integral from x/2 to 1-x/2 of 2-x-2y

Lösung

\int_{\frac{x}{2}}^{1 - \frac{x}{2}} 2 - x - 2 y d y

x^{2} - 2 x + 1

Schritte zur Lösung

\int_{\frac{x}{2}}^{1 - \frac{x}{2}} 2 - x - 2 y d y

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{\frac{x}{2}}^{1 - \frac{x}{2}} 2 d y - \int_{\frac{x}{2}}^{1 - \frac{x}{2}} x d y - \int_{\frac{x}{2}}^{1 - \frac{x}{2}} 2 y d y

\int_{\frac{x}{2}}^{1 - \frac{x}{2}} 2 d y = 2 - 2 x

\int_{\frac{x}{2}}^{1 - \frac{x}{2}} x d y = x - x^{2}

\int_{\frac{x}{2}}^{1 - \frac{x}{2}} 2 y d y = 1 - x

= 2 - 2 x - (x - x^{2}) - (1 - x)

Vereinfache

2 - 2 x - (x - x^{2}) - (1 - x) : x^{2} - 2 x + 1

= x^{2} - 2 x + 1

\int_{0}^{1} - 3 x (- 2 x^{2} - 1)^{3} d x

\int_{- 1}^{0} 2 x^{17} e^{1 + x^{9}} d x

\int_{0}^{2} (x)^{2} d x

\int_{0}^{1} \frac{2 x ^{2} - 5}{x} d x

\int_{- \infty}^{0} e^{2 x + 1} d x