integral from 0 to 2 of xsqrt(1+4x^2)

Lösung

\int_{0}^{2} x 1 + 4 x^{2} d x

\frac{17 17 - 1}{12}

Dezimale

5.75773 \dots

Schritte zur Lösung

\int_{0}^{2} x 1 + 4 x^{2} d x

Wende U-Substitution an

= \int_{1}^{17} \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}^{17} u d u

Wende die Potenzregel an

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

Berechne die Grenzen:

\frac{34 17}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{34 17}{3} - \frac{2}{3})

Vereinfache

= \frac{17 17 - 1}{12}

\int_{2}^{\infty} \frac{5}{4 + x ^{2}} d x

\int_{0}^{2} (x^{3})^{2} d x

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

\int_{0}^{3} 6 x - x^{2} d x

\int_{1}^{2} x^{2} + \frac{1}{x ^{4}} d x