integral from 0 to 1 of xsqrt(5x^2+4)

Lösung

\int_{0}^{1} x 5 x^{2} + 4 d x

\frac{19}{15}

Dezimale

1.26666 \dots

Schritte zur Lösung

\int_{0}^{1} x 5 x^{2} + 4 d x

Wende U-Substitution an

= \int_{4}^{9} \frac{u}{10} d u

Entferne die Konstante:

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

= \frac{1}{10} \cdot \int_{4}^{9} u d u

Wende die Potenzregel an

= \frac{1}{10} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9}

Berechne die Grenzen:

\frac{38}{3}

= \frac{1}{10} \cdot \frac{38}{3}

Vereinfache

= \frac{19}{15}

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

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

\int_{0}^{1} 3 y^{2} d y

\int_{- 3}^{- 1} \frac{1}{x ^{2} - 6 x + 25} d x

\int_{- 5}^{5} (x^{2} - 4) d x