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

解答

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

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

求解步骤

\int_{x}^{5} 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)]_{x}^{5}

带入上下限计算后得:

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

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

\int_{0}^{2} 1 + 0.5 sin (π x) d x

\int_{1}^{x} e^{t^{3}} d t

\int_{2}^{\frac{5}{2}} \frac{1}{1 - x ^{2}} d x

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\int_{x^{4}}^{x^{5}} (- 4 x - 2)^{3} d x