integral from 1 to sin(x) of 4t^3

Lösung

\int_{1}^{s i n (x)} 4 t^{3} d t

sin^{4} (x) - 1

Schritte zur Lösung

\int_{1}^{s i n (x)} 4 t^{3} d t

Entferne die Konstante:

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

= 4 \cdot \int_{1}^{s i n (x)} t^{3} d t

Wende die Potenzregel an

= 4 [\frac{t ^{4}}{4}]_{1}^{s i n (x)}

Berechne die Grenzen:

\frac{sin ^{4} ( x )}{4} - \frac{1}{4}

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

Vereinfache

= sin^{4} (x) - 1

\int_{0}^{\frac{1}{4}} sin^{2} (π x) d x

\int_{4}^{e^{x}} (y^{4} sin (y)) d y

\int_{0}^{\frac{π}{2}} 4 x cos (2 x) d x

\int_{2}^{4} x^{2} - 4 x + 4 d x

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