integral from 0 to 1 of (8-12x)^2

Lösung

\int_{0}^{1} (8 - 12 x)^{2} d x

16

Schritte zur Lösung

\int_{0}^{1} (8 - 12 x)^{2} d x

Wende U-Substitution an

= \int_{8}^{- 4} - \frac{u ^{2}}{12} d u

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

= - \int_{- 4}^{8} - \frac{u ^{2}}{12} d u

Entferne die Konstante:

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

= - (- \frac{1}{12} \cdot \int_{- 4}^{8} u^{2} d u)

Wende die Potenzregel an

= - (- \frac{1}{12} [\frac{u ^{3}}{3}]_{- 4}^{8})

Vereinfache

= \frac{1}{12} [\frac{u ^{3}}{3}]_{- 4}^{8}

Berechne die Grenzen:

192

= \frac{1}{12} \cdot 192

Vereinfache

= 16

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