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

Lösung

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

\frac{40}{3}

Dezimale

13.33333 \dots

Schritte zur Lösung

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

\int_{1}^{3} 2 x^{2} d x = \frac{52}{3}

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

= \frac{52}{3} - 4

Vereinfache

\frac{52}{3} - 4 : \frac{40}{3}

= \frac{40}{3}

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

\int_{0}^{1} \frac{1}{( 1 + 3 y ) ^{2}} d y

\int_{0}^{4} \frac{5}{x} d x

\int_{2}^{7} (3 x^{2} - 6 x + 5) d x

\int_{1}^{x} lo g_{\frac{1}{2}} (t) d t