integral from 0 to 2 of t^3*sqrt(9t^4+1)

Lösung

\int_{0}^{2} t^{3} \cdot 9 t^{4} + 1 d t

\frac{145 145 - 1}{54}

Dezimale

32.31539 \dots

Schritte zur Lösung

\int_{0}^{2} t^{3} 9 t^{4} + 1 d t

Wende U-Substitution an

= \int_{1}^{145} \frac{u}{36} d u

Entferne die Konstante:

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

= \frac{1}{36} \cdot \int_{1}^{145} u d u

Wende die Potenzregel an

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{145}

Berechne die Grenzen:

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

= \frac{1}{36} (\frac{290 145}{3} - \frac{2}{3})

Vereinfache

= \frac{145 145 - 1}{54}

\int_{8}^{10} 2 x^{- 3} d x

\int_{0}^{2} \frac{e ^{x} + e ^{- x}}{2} d x

\int_{- 3}^{4} ∣ + 1 ∣ d x

\int_{0}^{2} 3 x^{2} - 1 d x

\int_{0}^{20} (2 x - 20) d x