integral from 0 to 1 of (6-9x)^2

Lösung

\int_{0}^{1} (6 - 9 x)^{2} d x

9

Schritte zur Lösung

\int_{0}^{1} (6 - 9 x)^{2} d x

Wende U-Substitution an

= \int_{6}^{- 3} - \frac{u ^{2}}{9} d u

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

= - \int_{- 3}^{6} - \frac{u ^{2}}{9} d u

Entferne die Konstante:

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

= - (- \frac{1}{9} \cdot \int_{- 3}^{6} u^{2} d u)

Wende die Potenzregel an

= - (- \frac{1}{9} [\frac{u ^{3}}{3}]_{- 3}^{6})

Vereinfache

= \frac{1}{9} [\frac{u ^{3}}{3}]_{- 3}^{6}

Berechne die Grenzen:

81

= \frac{1}{9} \cdot 81

Vereinfache

= 9

\int_{2}^{3} \frac{x ^{5} + 1}{x ^{6} + 6 x} d x

\int_{- 1}^{1} k (3 - x^{2}) d x

\int_{0}^{1} x^{8} \cdot e^{- x} d x

\int_{0}^{b} t e^{- s t} d t

\int_{0}^{\frac{π}{2}} e^{2 x} sin (3 x) d x