Question

这个定积分如何积？

(r^3)(1+r^2)^0.5 dr
r从0到4

Answer 1

x=r^2, dx = 2rdr, r:0->4, x:0->16.
S_{r:0->4}r^3(1+r^2)^(0.5)dr = S_{x:0->16}(1/2)x(1+x)^(0.5)dx
= (1/2)S_{x:0->16}(1+x)(1+x)^(0.5)dx - (1/2)S_{x:0->16}(1+x)^(0.5)dx
= (1/2)S_{x:0->16}(1+x)^(1.5)dx - (1/2)S_{x:0->16}(1+x)^(0.5)dx
= (1/2)*1/(2.5)(1+x)^(2.5)|_{x:0->16} - (1/2)*1/(1.5)(1+x)^(1.5)|_{x:0->16}

= (1/5)[(1+16)^(2.5) - 1] - (1/3)[(1+16)^(1.5) - 1]
= (1/5)(17)^(5/2) - (1/3)(17)^(3/2) - 1/5 + 1/3
= (289/5)(17)^(1/2) - (289/3)/(17)^(1/2) + 2/15
= [289/5 - 17/3](17)^(1/2) + 2/15
= (782/15)(17)^(1/2) + 2/15
= [782(17)^(1/2) + 2]/15

Answer 2