integral from 0 to 1 of (x^4+3x^3+1)

Lösung

\int_{0}^{1} (x^{4} + 3 x^{3} + 1) d x

\frac{39}{20}

Dezimale

1.95

Schritte zur Lösung

\int_{0}^{1} x^{4} + 3 x^{3} + 1 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} x^{4} d x + \int_{0}^{1} 3 x^{3} d x + \int_{0}^{1} 1 d x

\int_{0}^{1} x^{4} d x = \frac{1}{5}

\int_{0}^{1} 3 x^{3} d x = \frac{3}{4}

\int_{0}^{1} 1 d x = 1

= \frac{1}{5} + \frac{3}{4} + 1

Vereinfache

\frac{1}{5} + \frac{3}{4} + 1 : \frac{39}{20}

= \frac{39}{20}

\int_{- 5}^{4} 20 - x - x^{2} d x

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

\int_{- 1}^{14} e^{- \frac{x}{2}} d x

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

\int_{0}^{1} x 1 - x d x