Question

12题求解，

Answer 1

(2)
∫(0->x) t(f(2x-t)dt = e^x -1
f(1) = 1
To find: ∫(1->2) f(x) dx
solution :
let
u= 2x-t
du =-dt
t=0, u=2x
t=x, u=x
∫(0->x) t(f(2x-t)dt
=∫(2x->x) (2x-u) f(u) (-du)
=∫(x->2x) (2x-u) f(u) du
=∫(x->2x) (2x-t) f(t) dt
∫(0->x) t(f(2x-t)dt = e^x -1
∫(x->2x) (2x-t) f(t) dt = e^x -1
2x∫(x->2x) f(t) dt - ∫(x->2x) tf(t) dt = e^x -1
d/dx {2x∫(x->2x) f(t) dt - ∫(x->2x) tf(t) dt} = d/dx {e^x -1 }
2∫(x->2x) f(t) dt + 2x[ 2f(2x) - f(x) ] - [ 4xf(2x) - xf(x) ] = e^x
x=1
2∫(1->2) f(t) dt + 2[ 2f(2) - f(1) ] - [ 4f(2) - f(1) ] = e
2∫(1->2) f(t) dt -3f(1) = e
2∫(1->2) f(x) dx -3 = e
∫(1->2) f(x) dx =(1/2) (e+3)