integral from 0 to 1 of (9x)/((2x+1)^3)

Lösung

\int_{0}^{1} \frac{9 x}{( 2 x + 1 ) ^{3}} d x

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

\int_{0}^{1} \frac{9 x}{( 2 x + 1 ) ^{3}} d x

Entferne die Konstante:

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

= 9 \cdot \int_{0}^{1} \frac{x}{( 2 x + 1 ) ^{3}} d x

Wende U-Substitution an

= 9 \cdot \int_{1}^{3} \frac{u - 1}{4 u ^{3}} d u

Entferne die Konstante:

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

= 9 \cdot \frac{1}{4} \cdot \int_{1}^{3} \frac{u - 1}{u ^{3}} d u

Multipliziere aus

\frac{u - 1}{u ^{3}} : \frac{1}{u ^{2}} - \frac{1}{u ^{3}}

= 9 \cdot \frac{1}{4} \cdot \int_{1}^{3} \frac{1}{u ^{2}} - \frac{1}{u ^{3}} 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

= 9 \cdot \frac{1}{4} (\int_{1}^{3} \frac{1}{u ^{2}} d u - \int_{1}^{3} \frac{1}{u ^{3}} d u)

\int_{1}^{3} \frac{1}{u ^{2}} d u = \frac{2}{3}

\int_{1}^{3} \frac{1}{u ^{3}} d u = \frac{4}{9}

= 9 \cdot \frac{1}{4} (\frac{2}{3} - \frac{4}{9})

Vereinfache

9 \cdot \frac{1}{4} (\frac{2}{3} - \frac{4}{9}) : \frac{1}{2}

= \frac{1}{2}

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