integral from 0 to 2 of ysqrt(x+2)

Lösung

\int_{0}^{2} y x + 2 d x

\frac{( 16 - 4 2 ) y}{3}

Schritte zur Lösung

\int_{0}^{2} y x + 2 d x

Entferne die Konstante:

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

= y \cdot \int_{0}^{2} 2 + x d x

Wende U-Substitution an

= y \cdot \int_{2}^{4} u d u

Wende die Potenzregel an

= y [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{4}

Berechne die Grenzen:

\frac{16}{3} - \frac{4 2}{3}

= y (\frac{16}{3} - \frac{4 2}{3})

Vereinfache

= \frac{( 16 - 4 2 ) y}{3}

\int_{0}^{4} (\frac{x}{1 + x ^{4}}) d x

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

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