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Beliebt Rechnen >

integral from 0 to 2 of 3tsqrt(4-t^2)

Lösung

\int_{0}^{2} 3 t 4 - t^{2} d t

Lösung

8

Schritte zur Lösung

\int_{0}^{2} 3 t 4 - t^{2} d t

Entferne die Konstante:

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

= 3 \cdot \int_{0}^{2} t 4 - t^{2} d t

Wende U-Substitution an

= 3 \cdot \int_{4}^{0} - \frac{u}{2} d u

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

= 3 (- \int_{0}^{4} - \frac{u}{2} d u)

Entferne die Konstante:

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

= 3 (- (- \frac{1}{2} \cdot \int_{0}^{4} u d u))

Wende die Potenzregel an

= 3 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}))

Vereinfache

3 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4})) : \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

Berechne die Grenzen:

\frac{16}{3}

= \frac{3}{2} \cdot \frac{16}{3}

Vereinfache

= 8

Graph

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