lim (1/x^2)*∫[(1+2t)*e^(t-x^2)]dt=?注:积分范围是0-》x^2 lim的下面是x-》00
问题描述:
lim (1/x^2)*∫[(1+2t)*e^(t-x^2)]dt=?注:积分范围是0-》x^2 lim的下面是x-》00
答
=lim (1/x^2)*∫[(1+2(t-x^2 + x^2))*e^(t-x^2)]d(t-x^2)令u=t-x^2,则u的范围是 -x^2 至 0.原式=lim (1/x^2)*∫[(1+2(u + x^2))*e^u ]du=lim (1/x^2)*{∫[(1 + 2x^2 + 2u )*e^u ]du }=lim (1/x^2)*{∫(1 + 2x^2)e^u ...你这个是不是先用换元法-----》令u=t-x^2再用基本积分公式与分部积分法(把x视为常数)-------》lim {(1 + 2x^2)∫e^u du + 2∫u *e^u du}/x^2再用洛比达法则----->-2 + 4·lim (2x) / [2x·e^(x^2)]问一下~~~~~~∫f(t-x)dt