integral from-27 to 0 of 1/(sqrt(9-x))

Lösung

\int_{- 27}^{0} \frac{1}{9 - x} d x

6

Schritte zur Lösung

\int_{- 27}^{0} \frac{1}{9 - x} d x

Wende U-Substitution an

= \int_{36}^{9} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{9}^{36} - \frac{1}{u} d u

Entferne die Konstante:

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

= - (- \int_{9}^{36} \frac{1}{u} d u)

Wende die Potenzregel an

= - (- [2 u^{\frac{1}{2}}]_{9}^{36})

Vereinfache

- (- [2 u^{\frac{1}{2}}]_{9}^{36}) : [2 u]_{9}^{36}

= [2 u]_{9}^{36}

Berechne die Grenzen:

6

= 6

x \to 0 lim ((\frac{1}{x})^{l n (x)})

x \to - 1 lim (- x^{2} + 5 x - 1)

\int \frac{1}{( x ^{2} + 100 ) ^{\frac{3}{2}}} d x

in v erse l a pl a ce \frac{( s + 3 )}{s ^{2} + 6 s + 13}

\int \frac{13 x ^{3}}{9 + x ^{2}} d x