integral from 0 to 1 of (sqrt(x))/(x+1)

Lösung

\int_{0}^{1} \frac{x}{x + 1} d x

2 (- \frac{π}{4} + 1)

Dezimale

0.42920 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{x}{x + 1} d x

Wende U-Substitution an

= \int_{0}^{1} \frac{2 u ^{2}}{u ^{2} + 1} d u

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{1} \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{1} \frac{1}{u ^{2} + 1} d u = \frac{π}{4}

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

= 2 (- \frac{π}{4} + 1)

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