integral from 0 to 1 of tsqrt(9t^2+16)

Lösung

\int_{0}^{1} t 9 t^{2} + 16 d t

\frac{61}{27}

Dezimale

2.25925 \dots

Schritte zur Lösung

\int_{0}^{1} t 9 t^{2} + 16 d t

Wende U-Substitution an

= \int_{16}^{25} \frac{u}{18} d u

Entferne die Konstante:

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

= \frac{1}{18} \cdot \int_{16}^{25} u d u

Wende die Potenzregel an

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{25}

Berechne die Grenzen:

\frac{122}{3}

= \frac{1}{18} \cdot \frac{122}{3}

Vereinfache

= \frac{61}{27}

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