Question

求这道定积分，求过程谢谢

9小题和10小题

Answer 1

(9)

let

u=x^2-t^2

du=-2t dt

t=0, u=x^2

t=x, u=0

∫(0->x) tf(x^2-t^2) dt

= -(1/2)∫(x^2->0) f(u) du

=(1/2)∫(0->x^2) f(t) dt

d/dx ∫(0->x) tf(x^2-t^2) dt

=d/dx [ (1/2)∫(0->x^2) f(t) dt ]

=(1/2)(2x)f(x^2)

=xf(x^2)

ans :A

(10)

F(x)= ∫(x->x+2π)  e^(sint). sint dt

F'(x) 

= e^[sin(x+2π)] .sin(x+2π)  - e^(sinx) .sinx

=e^(sinx) .sinx  - e^(sinx) .sinx

=0

=>F(x) =C

let

u= π -t

du = -dt

t=π, u=0

t=2π, u=π

F(0) 

=∫(0->2π)  e^(sint). sint dt

=∫(0->π)  e^(sint). sint dt  + ∫(π->2π)  e^(sint). sint dt

=∫(0->π)  e^(sint). sint dt  - ∫(0->π)  e^(-sinu). sinu du

=∫(0->π)  e^(sint). sint dt  - ∫(0->π)  e^(-sint). sint dt

=∫(0->π) sint . [ e^(sint) - e^(-sint) ] dt 

>0

=> C >0

ans : A 正常数