integral from 0 to 1 of t^3(1+t^4)^3

Lösung

\int_{0}^{1} t^{3} (1 + t^{4})^{3} d t

\frac{15}{16}

Dezimale

0.9375

Schritte zur Lösung

\int_{0}^{1} t^{3} (1 + t^{4})^{3} d t

Multipliziere aus

t^{3} (1 + t^{4})^{3} : t^{3} + 3 t^{7} + 3 t^{11} + t^{15}

= \int_{0}^{1} t^{3} + 3 t^{7} + 3 t^{11} + t^{15} 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_{0}^{1} t^{3} d t + \int_{0}^{1} 3 t^{7} d t + \int_{0}^{1} 3 t^{11} d t + \int_{0}^{1} t^{15} d t

\int_{0}^{1} t^{3} d t = \frac{1}{4}

\int_{0}^{1} 3 t^{7} d t = \frac{3}{8}

\int_{0}^{1} 3 t^{11} d t = \frac{1}{4}

\int_{0}^{1} t^{15} d t = \frac{1}{16}

= \frac{1}{4} + \frac{3}{8} + \frac{1}{4} + \frac{1}{16}

Vereinfache

\frac{1}{4} + \frac{3}{8} + \frac{1}{4} + \frac{1}{16} : \frac{15}{16}

= \frac{15}{16}

t an g e n t f (x) = 7 x - x, (4, 26)

\frac{d}{d x} (\frac{2 sin ( x )}{sec ( x )})

in v erse l a pl a ce \frac{2}{( s - 4 ) ^{2}}

\frac{d}{d x} (\frac{1}{1 + e ^{\frac{1}{x}}})

x \to 3 - lim (\frac{∣ x ∣}{x})