integral from 0 to 1 of (sqrt(x)(x+1))

Lösung

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

\frac{16}{15}

Dezimale

1.06666 \dots

Schritte zur Lösung

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

Multipliziere aus

x (x + 1) : x^{\frac{3}{2}} + x

= \int_{0}^{1} x^{\frac{3}{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}^{1} x^{\frac{3}{2}} d x + \int_{0}^{1} x d x

\int_{0}^{1} x^{\frac{3}{2}} d x = \frac{2}{5}

\int_{0}^{1} x d x = \frac{2}{3}

= \frac{2}{5} + \frac{2}{3}

Vereinfache

\frac{2}{5} + \frac{2}{3} : \frac{16}{15}

= \frac{16}{15}

d er i v a t i v e - \frac{2}{5 x}

t an g e n t f (x) = 3 x + 6 x, (1, 9)

\int_{5}^{\infty} e^{- \frac{x}{3}} d x

\int \frac{2 x}{9 - x ^{2}} d x

p a r i t y - x + tan (x)