integral from 0 to 1 of vsqrt(1-v^2)

Lösung

\int_{0}^{1} v 1 - v^{2} d v

\frac{1}{3}

Dezimale

0.33333 \dots

Schritte zur Lösung

\int_{0}^{1} v 1 - v^{2} d v

Wende U-Substitution an

= \int_{1}^{0} - u^{2} d u

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

= - \int_{0}^{1} - u^{2} d u

Entferne die Konstante:

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

= - (- \int_{0}^{1} u^{2} d u)

Wende die Potenzregel an

= - (- [\frac{u ^{3}}{3}]_{0}^{1})

Vereinfache

= [\frac{u ^{3}}{3}]_{0}^{1}

Berechne die Grenzen:

\frac{1}{3}

= \frac{1}{3}

\int_{0}^{5} \frac{x}{x ^{2} + 5} d x

\int_{2}^{5} (x^{3} + x - 4) d x

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

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

\int_{c}^{d} x d x