integral from 0 to 1 of (4t-1)^2

Lösung

\int_{0}^{1} (4 t - 1)^{2} d t

\frac{7}{3}

Dezimale

2.33333 \dots

Schritte zur Lösung

\int_{0}^{1} (4 t - 1)^{2} d t

Multipliziere aus

(4 t - 1)^{2} : 16 t^{2} - 8 t + 1

= \int_{0}^{1} 16 t^{2} - 8 t + 1 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_{0}^{1} 16 t^{2} d t - \int_{0}^{1} 8 t d t + \int_{0}^{1} 1 d t

\int_{0}^{1} 16 t^{2} d t = \frac{16}{3}

\int_{0}^{1} 8 t d t = 4

\int_{0}^{1} 1 d t = 1

= \frac{16}{3} - 4 + 1

Vereinfache

\frac{16}{3} - 4 + 1 : \frac{7}{3}

= \frac{7}{3}

\int_{0}^{4} π (x + 3)^{2} d x

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