设f(t)=e^(-x^2)从1到t的积分 求t^2*f(t)从0到1的积分，详细过程 谢谢 设f(x)连续，则(d/dx)积分(从x^2到e^(-x))...
\u5df2\u77e5f(x)\u7b49\u4e8ee^(-t^2)\u4ece0\u5230x^2\u7684\u5b9a\u79ef\u5206,\u6c42xf(x)\u4ece0\u52301\u7684\u5b9a\u79ef\u5206\u770b
\u222b[1\u2192cosx] (t²-e^x)f(t) dt
=\u222b[1\u2192cosx] t²f(t) dt - e^x\u222b[1\u2192cosx] f(t) dt
\u56e0\u6b64\u6c42\u5bfc\u540e\u4e3a
cos²xf(cosx)*(-sinx) - \u222b[1\u2192cosx] f(t) dt - e^xf(cosx)(-sinx)
=-sinxcos²xf(cosx) - \u222b[1\u2192cosx] f(t) dt + e^xsinxf(cosx)
所以∫(1,0) t^2*f(t)dt
=∫(1,0) t^2 df'(t)
=∫(1,0)t^2 de^(-t^2)
=t^2*e^(-t^2) /(1,0)-∫(1,0) e^(-t^2) *dt^2
=1/e +e^(-t^2) /(1,0)
=1/e +1/e -1
=2/e -1
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