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

Lösung

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

\frac{7}{3}

Dezimale

2.33333 \dots

Schritte zur Lösung

\int_{0}^{1} x^{2} - 4 x + 4 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} x^{2} d x - \int_{0}^{1} 4 x d x + \int_{0}^{1} 4 d x

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

\int_{0}^{1} 4 x d x = 2

\int_{0}^{1} 4 d x = 4

= \frac{1}{3} - 2 + 4

Vereinfache

\frac{1}{3} - 2 + 4 : \frac{7}{3}

= \frac{7}{3}

\int_{0}^{u} 11 + 12 t^{2} d t

\int_{4}^{5} x^{2} - 4 x d x

\int_{0}^{9} (x - \frac{1}{2}) d x

\int_{0}^{2 π} (cos^{2} (t)) d t

\int_{1}^{10} x (x - 1) d x