integral from 0 to pi/2 of 2sin(x)cos(x)

Lösung

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

1

Schritte zur Lösung

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

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{\frac{π}{2}} sin (x) cos (x) d x

Wende U-Substitution an

= 2 \cdot \int_{0}^{1} u d u

Wende die Potenzregel an

= 2 [\frac{u ^{2}}{2}]_{0}^{1}

Berechne die Grenzen:

\frac{1}{2}

= 2 \cdot \frac{1}{2}

Vereinfache

= 1

\int_{x^{2}}^{x^{4}} u d u

\int_{1}^{4} (x^{2} + 2 x - 5) d x

\int_{0}^{3} - x^{2} + 2 x + 3 d x

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

\int_{a}^{b} 0 d x