integral from-infinity to infinity of 1/(2x^2-8x+40)

Lösung

\int_{- \infty}^{\infty} \frac{1}{2 x ^{2} - 8 x + 40} d x

\frac{π}{8}

Dezimale

0.39269 \dots

Schritte zur Lösung

\int_{- \infty}^{\infty} \frac{1}{2 x ^{2} - 8 x + 40} d x

Berechne das unbestimmte Integral:

\int \frac{1}{2 x ^{2} - 8 x + 40} d x = \frac{1}{8} arctan (\frac{x - 2}{4}) + C

Berechne die Grenzen:

\int_{- \infty}^{\infty} \frac{1}{2 x ^{2} - 8 x + 40} d x = \frac{π}{16} - (- \frac{π}{16})

= \frac{π}{16} - (- \frac{π}{16})

Vereinfache

= \frac{π}{8}

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