integral from 0 to 3 of tsqrt(1+4t^2)

Lösung

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

\frac{37 37 - 1}{12}

Dezimale

18.67185 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{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_{1}^{37} u d u

Wende die Potenzregel an

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

Berechne die Grenzen:

\frac{74 37}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{74 37}{3} - \frac{2}{3})

Vereinfache

= \frac{37 37 - 1}{12}

\int_{0}^{0.5} \frac{1}{x ln ^{2} ∣ x ∣} d x

\int_{0}^{1} 1 x + 7 d x

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

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

\int_{\frac{1}{2} π}^{π} 2 sin (x) d x