integral from 0 to 1 of (4x-3)^2

Lösung

\int_{0}^{1} (4 x - 3)^{2} d x

\frac{7}{3}

Dezimale

2.33333 \dots

Schritte zur Lösung

\int_{0}^{1} (4 x - 3)^{2} d x

Multipliziere aus

(4 x - 3)^{2} : 16 x^{2} - 24 x + 9

= \int_{0}^{1} 16 x^{2} - 24 x + 9 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_{0}^{1} 16 x^{2} d x - \int_{0}^{1} 24 x d x + \int_{0}^{1} 9 d x

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

\int_{0}^{1} 24 x d x = 12

\int_{0}^{1} 9 d x = 9

= \frac{16}{3} - 12 + 9

Vereinfache

\frac{16}{3} - 12 + 9 : \frac{7}{3}

= \frac{7}{3}

\int_{0}^{2} \frac{3}{2} x^{2} d x

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\int_{0}^{y} \frac{1}{1 - x} d x