de-CH
utf-8
math
Integration durch Substitution
i-08-01
expression
randRange(2,9) [ ["\\sin(x)", "\\cos(x)", "\\sin(x)", "\\cos(x)", "\\sin(\\sin(x))", "\\cos(x)\\cos(\\sin(x))", ], ["\\sin(x)", "\\cos(x)", "\\cos(x)", "-\\sin(x)", "\\sin(\\cos(x))", "-\\cos(\\cos(x))\\sin(x)", ], ["\\sin(x)", "\\cos(x)", "e^x", "e^x", "\\sin(e^x)", "\\cos(e^x)e^x", ], ["\\sin(x)", "\\cos(x)", "x^"+n, n+"x^{"+(n-1)+"}", "\\sin(x^"+n+")", n+"x^{"+(n-1)+"}\\cos(x^"+n+")", ], ["\\sin(x)", "\\cos(x)", "1/x", "-1/x^2", "\\sin(1/x)", "-\\cos(1/x)/x^2", ], ["\\cos(x)", "-\\sin(x)", "\\sin(x)", "\\cos(x)", "\\cos(\\sin(x))", "-\\cos(x)\\sin(\\sin(x))", ], ["\\cos(x)", "-\\sin(x)", "\\cos(x)", "-\\sin(x)", "\\cos(\\cos(x))", "\\sin(x)\\sin(\\cos(x))", ], ["\\cos(x)", "-\\sin(x)", "e^x", "e^x", "\\cos(e^x)", "-e^x\\sin(e^x)", ], ["\\cos(x)", "-\\sin(x)", "x^"+n, n+"x^{"+(n-1)+"}", "\\cos(x^"+n+")", "-"+n+"x^{"+(n-1)+"}\\sin(x^"+n+")", ], ["\\cos(x)", "-\\sin(x)", "1/x", "-1/x^2", "\\cos(1/x)", "\\sin(1/x)/x^2", ], ["e^x", "e^x", "\\sin(x)", "\\cos(x)", "e^{\\sin(x)}", "\\cos(x)e^{\\sin(x)}", ], ["e^x", "e^x", "\\cos(x)", "-\\sin(x)", "e^{\\cos(x)}", "-e^{\\cos(x)}\\sin(x)", ], ["e^x", "e^x", "e^x", "e^x", "e^{e^x}", "e^x e^{e^x}", ], ["e^x", "e^x", "x^"+n, n+"x^{"+(n-1)+"}", "e^{x^"+n+"}", n+"x^{"+(n-1)+"}e^{x^"+n+"}", ], ["e^x", "e^x", "1/x", "-1/x^2", "e^{1/x}", "-e^{1/x}/x^2", ], ["x^"+n, n+"x^{"+(n-1)+"}", "\\sin(x)", "\\cos(x)", "\\sin^"+n+"(x)", n+"\\sin^{"+(n-1)+"}(x)\\cos(x)", ], ["x^"+n, n+"x^{"+(n-1)+"}", "\\cos(x)", "-\\sin(x)", "\\cos^"+n+"(x)", "-"+n+"\\cos^{"+(n-1)+"}(x)\\sin(x)", ], ["x^"+n, n+"x^{"+(n-1)+"}", "e^x", "e^x", "(e^x)^"+n, n+"(e^x)^{"+(n-1)+"}e^x", ], ] randRange(0,functionBank.length-1) functionBank[fNum]
Bestimmen Sie \displaystyle \int f[5] dx.

Verwenden Sie C als Integrationskonstante.

f[4] + C

Bei der Integration durch Substitution verwenden wir die Gleichung

\displaystyle \int f\left(g(x)\right)g'(x) dx = F(g(x)) + C,

wobei F eine Stammfunktion von f ist.

Es geht also darum, Funktionen f und g so zu identifizieren, dass f\left(g(x)\right)g'(x) gleich dem Integranden f[5] ist und wir eine Stammfunktion F von f bestimmen können.

Hier eignen sich f(x) = f[1] und g(x) = f[2] mit g'(x) = f[3].

Es sind dann F(x) = f[0] eine Stammfunktion von f und F(g(x)) = f[4].

Damit ergibt sich

\displaystyle \int f[5] dx = f[4] + C.