integral from 0 to y of (x+y)^2

Lösung

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

2 y^{3} + \frac{y ^{3}}{3}

Schritte zur Lösung

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

Multipliziere aus

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

= \int_{0}^{y} 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_{0}^{y} y^{2} d x + \int_{0}^{y} 2 y x d x + \int_{0}^{y} x^{2} d x

\int_{0}^{y} y^{2} d x = y^{3}

\int_{0}^{y} 2 y x d x = y^{3}

\int_{0}^{y} x^{2} d x = \frac{y ^{3}}{3}

= y^{3} + y^{3} + \frac{y ^{3}}{3}

Vereinfache

= 2 y^{3} + \frac{y ^{3}}{3}

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

\int_{2}^{4} \frac{1}{x ( ln ( x ) ) ^{3}} d x

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

\int_{0}^{1} (x^{2} + 6) e^{5 x} d x

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