integral from 0 to 5 of sin(t^2)

Lösung

\int_{0}^{5} sin (t^{2}) d t

\frac{π}{2} § (\frac{2}{π} \cdot 5) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

Schritte zur Lösung

\int_{0}^{5} sin (t^{2}) d t

Das ist ein nicht elementares Integral :

\int sin (x^{2}) d x = \frac{π}{2} § (\frac{2}{π} x)

= [\frac{π}{2} § (\frac{2}{π} t)]_{0}^{5}

Berechne die Grenzen:

\frac{π}{2} § (\frac{2}{π} \cdot 5) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

= \frac{π}{2} § (\frac{2}{π} \cdot 5) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

\int_{3}^{6} 3 d x

\int_{0}^{1} x^{2} (4 - x^{3})^{3} d x

\int_{0}^{1} \frac{2 x + 1}{x ^{2} + 1} d x

\int_{2}^{5} \frac{k}{x ^{3}} d x

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