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

Lösung

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

8 e^{8} + \frac{- 12 e ^{8} - 4}{4}

Dezimale

14903.78993 \dots

Schritte zur Lösung

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

Multipliziere aus

(x^{2} - x + 1) e^{2 x} : e^{2 x} x^{2} - e^{2 x} x + e^{2 x}

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

\int_{0}^{4} e^{2 x} x^{2} d x = 8 e^{8} + \frac{- 7 e ^{8} - 1}{4}

\int_{0}^{4} e^{2 x} x d x = \frac{7 e ^{8} + 1}{4}

\int_{0}^{4} e^{2 x} d x = \frac{e ^{8} - 1}{2}

= 8 e^{8} + \frac{- 7 e ^{8} - 1}{4} - \frac{7 e ^{8} + 1}{4} + \frac{e ^{8} - 1}{2}

Vereinfache

8 e^{8} + \frac{- 7 e ^{8} - 1}{4} - \frac{7 e ^{8} + 1}{4} + \frac{e ^{8} - 1}{2} : 8 e^{8} + \frac{- 12 e ^{8} - 4}{4}

= 8 e^{8} + \frac{- 12 e ^{8} - 4}{4}

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

\int_{- \infty}^{0} x 5^{- x} d x

\int_{0}^{1} \frac{1}{x ^{3} + 1} d x

\int_{1}^{2} \frac{x ^{2} + 4}{x} d x

\int_{0}^{\frac{7}{12}} 20 x d x