integral from 0 to 1 of (y^4+y^8)

Lösung

\int_{0}^{1} (y^{4} + y^{8}) d y

\frac{14}{45}

Dezimale

0.31111 \dots

Schritte zur Lösung

\int_{0}^{1} y^{4} + y^{8} d y

Wende die Summenregel an:

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

= \int_{0}^{1} y^{4} d y + \int_{0}^{1} y^{8} d y

\int_{0}^{1} y^{4} d y = \frac{1}{5}

\int_{0}^{1} y^{8} d y = \frac{1}{9}

= \frac{1}{5} + \frac{1}{9}

Vereinfache

\frac{1}{5} + \frac{1}{9} : \frac{14}{45}

= \frac{14}{45}

\int_{2}^{7} \frac{10 x}{x ^{2} + 1} d x

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\int_{3}^{7} 15 ln (x + 5) d x