第1个回答 2022-05-11
(1)
∫(0->3) √(x+1)dx
=(2/3)(x+1)^(3/2)|(0->3)
=(2/3)(8-1)
=14/3
ans : C
(2)
f(x) =x(cosx)^3/(x^2+1)
f(-x) =-f(x)
=>
∫(-5->5) x(cosx)^3/(x^2+1) dx =0
ans :D
(3)
y=∫(3->x^2) tf(t^2) dt
y'
=x^2.f(x^4) .(x^2)'
=x^2.f(x^4) .(2x)
=2x^3.f(x^4)
ans :A
(4)
∫(2->+无穷) dx/x^p
=[-1/(p-1)][ 1/x^(p-1)]|(2->+无穷)
收敛
p-1>0
p>1
ans :B
(5)
lim(x,y)->(0,0) sin(xy)/y = 0
lim(n->无穷)∫(0->1) x^(n-1) dx
=lim(n->无穷) (1/n) x^n |(0->1)
=lim(n->无穷) (1/n)
=0
ans :B本回答被提问者采纳
∫(0->3) √(x+1)dx
=(2/3)(x+1)^(3/2)|(0->3)
=(2/3)(8-1)
=14/3
ans : C
(2)
f(x) =x(cosx)^3/(x^2+1)
f(-x) =-f(x)
=>
∫(-5->5) x(cosx)^3/(x^2+1) dx =0
ans :D
(3)
y=∫(3->x^2) tf(t^2) dt
y'
=x^2.f(x^4) .(x^2)'
=x^2.f(x^4) .(2x)
=2x^3.f(x^4)
ans :A
(4)
∫(2->+无穷) dx/x^p
=[-1/(p-1)][ 1/x^(p-1)]|(2->+无穷)
收敛
p-1>0
p>1
ans :B
(5)
lim(x,y)->(0,0) sin(xy)/y = 0
lim(n->无穷)∫(0->1) x^(n-1) dx
=lim(n->无穷) (1/n) x^n |(0->1)
=lim(n->无穷) (1/n)
=0
ans :B本回答被提问者采纳
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