integral from 0 to 1 of x^2sqrt(7x+8)

Lösung

\int_{0}^{1} x^{2} 7 x + 8 d x

\frac{17250 15 - 16384 2}{36015}

Dezimale

1.21167 \dots

Schritte zur Lösung

\int_{0}^{1} x^{2} 7 x + 8 d x

Wende U-Substitution an

= \int_{8}^{15} \frac{u ( u - 8 ) ^{2}}{343} d u

Entferne die Konstante:

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

= \frac{1}{343} \cdot \int_{8}^{15} u (u - 8)^{2} d u

Multipliziere aus

u (u - 8)^{2} : u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 u

= \frac{1}{343} \cdot \int_{8}^{15} u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 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}{343} (\int_{8}^{15} u^{\frac{5}{2}} d u - \int_{8}^{15} 16 u^{\frac{3}{2}} d u + \int_{8}^{15} 64 u d u)

\int_{8}^{15} u^{\frac{5}{2}} d u = \frac{6750 15 - 2048 2}{7}

\int_{8}^{15} 16 u^{\frac{3}{2}} d u = 16 (9015 - \frac{256 2}{5})

\int_{8}^{15} 64 u d u = 64 (1015 - \frac{32 2}{3})

= \frac{1}{343} (\frac{6750 15 - 2048 2}{7} - 16 (9015 - \frac{256 2}{5}) + 64 (1015 - \frac{32 2}{3}))

Vereinfache

\frac{1}{343} (\frac{6750 15 - 2048 2}{7} - 16 (9015 - \frac{256 2}{5}) + 64 (1015 - \frac{32 2}{3})) : \frac{17250 15 - 16384 2}{36015}

= \frac{17250 15 - 16384 2}{36015}

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