integral from 0 to 4 of y/(sqrt(25-y^2))

Lösung

\int_{0}^{4} \frac{y}{25 - y ^{2}} d y

2

Schritte zur Lösung

\int_{0}^{4} \frac{y}{25 - y ^{2}} d y

Wende U-Substitution an

= \int_{25}^{9} - \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{9}^{25} - \frac{1}{2 u} d u

Entferne die Konstante:

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

= - (- \frac{1}{2} \cdot \int_{9}^{25} \frac{1}{u} d u)

Wende die Potenzregel an

= - (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{9}^{25})

Vereinfache

- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{9}^{25}) : \frac{1}{2} [2 u]_{9}^{25}

= \frac{1}{2} [2 u]_{9}^{25}

Berechne die Grenzen:

4

= \frac{1}{2} \cdot 4

Vereinfache

= 2

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