integral from 0 to 1 of sqrt(4x^2+9x^4)

Lösung

\int_{0}^{1} 4 x^{2} + 9 x^{4} d x

\frac{13 13 - 8}{27}

Dezimale

1.43970 \dots

Schritte zur Lösung

\int_{0}^{1} 4 x^{2} + 9 x^{4} d x

Faktorisiere

4 x^{2} + 9 x^{4} : x^{2} 4 + 9 x^{2}

= \int_{0}^{1} x^{2} 4 + 9 x^{2} d x

x^{2} = (x),

angenommen

x \geq 0

= \int_{0}^{1} x 4 + 9 x^{2} d x

Wende U-Substitution an

= \int_{4}^{13} \frac{u}{18} d u

Entferne die Konstante:

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

= \frac{1}{18} \cdot \int_{4}^{13} u d u

Wende die Potenzregel an

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

Berechne die Grenzen:

\frac{26 13}{3} - \frac{16}{3}

= \frac{1}{18} (\frac{26 13}{3} - \frac{16}{3})

Vereinfache

= \frac{13 13 - 8}{27}

\frac{d}{d x} (\frac{- 4 x}{( x ^{2} - 1 ) ^{2}})

\int 2 e^{9 x} d x

\frac{\partial}{\partial y} (\frac{2 x ^{2} y}{4 + x ^{2} y ^{2}})

\int sec^{2} (2 θ) d θ

\frac{\partial}{\partial z} (9 x sin (y - z))