integral from 0 to 24/25 of sqrt(1+5x)

Lösung

\int_{0}^{\frac{24}{25}} 1 + 5 x d x

\frac{58 29 - 10 5}{75 5}

Dezimale

1.72909 \dots

Schritte zur Lösung

\int_{0}^{\frac{24}{25}} 1 + 5 x d x

Wende U-Substitution an

= \int_{1}^{\frac{29}{5}} \frac{u}{5} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int_{1}^{\frac{29}{5}} u d u

Wende die Potenzregel an

= \frac{1}{5} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{29}{5}}

Berechne die Grenzen:

\frac{58 29}{15 5} - \frac{2}{3}

= \frac{1}{5} (\frac{58 29}{15 5} - \frac{2}{3})

Vereinfache

= \frac{58 29 - 10 5}{75 5}

\int_{- 1}^{1} \frac{8}{x ^{2} + 4} d x

\int_{x^{2}}^{x^{7}} (2 t - 1)^{3} d t

\int_{2}^{1} (2 x^{3} - 2 x^{2} - 4 x) d x

\int_{- l n (6)}^{l n (6)} cosh (x) d x

\int_{- 5}^{5} 5 + 25 - x^{2} d x