integral from 1 to 4 of (x^3+1/(4x^3))

Lösung

\int_{1}^{4} (x^{3} + \frac{1}{4 x ^{3}}) d x

\frac{8175}{128}

Dezimale

63.8671875

Schritte zur Lösung

\int_{1}^{4} x^{3} + \frac{1}{4 x ^{3}} d x

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}^{4} x^{3} d x + \int_{1}^{4} \frac{1}{4 x ^{3}} d x

\int_{1}^{4} x^{3} d x = \frac{255}{4}

\int_{1}^{4} \frac{1}{4 x ^{3}} d x = \frac{15}{128}

= \frac{255}{4} + \frac{15}{128}

Vereinfache

\frac{255}{4} + \frac{15}{128} : \frac{8175}{128}

= \frac{8175}{128}

n = 0 \sum \infty \frac{n ^{6}}{2 ^{n}}

n = 1 \sum \infty - 1^{n}

\frac{d}{d x} (\frac{2 x ^{3} - 7}{x ^{2}})

t an g e n t y = 3 x^{2} - x^{3}, (1, 2)

\int - \frac{1}{4} csc (8 t) d t