integral from 0 to 1 of (2-x)^4

Lösung

\int_{0}^{1} (2 - x)^{4} d x

\frac{31}{5}

Dezimale

6.2

Schritte zur Lösung

\int_{0}^{1} (2 - x)^{4} d x

Wende U-Substitution an

= \int_{2}^{1} - u^{4} d u

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

= - \int_{1}^{2} - u^{4} d u

Entferne die Konstante:

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

= - (- \int_{1}^{2} u^{4} d u)

Wende die Potenzregel an

= - (- [\frac{u ^{5}}{5}]_{1}^{2})

Vereinfache

= [\frac{u ^{5}}{5}]_{1}^{2}

Berechne die Grenzen:

\frac{31}{5}

= \frac{31}{5}

\int_{- 1}^{1} (x^{3} - x^{5}) d x

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

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

\int_{- 2}^{2} \frac{8}{4 + x ^{2}} d x

\int_{2}^{4} (x^{2} + 2 x) d x