integral from 0 to 2-x of 2-x-y

Lösung

\int_{0}^{2 - x} 2 - x - y d y

x^{2} - 4 x - \frac{( - x + 2 ) ^{2}}{2} + 4

Schritte zur Lösung

\int_{0}^{2 - x} 2 - x - 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_{0}^{2 - x} 2 d y - \int_{0}^{2 - x} x d y - \int_{0}^{2 - x} y d y

\int_{0}^{2 - x} 2 d y = 2 (- x + 2)

\int_{0}^{2 - x} x d y = x (- x + 2)

\int_{0}^{2 - x} y d y = \frac{( - x + 2 ) ^{2}}{2}

= 2 (- x + 2) - x (- x + 2) - \frac{( - x + 2 ) ^{2}}{2}

Vereinfache

2 (- x + 2) - x (- x + 2) - \frac{( - x + 2 ) ^{2}}{2} : x^{2} - 4 x - \frac{( - x + 2 ) ^{2}}{2} + 4

= x^{2} - 4 x - \frac{( - x + 2 ) ^{2}}{2} + 4

\int_{0}^{2} 4 x (x^{2} + 1)^{3} d x

\int_{1}^{2} e^{x - 1} - e d x

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

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

\int_{0}^{6} x + 8 d x