integral from 0 to 1 of 6tsqrt(t^2+1)

Lösung

\int_{0}^{1} 6 t t^{2} + 1 d t

2 (22 - 1)

Dezimale

3.65685 \dots

Schritte zur Lösung

\int_{0}^{1} 6 t t^{2} + 1 d t

Entferne die Konstante:

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

= 6 \cdot \int_{0}^{1} t t^{2} + 1 d t

Wende U-Substitution an

= 6 \cdot \int_{1}^{2} u^{2} d u

Wende die Potenzregel an

= 6 [\frac{u ^{3}}{3}]_{1}^{2}

Berechne die Grenzen:

\frac{2 2}{3} - \frac{1}{3}

= 6 (\frac{2 2}{3} - \frac{1}{3})

Vereinfache

= 2 (22 - 1)

\int_{0}^{11} ∣ x - 7 ∣ d x

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