integral from 1 to 2t of usqrt(9+u^2)

Lösung

\int_{1}^{2 t} u 9 + u^{2} d u

\frac{( 9 + 4 t ^{2} ) ^{\frac{3}{2}} - 10 10}{3}

Schritte zur Lösung

\int_{1}^{2 t} u 9 + u^{2} d u

Wende U-Substitution an

= \int_{10}^{9 + 4 t^{2}} v^{2} d v

Wende die Potenzregel an

= [\frac{v ^{3}}{3}]_{10}^{9 + 4 t^{2}}

Berechne die Grenzen:

\frac{( 9 + 4 t ^{2} ) ^{\frac{3}{2}}}{3} - \frac{10 10}{3}

= \frac{( 9 + 4 t ^{2} ) ^{\frac{3}{2}}}{3} - \frac{10 10}{3}

Vereinfache

= \frac{( 9 + 4 t ^{2} ) ^{\frac{3}{2}} - 10 10}{3}

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