integral from 0 to 1-x of (1-x-y)

Lösung

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

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

Schritte zur Lösung

\int_{0}^{1 - x} 1 - 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}^{1 - x} 1 d y - \int_{0}^{1 - x} x d y - \int_{0}^{1 - x} y d y

\int_{0}^{1 - x} 1 d y = - x + 1

\int_{0}^{1 - x} x d y = x (- x + 1)

\int_{0}^{1 - x} y d y = \frac{( - x + 1 ) ^{2}}{2}

= - x + 1 - x (- x + 1) - \frac{( - x + 1 ) ^{2}}{2}

Vereinfache

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

= x^{2} - 2 x - \frac{( - x + 1 ) ^{2}}{2} + 1

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