integral from-3 to-1 of (t+1)/(t^3)

Lösung

\int_{- 3}^{- 1} \frac{t + 1}{t ^{3}} d t

\frac{2}{9}

Dezimale

0.22222 \dots

Schritte zur Lösung

\int_{- 3}^{- 1} \frac{t + 1}{t ^{3}} d t

Multipliziere aus

\frac{t + 1}{t ^{3}} : \frac{1}{t ^{2}} + \frac{1}{t ^{3}}

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

Wende die Summenregel an:

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

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

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

\int_{- 3}^{- 1} \frac{1}{t ^{3}} d t = - \frac{4}{9}

= \frac{2}{3} - \frac{4}{9}

Vereinfache

\frac{2}{3} - \frac{4}{9} : \frac{2}{9}

= \frac{2}{9}

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