integral from-2 to 1 of [(2-y)-y^2]

Lösung

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

\frac{9}{2}

Dezimale

4.5

Schritte zur Lösung

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

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

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

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

= 6 - (- \frac{3}{2}) - 3

Vereinfache

6 - (- \frac{3}{2}) - 3 : \frac{9}{2}

= \frac{9}{2}

\int_{0}^{2} (4 - x^{2})^{3} d x

\int_{- 2}^{0} \frac{1}{1 - 4 x} d x

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

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

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