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

Lösung

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

\frac{19}{3}

Dezimale

6.33333 \dots

Schritte zur Lösung

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

Multipliziere aus

(x + 2)^{2} : x^{2} + 4 x + 4

= \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{19}{3}

= \frac{19}{3}

\int_{0}^{1} \frac{8}{1 + x ^{2}} d x

\int_{\frac{π}{2}}^{π} (1 + cos (x))^{2} d x

\int_{0}^{2} - 5 - x^{2} d x

\int_{- 6}^{6} x^{3} d x

\int_{0}^{2 π} \frac{sin ( x )}{x} d x