integral from 1 to x^2 of 16xy^3

Lösung

\int_{1}^{x^{2}} 16 x y^{3} d y

4 x (x^{8} - 1)

Schritte zur Lösung

\int_{1}^{x^{2}} 16 x y^{3} d y

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 x \cdot \int_{1}^{x^{2}} y^{3} d y

Wende die Potenzregel an

= 16 x [\frac{y ^{4}}{4}]_{1}^{x^{2}}

Berechne die Grenzen:

\frac{x ^{8}}{4} - \frac{1}{4}

= 16 x (\frac{x ^{8}}{4} - \frac{1}{4})

Vereinfache

= 4 x (x^{8} - 1)

\int_{0}^{3} (t^{2} + 2 t - 3) d t

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

\int_{- 1}^{2} ∣ 2 x - 1 ∣ d x

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

\int_{- 7}^{1} (- x^{2} - 3 x) d x