integral from |y| to 1 of (x+y)^2

Lösung

\int_{∣ y ∣}^{1} (x + y)^{2} d x

y^{2} - y^{2} ∣ y ∣ + y (1 - y^{2}) + \frac{1 - ∣ y ∣ ^{3}}{3}

Schritte zur Lösung

\int_{∣ y ∣}^{1} (x + y)^{2} d x

Multipliziere aus

(y + x)^{2} : y^{2} + 2 y x + x^{2}

= \int_{∣ y ∣}^{1} y^{2} + 2 y x + x^{2} 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_{∣ y ∣}^{1} y^{2} d x + \int_{∣ y ∣}^{1} 2 y x d x + \int_{∣ y ∣}^{1} x^{2} d x

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

\int_{∣ y ∣}^{1} 2 y x d x = y (1 - y^{2})

\int_{∣ y ∣}^{1} x^{2} d x = \frac{1 - ∣ y ∣ ^{3}}{3}

= y^{2} - y^{2} ∣ y ∣ + y (1 - y^{2}) + \frac{1 - ∣ y ∣ ^{3}}{3}

\int_{2}^{6} \frac{4}{x} - \frac{8}{x ^{2}} d x

\int_{0}^{2 π} 1 + sin^{2} (x) d x

\int_{1}^{e} \frac{( 3 + ln ( x ) ) ^{4}}{x} d x

\int_{0}^{4 - y^{2}} (2 - z) d z

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