integral from 0 to x of sqrt(1+36t)

Lösung

\int_{0}^{x} 1 + 36 t d t

\frac{( 36 x + 1 ) ^{\frac{3}{2}} - 1}{54}

Schritte zur Lösung

\int_{0}^{x} 1 + 36 t d t

Wende U-Substitution an

= \int_{1}^{1 + 36 x} \frac{u}{36} d u

Entferne die Konstante:

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

= \frac{1}{36} \cdot \int_{1}^{1 + 36 x} u d u

Wende die Potenzregel an

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{1 + 36 x}

Berechne die Grenzen:

\frac{2}{3} (1 + 36 x)^{\frac{3}{2}} - \frac{2}{3}

= \frac{1}{36} (\frac{2}{3} (1 + 36 x)^{\frac{3}{2}} - \frac{2}{3})

Vereinfache

= \frac{( 36 x + 1 ) ^{\frac{3}{2}} - 1}{54}

\int_{1}^{n} ln (x) d x

\frac{d}{d x} (\frac{9 + 4 x - 6 x}{x})

\frac{\partial}{\partial x} (ln (x \cdot cos (y)))

\int \frac{1}{( x ^{2} - 441 ) ^{\frac{3}{2}}} d x

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{- 2 t}