integral from 0 to 1 of (x^2+2)e^{4x}

Lösung

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

\frac{21 e ^{4} - 17}{32}

Dezimale

35.29878 \dots

Schritte zur Lösung

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

Multipliziere aus

(x^{2} + 2) e^{4 x} : e^{4 x} x^{2} + 2 e^{4 x}

= \int_{0}^{1} e^{4 x} x^{2} + 2 e^{4 x} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} e^{4 x} x^{2} d x + \int_{0}^{1} 2 e^{4 x} d x

\int_{0}^{1} e^{4 x} x^{2} d x = \frac{5 e ^{4} - 1}{32}

\int_{0}^{1} 2 e^{4 x} d x = \frac{e ^{4} - 1}{2}

= \frac{5 e ^{4} - 1}{32} + \frac{e ^{4} - 1}{2}

Vereinfache

\frac{5 e ^{4} - 1}{32} + \frac{e ^{4} - 1}{2} : \frac{21 e ^{4} - 17}{32}

= \frac{21 e ^{4} - 17}{32}

\int_{0}^{π} 1 + cos (x) d x

\int_{0}^{0.5} e^{- x^{2}} d x

\int_{0}^{1} x^{2} + y^{2} d x

\int_{0}^{1} \frac{1}{x ^{2} - 9} d x

\int_{0}^{π} 2 x sin (x) d x