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

解答

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

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

求解步骤

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

这是非初等函数积分:

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

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

带入上下限计算后得:

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

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