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

Lösung

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

\frac{22}{3}

Dezimale

7.33333 \dots

Schritte zur Lösung

\int_{1}^{3} 4 z - z^{2} d z

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} 4 z d z - \int_{1}^{3} z^{2} d z

\int_{1}^{3} 4 z d z = 16

\int_{1}^{3} z^{2} d z = \frac{26}{3}

= 16 - \frac{26}{3}

Vereinfache

16 - \frac{26}{3} : \frac{22}{3}

= \frac{22}{3}

y^{^{'}} + 4 y = 2 x

\int \frac{x ^{2}}{( 169 + x ^{2} ) ^{2}} d x

n = 0 \sum \infty n z^{n}

\int \frac{( 2 x + 1 )}{( x + 1 ) ( x - 2 ) ( x - 3 )} d x

\int 7 - 2 x d x