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

Lösung

\int_{0}^{2} 2 π x 1 + 4 x^{2} d x

\frac{π ( 17 17 - 1 )}{6}

Dezimale

36.17690 \dots

Schritte zur Lösung

\int_{0}^{2} 2 π x 1 + 4 x^{2} d x

Entferne die Konstante:

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

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

Wende U-Substitution an

= 2 π \cdot \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

= 2 π \frac{1}{8} \cdot \int_{1}^{17} u d u

Wende die Potenzregel an

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

Vereinfache

2 π \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17} : \frac{π}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

= \frac{π}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Berechne die Grenzen:

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

= \frac{π}{4} (\frac{34 17}{3} - \frac{2}{3})

Vereinfache

= \frac{π ( 17 17 - 1 )}{6}

\int_{0}^{l n (x)} e^{- y} d y

\int_{0}^{\frac{x}{4}} tan^{3} (x) d x

\int_{- 1}^{0} x^{3} - 6 x d x

\int_{1}^{4} \frac{7}{x ln ( 7 x )} d x

\int_{- 2}^{0} - x^{3} + 4 x d x