integral from 0 to 1 of tsqrt(4+5t)

Lösung

\int_{0}^{1} t 4 + 5 t d t

\frac{506}{375}

Dezimale

1.34933 \dots

Schritte zur Lösung

\int_{0}^{1} t 4 + 5 t d t

Wende U-Substitution an

= \int_{4}^{9} \frac{u ( u - 4 )}{25} d u

Entferne die Konstante:

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

= \frac{1}{25} \cdot \int_{4}^{9} u (u - 4) d u

Multipliziere aus

u (u - 4) : u^{\frac{3}{2}} - 4 u

= \frac{1}{25} \cdot \int_{4}^{9} u^{\frac{3}{2}} - 4 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}{25} (\int_{4}^{9} u^{\frac{3}{2}} d u - \int_{4}^{9} 4 u d u)

\int_{4}^{9} u^{\frac{3}{2}} d u = \frac{422}{5}

\int_{4}^{9} 4 u d u = \frac{152}{3}

= \frac{1}{25} (\frac{422}{5} - \frac{152}{3})

Vereinfache

\frac{1}{25} (\frac{422}{5} - \frac{152}{3}) : \frac{506}{375}

= \frac{506}{375}

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