integral from 1 to 2 of ((1+t^2)/(t^4))

Lösung

\int_{1}^{2} (\frac{1 + t ^{2}}{t ^{4}}) d t

\frac{19}{24}

Dezimale

0.79166 \dots

Schritte zur Lösung

\int_{1}^{2} \frac{1 + t ^{2}}{t ^{4}} d t

Multipliziere aus

\frac{1 + t ^{2}}{t ^{4}} : \frac{1}{t ^{4}} + \frac{1}{t ^{2}}

= \int_{1}^{2} \frac{1}{t ^{4}} + \frac{1}{t ^{2}} 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_{1}^{2} \frac{1}{t ^{4}} d t + \int_{1}^{2} \frac{1}{t ^{2}} d t

\int_{1}^{2} \frac{1}{t ^{4}} d t = \frac{7}{24}

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

= \frac{7}{24} + \frac{1}{2}

Vereinfache

\frac{7}{24} + \frac{1}{2} : \frac{19}{24}

= \frac{19}{24}

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