integral from 0 to 2 of 9y^2sqrt(y^3+1)

Lösung

\int_{0}^{2} 9 y^{2} y^{3} + 1 d y

52

Schritte zur Lösung

\int_{0}^{2} 9 y^{2} y^{3} + 1 d y

Entferne die Konstante:

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

= 9 \cdot \int_{0}^{2} y^{2} y^{3} + 1 d y

Wende U-Substitution an

= 9 \cdot \int_{1}^{9} \frac{u}{3} d u

Entferne die Konstante:

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

= 9 \cdot \frac{1}{3} \cdot \int_{1}^{9} u d u

Wende die Potenzregel an

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

Vereinfache

9 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9} : 3 [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

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

Berechne die Grenzen:

\frac{52}{3}

= 3 \cdot \frac{52}{3}

Vereinfache

= 52

\int_{1}^{2} \frac{1}{x} d x

\int_{0}^{3} (x - 3) e^{- x} d x

\int_{0}^{1} \frac{5 r}{( 4 + r ^{2} ) ^{2}} d r

\int_{1}^{1} 1

\int_{- 5}^{1} (x^{2} + 4 x - 5) d x