integration

(a) ∫ f(x) dx (0 -> k)

Let u = k - x, when x = k, u = 0 ; when x = 0, u = k

du = -dx

= - ∫ f(k - u) du (k -> 0)

= ∫ f(k - x) dx (0 -> k) because u is dummy

(b) Let f(x) = x^2 / [x^2+ ( x - k )^2]

By (a) ∫ f(x) dx (0 -> k) = ∫ f(k - x) dx (0 -> k)

∫ x^2 / [x^2+ ( x - k )^2] dx = ∫ (x - k)^2 / [(x - k)^2+ x^2] dx

2 ∫ x^2 / [x^2+ ( x - k )^2] dx = ∫ [x^2 + (x - k)^2] / [(x - k)^2+ x^2] dx

∫ x^2 / [x^2+ ( x - k )^2] dx = ∫ dx/2

∫ x^2 / [x^2+ ( x - k )^2] dx = k/2

intergrate ( k to 0 ) f(x) dx
=intergrate ( 0 to k ) f(k-t) d(k-t) (x=k-t)
= - intergrate ( 0 to k ) f(k-t) dt
=intergrate ( k to 0 ) f(k-t) dt
= intergrate ( k to 0 ) f(k-x) dx

2012-03-05 11:39:12 補充：
另外一題soln:
http://www.shareapic.net/View-26157349-integration.html
自已入去題啦