integral from 0 to 1 of (7x^2-4x+6)

Lösung

\int_{0}^{1} (7 x^{2} - 4 x + 6) d x

\frac{19}{3}

Dezimale

6.33333 \dots

Schritte zur Lösung

\int_{0}^{1} 7 x^{2} - 4 x + 6 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} 7 x^{2} d x - \int_{0}^{1} 4 x d x + \int_{0}^{1} 6 d x

\int_{0}^{1} 7 x^{2} d x = \frac{7}{3}

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

\int_{0}^{1} 6 d x = 6

= \frac{7}{3} - 2 + 6

Vereinfache

\frac{7}{3} - 2 + 6 : \frac{19}{3}

= \frac{19}{3}

\int_{- 7}^{1} x 2 - x d x

\int_{- 1}^{2} [(x^{2}) - (x + 2)] d x

\int_{0}^{28} 1 + \frac{9}{4} x d x

\int_{2}^{5} \frac{x}{x + 1} d x

\int_{3}^{5} (x - 3)^{4} (5 - x)^{2} d x