integral from 0 to 1 of x^2-2x+2

Lösung

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

\frac{4}{3}

Dezimale

1.33333 \dots

Schritte zur Lösung

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

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

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

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

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

Vereinfache

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

= \frac{4}{3}

\int_{x^{2}}^{x^{4}} (t + t^{2})^{2} d t

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

\int_{0}^{9} 9 - y d y

\int_{0}^{54} x^{2} (\frac{1}{1458} x) d x

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