integral from 2 to 5 of 1/(sqrt(8x+9))

Lösung

\int_{2}^{5} \frac{1}{8 x + 9} d x

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

\int_{2}^{5} \frac{1}{8 x + 9} d x

Wende U-Substitution an

= \int_{25}^{49} \frac{1}{8 u} d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{25}^{49} \frac{1}{u} d u

Wende die Potenzregel an

= \frac{1}{8} [2 u^{\frac{1}{2}}]_{25}^{49}

Wende Radikal Regel an:

a^{\frac{1}{n}} = n a

= \frac{1}{8} [2 u]_{25}^{49}

Berechne die Grenzen:

4

= \frac{1}{8} \cdot 4

Vereinfache

= \frac{1}{2}

\int_{0}^{\infty} e^{4 x} d x

\int_{0}^{1} 4 + x^{2} d x

\int_{100}^{150} \frac{100}{x ^{2}} d x

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

\int_{0}^{y} 6 e^{- x} e^{- 2 y} d x