integral from-x to x^2+1 of (x^3-x)

Lösung

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

\frac{x ^{8} + 4 x ^{6} + 5 x ^{4} + 4 x ^{2} + 1}{4} - \frac{x ^{4} + x ^{2} + 1}{2}

Schritte zur Lösung

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

\int_{- x}^{x^{2} + 1} x^{3} d x = \frac{x ^{8} + 4 x ^{6} + 5 x ^{4} + 4 x ^{2} + 1}{4}

\int_{- x}^{x^{2} + 1} x d x = \frac{x ^{4} + x ^{2} + 1}{2}

= \frac{x ^{8} + 4 x ^{6} + 5 x ^{4} + 4 x ^{2} + 1}{4} - \frac{x ^{4} + x ^{2} + 1}{2}

\int_{- 1}^{2} (x^{3} - 2 x^{2} - 5 x + 6) d x

\int_{0}^{3} x 1 + x^{2} d x

\int_{2}^{3} sin (\frac{π}{3} x) d x

\int_{0}^{1} y sin (x y) d x

\int_{1}^{2} t sin (π t) d t