定積分數學題

Let 1 + x = u, so dx = du, x = u - 1.
When x = 1, u = 2. When x = 0, u = 1.
So the integral changes to
S [(u - 1) sqrt u] du from 2 to 1.
= S (u sqrt u - sqrt u) du
= S [u^(3/2) - u^(1/2)] du
= (2/5)u^(5/2) - (2/3)u^(3/2) from 2 to 1.
= {(2/5)[2^(5/2)] - (2/3)[2^(3/2)]} - {(2/5)[1^(5/2) - (2/3)[1^(3/2)]}
= (8 sqrt 2)/5 - (4 sqrt 2)/3 - 2/5 + 2/3
= (24 sqrt 2 - 20 sqrt 2 - 6 + 10)/15
= (4 sqrt 2 + 4)/15
= 4( 1 + sqrt 2)/15.