integral from 1 to 2 of 4x^3-3x^2+4x+2

Lösung

\int_{1}^{2} 4 x^{3} - 3 x^{2} + 4 x + 2 d x

16

Schritte zur Lösung

\int_{1}^{2} 4 x^{3} - 3 x^{2} + 4 x + 2 d x

Wende die Summenregel an:

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

= \int_{1}^{2} 4 x^{3} d x - \int_{1}^{2} 3 x^{2} d x + \int_{1}^{2} 4 x d x + \int_{1}^{2} 2 d x

\int_{1}^{2} 4 x^{3} d x = 15

\int_{1}^{2} 3 x^{2} d x = 7

\int_{1}^{2} 4 x d x = 6

\int_{1}^{2} 2 d x = 2

= 15 - 7 + 6 + 2

Vereinfache

= 16

\int_{3}^{x} (t^{2} + 1) d t

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\int_{0}^{\frac{1}{15}} x ln (15 x) d x

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