Integration

Oh, you got several answers here, let's me try it also..
Q1
∫[4,5] ∫[3,v] ∫[2,t] (x+1)dx dt dv

step 1,
∫[2,t] (x+1)dx
x[2,t] (x^2)/2+x
(t^2)/2+t-(2^2)/2-2
(t^2)/2+t-4

alternatively:
(x+1)d(x+1)*dx/dx+1
x[2,t] (x+1)^2/2
(t+1)^2/2-(2+1)^2/2
(t^2)/2+t+1/2-9/2
(t^2)/2+t-4

step2,
∫[3,v] (t^2)/2+t-4 dt
t[3,v] (t^3)/6+(t^2)/2-4t
(v^3)/6+(v^2)/2-4v-((3^3)/6+(3^2)/2-4*3)
(v^3)/6+(v^2)/2-4v-(27/6+9/2-12)
(v^3)/6+(v^2)/2-4v-(54/6-12)
(v^3)/6+(v^2)/2-4v-(9-12)
(v^3)/6+(v^2)/2-4v+3

step3,
∫[4,5] (v^3)/6+(v^2)/2-4v+3 dv
v[4,5] (v^4)/24+(v^3)/6-2(v^2)+3v
(5^4)/24+(5^3)/6-2(5^2)+3*5-((4^4)/24+(4^3)/6-2(4^2)+3*4)
625/24+125/6-50+15-(256/24+64/6-32+12)
625/24+500/24-35-(256/24+256/24-20)
1125/24-35-512/24+20
1125/24-15-512/24
613/24-15
613/24-360/24
253/24

Q2
Area =
∫[2pi,3pi] sinx dx + Triangle
x[2pi,3pi] -cosx + pi*2/2
-cos(3pi)+cos(2pi) + pi
1+1+pi
2+pi

2008-06-29 23:41:44 補充：
Audrey Hepburn the Elegant answer step 4 should be
(t+1)^3/3-9(t+1)
cause chain rule is need if don't break the brackets (so I break it...)

2008-06-29 23:43:06 補充：
hawk_wing_1999 answer step 2 should be
(t^2)/2+t-2-2

004 is correct Q1 = 253/24

For the sake of convenience, I don't type in the range
∫∫∫(x+1)dxdtdv
=∫∫[t/2+t-2-2]dtdv
=∫∫[t+t-4]dtdv
=∫[(1/3)v+(1/2)v-4v-9-9/2+12]dv
=∫[(1/3)v+(1/2)v-4v-3/2]dv
=∫[(1/12)v4+(1/6)v-2v-(3/2)v]
=-119/40