integral from 1 to 5 of xsqrt(2x-1)

Lösung

\int_{1}^{5} x 2 x - 1 d x

\frac{428}{15}

Dezimale

28.53333 \dots

Schritte zur Lösung

\int_{1}^{5} x 2 x - 1 d x

Wende U-Substitution an

= \int_{1}^{9} \frac{u ( u + 1 )}{4} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{1}^{9} u (u + 1) d u

Multipliziere aus

u (u + 1) : u^{\frac{3}{2}} + u

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

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

\int_{1}^{9} u^{\frac{3}{2}} d u = \frac{484}{5}

\int_{1}^{9} u d u = \frac{52}{3}

= \frac{1}{4} (\frac{484}{5} + \frac{52}{3})

Vereinfache

\frac{1}{4} (\frac{484}{5} + \frac{52}{3}) : \frac{428}{15}

= \frac{428}{15}

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

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

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\int_{- 5}^{0} y + 9 d y