integral from 0 to 3 of tsqrt(1+16t^2)

Lösung

\int_{0}^{3} t 1 + 16 t^{2} d t

\frac{145 145 - 1}{48}

Dezimale

36.35481 \dots

Schritte zur Lösung

\int_{0}^{3} t 1 + 16 t^{2} d t

Wende U-Substitution an

= \int_{1}^{145} \frac{u}{32} d u

Entferne die Konstante:

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

= \frac{1}{32} \cdot \int_{1}^{145} u d u

Wende die Potenzregel an

= \frac{1}{32} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{145}

Berechne die Grenzen:

\frac{290 145}{3} - \frac{2}{3}

= \frac{1}{32} (\frac{290 145}{3} - \frac{2}{3})

Vereinfache

= \frac{145 145 - 1}{48}

\int_{1}^{3} \frac{1}{1 + x ^{2}} d x

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