Question

两道定积分题目

Answer 1

(1)
let
u=x^2-t^2
du =-2tdt
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(u)du

d/dx {∫(0->x) tf(x^2-t^2)dt}
=d/dx {(1/2)∫(0->x^2) f(u)du}
=2xf(x^2)

(2)
f(x) = ∫(0->x) sint/(x-t) dt

To find: ∫(0->π) f(x) dt
solution:
let
u=x-t
du =-dt
t=0, u=x
t=x, u=0

f(x) =∫(0->x) sint/(x-t) dt
f'(x) = lim(x->0) sinx/x =1
=>
f(x) = x+C

f(x) =∫(0->x) sint/(x-t) dt
f(0) =0
=> C=0
ie
f(x) =x

∫(0->π) f(x) dt
=∫(0->π) x dt
=(1/2)π^2