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

Lösung

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

\frac{125}{6}

Dezimale

20.83333 \dots

Schritte zur Lösung

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

\int_{- 1}^{4} 3 x d x = \frac{45}{2}

\int_{- 1}^{4} 4 d x = 20

\int_{- 1}^{4} x^{2} d x = \frac{65}{3}

= \frac{45}{2} + 20 - \frac{65}{3}

Vereinfache

\frac{45}{2} + 20 - \frac{65}{3} : \frac{125}{6}

= \frac{125}{6}

\int_{1}^{9} 2 x + \frac{1}{x} d x

\int_{0}^{2} (8 r^{2} - 2 r^{4}) d r

\int_{- \frac{2 π}{3}}^{\frac{π}{3}} sin (x) d x

\int_{- 1}^{1} x^{2} ln ∣ x ∣ d x

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