integral from 2 to x of (t^3+4t-2)

Lösung

\int_{2}^{x} (t^{3} + 4 t - 2) d t

2 x^{2} - 2 x + \frac{x ^{4}}{4} - 8

Schritte zur Lösung

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

Wende die Summenregel an:

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

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

\int_{2}^{x} t^{3} d t = \frac{x ^{4}}{4} - 4

\int_{2}^{x} 4 t d t = 4 (\frac{x ^{2}}{2} - 2)

\int_{2}^{x} 2 d t = 2 x - 4

= \frac{x ^{4}}{4} - 4 + 4 (\frac{x ^{2}}{2} - 2) - (2 x - 4)

Vereinfache

\frac{x ^{4}}{4} - 4 + 4 (\frac{x ^{2}}{2} - 2) - (2 x - 4) : 2 x^{2} - 2 x + \frac{x ^{4}}{4} - 8

= 2 x^{2} - 2 x + \frac{x ^{4}}{4} - 8

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