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

Lösung

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

\frac{75}{2}

Dezimale

37.5

Schritte zur Lösung

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

\int_{1}^{4} 2 x^{2} d x = 42

\int_{1}^{4} x d x = \frac{15}{2}

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

= 42 - \frac{15}{2} + 3

Vereinfache

42 - \frac{15}{2} + 3 : \frac{75}{2}

= \frac{75}{2}

\int_{0}^{\frac{L}{2}} x^{2} d x

\int_{- 1}^{0} \frac{x}{4 - x ^{2}} d x

\int_{0}^{2} 12 x^{3} 9 + x^{4} d x

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

\int_{0}^{1} \frac{1}{x ^{0.9}} d x