integral from 0 to 1 of x(x^2+6)

Lösung

\int_{0}^{1} x (x^{2} + 6) d x

\frac{13}{4}

Dezimale

3.25

Schritte zur Lösung

\int_{0}^{1} x (x^{2} + 6) d x

Multipliziere aus

x (x^{2} + 6) : x^{3} + 6 x

= \int_{0}^{1} x^{3} + 6 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} x^{3} d x + \int_{0}^{1} 6 x d x

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

\int_{0}^{1} 6 x d x = 3

= \frac{1}{4} + 3

Vereinfache

\frac{1}{4} + 3 : \frac{13}{4}

= \frac{13}{4}

\int_{- 1}^{3} (3 x - 8) (x + 3) d x

\int_{0}^{4} 5 x^{2} d x

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

\int_{- 1}^{1} - 3 x^{2} + 3 d x

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