integral from 0 to 2 of sqrt(4-x)

Lösung

\int_{0}^{2} 4 - x d x

\frac{16 - 4 2}{3}

Dezimale

3.44771 \dots

Schritte zur Lösung

\int_{0}^{2} 4 - x d x

Wende U-Substitution an

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

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

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

Entferne die Konstante:

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

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

Wende die Potenzregel an

= - (- [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{4})

Vereinfache

= [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{4}

Berechne die Grenzen:

\frac{16}{3} - \frac{4 2}{3}

= \frac{16}{3} - \frac{4 2}{3}

Vereinfache

= \frac{16 - 4 2}{3}

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