integral from 0 to pi of 5x^2sin(1/2 x)

Lösung

\int_{0}^{π} 5 x^{2} sin (\frac{1}{2} x) d x

5 (8 π - 16)

Dezimale

45.66370 \dots

Schritte zur Lösung

\int_{0}^{π} 5 x^{2} sin (\frac{1}{2} x) d x

Entferne die Konstante:

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

= 5 \cdot \int_{0}^{π} x^{2} sin (\frac{1}{2} x) d x

Vereinfache

\frac{1}{2} x : \frac{x}{2}

= 5 \cdot \int_{0}^{π} x^{2} sin (\frac{x}{2}) d x

Wende die partielle Integration an

= 5 [- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x]_{0}^{π}

\int - 4 x cos (\frac{x}{2}) d x = - 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))

= 5 [- 2 x^{2} cos (\frac{x}{2}) - (- 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2})))]_{0}^{π}

Vereinfache

= 5 [- 2 x^{2} cos (\frac{x}{2}) + 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))]_{0}^{π}

Berechne die Grenzen:

8 π - 16

= 5 (8 π - 16)

x \to 1 lim (9 x + 5)

t^{2} y^{^{''}} + t y^{^{'}} + 4 y = sin (2 ln (t))

\int - 5 tan^{5} (4 x) sec^{6} (4 x) d x

x \to 0 lim (\frac{tan ( 9 x )}{sin ( 4 x )})

\int \frac{x}{4 x + 4} d x