integral from 2 to 3 of tsqrt(1+4t^2)

Lösung

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

\frac{37 37 - 17 17}{12}

Dezimale

12.91411 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{17}^{37} \frac{u}{8} d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{17}^{37} u d u

Wende die Potenzregel an

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{17}^{37}

Berechne die Grenzen:

\frac{74 37 - 34 17}{3}

= \frac{1}{8} \cdot \frac{74 37 - 34 17}{3}

Vereinfache

= \frac{37 37 - 17 17}{12}

\int_{0}^{\frac{1}{10}} e^{- x^{2}} d x

\int_{2}^{5} 6 x - 12 d x

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\int_{- 1}^{4} 6 x^{2} - x^{3} - 5 x d x