integral from-4 to-2 of y^2+1/(y^3)

Lösung

\int_{- 4}^{- 2} y^{2} + \frac{1}{y ^{3}} d y

\frac{1783}{96}

Dezimale

18.57291 \dots

Schritte zur Lösung

\int_{- 4}^{- 2} y^{2} + \frac{1}{y ^{3}} 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_{- 4}^{- 2} y^{2} d y + \int_{- 4}^{- 2} \frac{1}{y ^{3}} d y

\int_{- 4}^{- 2} y^{2} d y = \frac{56}{3}

\int_{- 4}^{- 2} \frac{1}{y ^{3}} d y = - \frac{3}{32}

= \frac{56}{3} - \frac{3}{32}

Vereinfache

\frac{56}{3} - \frac{3}{32} : \frac{1783}{96}

= \frac{1783}{96}

\int_{- L}^{0} x^{2} cos (\frac{nπ x}{L}) d x

\int_{- 1}^{1} (y^{3} - y) d y

\int_{0}^{in f in t y} y^{5} e^{- 5 y} d y

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

\int_{0}^{2} 4 sec (\frac{π x}{6}) d x