integral from-1 to 1 of (x^4+3x^3+2x+1)

Lösung

\int_{- 1}^{1} (x^{4} + 3 x^{3} + 2 x + 1) d x

\frac{12}{5}

Dezimale

2.4

Schritte zur Lösung

\int_{- 1}^{1} x^{4} + 3 x^{3} + 2 x + 1 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}^{1} x^{4} d x + \int_{- 1}^{1} 3 x^{3} d x + \int_{- 1}^{1} 2 x d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} x^{4} d x = \frac{2}{5}

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

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

\int_{- 1}^{1} 1 d x = 2

= \frac{2}{5} + 0 + 0 + 2

Vereinfache

\frac{2}{5} + 0 + 0 + 2 : \frac{12}{5}

= \frac{12}{5}

\int_{0}^{π} 5 d x

\int_{\frac{x}{2}}^{\frac{1}{2}} e^{- y^{2}} d y

\int_{- 3}^{3} [(18 - x^{2}) - (x^{2})] d x

\int_{0}^{2} x cos (\frac{π}{2} y x) d x

\int_{1}^{4} - x - (2 - x) d x