integral from 6 to 4 of ((-1)/2 x+3)

Lösung

\int_{6}^{4} (\frac{- 1}{2} x + 3) d x

- 1

Schritte zur Lösung

\int_{6}^{4} \frac{- 1}{2} x + 3 d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{4}^{6} \frac{- 1}{2} x + 3 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_{4}^{6} \frac{- 1}{2} x d x + \int_{4}^{6} 3 d x)

\int_{4}^{6} \frac{- 1}{2} x d x = - 5

\int_{4}^{6} 3 d x = 6

= - (- 5 + 6)

Vereinfache

= - 1

x \to 0 lim (\frac{( ∣ x ∣ )}{x})

\int (ln (x))^{2} x d x

y^{^{''}} + 4 y = 3 csc^{2} (2 t)

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

\frac{\partial}{\partial x} (2 sin (x y))