-2}^{2} = $\\begin{multline*}-\\frac{4}{3} \\cdot \\left(2\\right)^{3}+1 \\cdot \\left(2\\right)^{2}-3 \\cdot 2-\\left(-\\frac{4}{3} \\cdot \\left(-2\\right)^{3}+1 \\cdot \\left(-2\\right)^{2}-3\\left(-2\\right)\\right) = \\\\\n-\\frac{4}{3} \\cdot 8+1 \\cdot 4-6+\\frac{4}{3} \\cdot \\left(-2\\right)^{3}-1 \\cdot \\left(-2\\right)^{2}+3\\left(-2\\right) = \\\\\n-\\frac{32}{3}+4-6+\\frac{4}{3}\\left(-8\\right)-1 \\cdot 4-6 = \\\\\n-\\frac{32}{3}+4-6-\\frac{32}{3}-4-6 = \\\\\n-\\frac{100}{3}\\\\\n\n\\end{multline*}"], ["$\\displaystyle \\int_{2}^{3}\\left(2x^{2}-3x+3\\right)\\mathrm{d}x$", "$\\displaystyle \\int_{2}^{3}\\left(2x^{2}-3x+3\\right)\\mathrm{d}x = \\frac{2}{3}x^{3}-\\frac{3}{2}x^{2}+3x \\,\\,\\Bigr\\rvert_{2}^{3} = $\\begin{multline*}\\frac{2}{3} \\cdot \\left(3\\right)^{3}-\\frac{3}{2} \\cdot \\left(3\\right)^{2}+3 \\cdot 3-\\left(\\frac{2}{3} \\cdot \\left(2\\right)^{3}-\\frac{3}{2} \\cdot \\left(2\\right)^{2}+3 \\cdot 2\\right) = \\\\\n\\frac{2}{3} \\cdot 27-\\frac{3}{2} \\cdot 9+9-\\frac{2}{3} \\cdot \\left(2\\right)^{3}+\\frac{3}{2} \\cdot \\left(2\\right)^{2}-3 \\cdot 2 = \\\\\n18-\\frac{27}{2}+9-\\frac{2}{3} \\cdot 8+\\frac{3}{2} \\cdot 4-6 = \\\\\n18-\\frac{27}{2}+9-\\frac{16}{3}+6-6 = \\\\\n\\frac{49}{6}\\\\\n\n\\end{multline*}"], ["$\\displaystyle \\int_{-3}^{-2}\\left(4x^{2}+2x-2\\right)\\mathrm{d}x$", "$\\displaystyle \\int_{-3}^{-2}\\left(4x^{2}+2x-2\\right)\\mathrm{d}x = \\frac{4}{3}x^{3}+x^{2}-2x \\,\\,\\Bigr\\rvert_{-3}^{-2} = $\\begin{multline*}\\frac{4}{3} \\cdot \\left(-2\\right)^{3}+1 \\cdot \\left(-2\\right)^{2}-2\\left(-2\\right)-\\left(\\frac{4}{3} \\cdot \\left(-3\\right)^{3}+1 \\cdot \\left(-3\\right)^{2}-2\\left(-3\\right)\\right) = \\\\\n\\frac{4}{3}\\left(-8\\right)+1 \\cdot 4+4-\\frac{4}{3} \\cdot \\left(-3\\right)^{3}-1 \\cdot \\left(-3\\right)^{2}+2\\left(-3\\right) = \\\\\n-\\frac{32}{3}+4+4-\\frac{4}{3}\\left(-27\\right)-1 \\cdot 9-6 = \\\\\n-\\frac{32}{3}+4+4+36-9-6 = \\\\\n\\frac{55}{3}\\\\\n\n\\end{multline*}"], ["$\\displaystyle \\int_{2}^{3}\\left(-4x^{2}-4x+2\\right)\\mathrm{d}x$", "$\\displaystyle \\int_{2}^{3}\\left(-4x^{2}-4x+2\\right)\\mathrm{d}x = -\\frac{4}{3}x^{3}-2x^{2}+2x \\,\\,\\Bigr\\rvert_{2}^{3} = $\\begin{multline*}-\\frac{4}{3} \\cdot \\left(3\\right)^{3}-2 \\cdot \\left(3\\right)^{2}+2 \\cdot 3-\\left(-\\frac{4}{3} \\cdot \\left(2\\right)^{3}-2 \\cdot \\left(2\\right)^{2}+2 \\cdot 2\\right) = \\\\\n-\\frac{4}{3} \\cdot 27-2 \\cdot 9+6+\\frac{4}{3} \\cdot \\left(2\\right)^{3}+2 \\cdot \\left(2\\right)^{2}-2 \\cdot 2 = \\\\\n-36-18+6+\\frac{4}{3} \\cdot 8+2 \\cdot 4-4 = \\\\\n-36-18+6+\\frac{32}{3}+8-4 = \\\\\n-\\frac{100}{3}\\\\\n\n\\end{multline*}"], ["$\\displaystyle \\int_{-3}^{-2}\\left(-4x^{2}-4x-2\\right)\\mathrm{d}x$", "$\\displaystyle \\int_{-3}^{-2}\\left(-4x^{2}-4x-2\\right)\\mathrm{d}x = -\\frac{4}{3}x^{3}-2x^{2}-2x \\,\\,\\Bigr\\rvert_{-3}^{-2} = $\\begin{multline*}-\\frac{4}{3} \\cdot \\left(-2\\right)^{3}-2 \\cdot \\left(-2\\right)^{2}-2\\left(-2\\right)-\\left(-\\frac{4}{3} \\cdot \\left(-3\\right)^{3}-2 \\cdot \\left(-3\\right)^{2}-2\\left(-3\\right)\\right) = \\\\\n-\\frac{4}{3}\\left(-8\\right)-2 \\cdot 4+4+\\frac{4}{3} \\cdot \\left(-3\\right)^{3}+2 \\cdot \\left(-3\\right)^{2}+2\\left(-3\\right) = \\\\\n\\frac{32}{3}-8+4+\\frac{4}{3}\\left(-27\\right)+2 \\cdot 9-6 = \\\\\n\\frac{32}{3}-8+4-36+18-6 = \\\\\n-\\frac{52}{3}\\\\\n\n\\end{multline*}"], ["$\\displaystyle \\int_{-2}^{3}\\left(-2x^{2}-3x+3\\right)\\mathrm{d}x$", "$\\displaystyle \\int_{-2}^{3}\\left(-2x^{2}-3x+3\\right)\\mathrm{d}x = -\\frac{2}{3}x^{3}-\\frac{3}{2}x^{2}+3x \\,\\,\\Bigr\\rvert_{-2}^{3} = $\\begin{multline*}-\\frac{2}{3} \\cdot \\left(3\\right)^{3}-\\frac{3}{2} \\cdot \\left(3\\right)^{2}+3 \\cdot 3-\\left(-\\frac{2}{3} \\cdot \\left(-2\\right)^{3}-\\frac{3}{2} \\cdot \\left(-2\\right)^{2}+3\\left(-2\\right)\\right) = \\\\\n-\\frac{2}{3} \\cdot 27-\\frac{3}{2} \\cdot 9+9+\\frac{2}{3} \\cdot \\left(-2\\right)^{3}+\\frac{3}{2} \\cdot \\left(-2\\right)^{2}-3\\left(-2\\right) = \\\\\n-18-\\frac{27}{2}+9+\\frac{2}{3}\\left(-8\\right)+\\frac{3}{2} \\cdot 4+6 = \\\\\n-18-\\frac{27}{2}+9-\\frac{16}{3}+6+6 = \\\\\n-\\frac{95}{6}\\\\\n\n\\end{multline*}"]],
" <br> ");
</JS>
<HTML>
<div id="exopolynome_integrieren"></div>

</HTML>
<hidden Lösungen>

<HTML>
<div id="solpolynome_integrieren"></div>
<div style='font-size:12px;color:gray;'>ruby polynome-integrieren.rb 1</div>
</HTML>

</hidden>