integral from-1 to 2 of-y^2+y+2

Lösung

\int_{- 1}^{2} - y^{2} + y + 2 d y

\frac{9}{2}

Dezimale

4.5

Schritte zur Lösung

\int_{- 1}^{2} - y^{2} + y + 2 d y

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 1}^{2} y^{2} d y + \int_{- 1}^{2} y d y + \int_{- 1}^{2} 2 d y

\int_{- 1}^{2} y^{2} d y = 3

\int_{- 1}^{2} y d y = \frac{3}{2}

\int_{- 1}^{2} 2 d y = 6

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

Vereinfache

- 3 + \frac{3}{2} + 6 : \frac{9}{2}

= \frac{9}{2}

\int_{0}^{4} x cos^{2} (x) d x

\int_{0}^{2} \frac{3}{2 + x ^{2}} d x

\int_{0}^{2} ∣ t - 1 ∣ d t

\int_{0}^{π} sin (v) d v

\int_{2}^{5} \frac{2}{3 x + 1} d x