integral from 0 to 1 of 50x(2x^2+1)^4

Lösung

\int_{0}^{1} 50 x (2 x^{2} + 1)^{4} d x

605

Schritte zur Lösung

\int_{0}^{1} 50 x (2 x^{2} + 1)^{4} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 50 \cdot \int_{0}^{1} x (2 x^{2} + 1)^{4} d x

Multipliziere mit der Konjugation von

2 x^{2} + 1 : \frac{4 x ^{4} - 1}{2 x ^{2} - 1}

= 50 \cdot \int_{0}^{1} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x

Bestimme unbestimmte (Singularitäts-)Punkte:

x = \frac{1}{2}

Wenn

b, a < b < c, f (b) =

unbestimmt

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= 50 (\int_{0}^{\frac{1}{2}} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x + \int_{\frac{1}{2}}^{1} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x)

\int_{0}^{\frac{1}{2}} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x = \frac{31}{20}

\int_{\frac{1}{2}}^{1} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x = \frac{211}{20}

= 50 (\frac{31}{20} + \frac{211}{20})

Vereinfache

50 (\frac{31}{20} + \frac{211}{20}) : 605

= 605

\int_{1}^{13} ln (t) d t

\int_{0}^{3} x^{5} d x

\int_{1}^{3} \frac{2 x}{( x ^{2} + 3 ) ^{2}} d x

\int_{3}^{11} x^{2} d x

\int_{1}^{2} x x^{2} + 1 d x