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

Lösung

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

- \frac{4}{3}

Dezimale

- 1.33333 \dots

Schritte zur Lösung

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \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_{0}^{4} (2 + 6 x) d x

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

\int_{- 1}^{2} x^{2} - 3 x d x

\int_{0}^{2 π} θ^{2} d θ

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