integral from 0 to 1 of x(x^2+2)^2

Lösung

\int_{0}^{1} x (x^{2} + 2)^{2} d x

\frac{19}{6}

Dezimale

3.16666 \dots

Schritte zur Lösung

\int_{0}^{1} x (x^{2} + 2)^{2} d x

Multipliziere aus

x (x^{2} + 2)^{2} : x^{5} + 4 x^{3} + 4 x

= \int_{0}^{1} x^{5} + 4 x^{3} + 4 x 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_{0}^{1} x^{5} d x + \int_{0}^{1} 4 x^{3} d x + \int_{0}^{1} 4 x d x

\int_{0}^{1} x^{5} d x = \frac{1}{6}

\int_{0}^{1} 4 x^{3} d x = 1

\int_{0}^{1} 4 x d x = 2

= \frac{1}{6} + 1 + 2

Vereinfache

\frac{1}{6} + 1 + 2 : \frac{19}{6}

= \frac{19}{6}

\int \frac{( x + 6 )}{x + 1} d x

\frac{\partial}{\partial y} (x^{4} - x^{2} y)

\int_{- 1}^{2} (x^{3} - x + 2) d x

d er i v a t i v e f (x) = (cot (x))^{4}

d er i v a t i v e g (u) = (\frac{u ^{3} - 1}{u ^{3} + 1})^{8}