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
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/22 01:22:03
xJA_@3ڴCPк=PEP!BQaDAA^((^(v+43ZNTWZ9{{ki42,'2d5ޭ^x|z[uA"v{/Sjm3t7r5S: Z/J:<
Π&Ez;{o'qp|zB 1zAc fe&a<+İ@AkE(Zj_IĚ;?g.))~ E<!Y۲J
bzXE=F4$(H-5̨ Rx"R(
A0@U1Wi*?us`P:[uq9數1҅Fl0YQ`][> h
定积分:(1) lim(x→a) 1/(x-a) ∫[a,x] f(t)dt(2) lim(x→∞) ∫[x,x+1] (sint)/t dt
lim x→0[∫上x下0 cos(t^2)dt]/x ; lim x→0[∫上x下0 ln(1+t)dt]/(xsinx)
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
lim(x→o) (∫(0,x^2) √(1+t^2)dt) /x^2
lim x-0 (∫0-x^2 (ln(1+2t)dt)/x^4
lim(n,0)x/(1-e^x^2)∫(0,x)e^t^2dt
lim(x→0)[x∫(x→1)(cost/t^2)dt]
lim(x→0)[∫(0,x)(e^(t^2)-1)dt]/x^3
lim(x→0)[∫(0,x)(e^(t^2)-1)dt]/x^3
lim(x→0)(x∫(x→1)cost/t^2dt)
lim(x→0)1/x∫[x,0](1+t^2)*e^(t^2-x^2)dt
f(t)=lim x→无穷大 [t(1+1/x)^2tx] 求f'(t)
lim x-0 ∫0-π (ln(1+2t)dt)/x^4
lim x-0 ∫0-π (ln(1+2t)dt)/x^4
lim(1+2t/x)^3x=e^6t x->无穷
解一道大一极限题 lim(x→1)(1-x^2)/sinπxt=1-x,t-->0lim(2t-t^2)/sin(π-πt)=lim(2-t)t/sinπt=lim(2-t)t/πt=2/πlim(2-t)t/sinπt=lim(2-t)t/πt=2/π这部怎么来的
设f(t)=lim(x→∞)t(1+2/x)^(x-t),求f'(t)
lim→0[∫(上限x,下限0)(1+t^2)e^t^2dt]/xe^x^2