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

Lösung

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

\frac{748}{15}

Dezimale

49.86666 \dots

Schritte zur Lösung

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

\int_{- 2}^{2} 4 x^{4} d x = \frac{256}{5}

\int_{- 2}^{2} x^{2} d x = \frac{16}{3}

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

= \frac{256}{5} - \frac{16}{3} + 4

Vereinfache

\frac{256}{5} - \frac{16}{3} + 4 : \frac{748}{15}

= \frac{748}{15}

\int_{0}^{\frac{π}{3}} \frac{1}{1 - sin ( x )} d x

\int_{0}^{π} cos^{3} (θ) d θ

\int_{2}^{5} \frac{1}{( x - 2 ) ^{2}} d x

\int_{- π}^{π} sin (5 x) cos (3 x) d x

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