請求解答Definite Integration兩條問題

Q1b
∫[-7,4] [1-f(t)-g(t)] dt
=∫[-7,4] dt - ∫[-7,4] f(t)dt - ∫[-7,4] g(t)dt
=11- ∫[-7,4] f(-t)dt -∫[-7,4] g(-t)dt
=11+ ∫[-7,4] f(-t)d(-t) +∫[-7,4] g(-t)d(-t)
=11+∫[7,-4]f(u)du+∫[7,-4]g(u)du [Let u=-t]
=11-∫[-4,7]f(u)du-∫[-4,7]g(u)du
=11-2-5
=4

Q2c
x=a is the vertical line divides the area of the region.
0<a<2.
∫[0,a] [(x^3-6x^2+12x+2)-(x^3+2)] dx=∫[a,2][(x^3-6x^2+12x+2)-(x^3+2)]dx
∫[0,a] (12x-6x^2)dx=∫[a,2] (12x-6x^2)dx
[6x^2-2x^3] | [0,a]=[6x^2-2x^3] | [a,2]
6a^2-2a^3=8-(6a^2-2a^3)
a^3-3a^2+2=0

1(a) ∫ [f(t)+g(t)+t] dt
=∫f(t)dt+∫g(t)dt+∫tdt
=2+5+t^2/2 |[-4,7]
=47/2(b) ∫ [1-f(t)-g(t)] dt
=∫dt-∫f(t)dt-∫g(t)dt
=11-∫f(-t)dt-∫g(-t)dt (Let y=-t)
=11+∫f(y)dy+∫g(y)dy (Notice that the lower and upper limit)
=11-2-5
=42(a) x^3+2=x^3-6x^2+12x+2
6x^2-12x=0
x^2-2x=0
x=0 or 2
Intersection point (0,2) and (2,0)(b) Area
=∫ (x^3-6x^2+12x+2)-(x^3+2) dx [from 0 to 2]
=6 ∫ (2x-x^2) dx
=6x^2-2x^3 |[0,2]
=24-16
=8(c) The area between the y axis and x=a is equal to
∫ (x^3-6x^2+12x+2)-(x^3+2) dx [from 0 to a]
=6 ∫ (2x-x^2) dx
=6x^2-2x^3 |[0,a]
=6a^2-2a^3Swt it equal to 4, 6a^2-2a^3=4
3a^2-a^3=2
a^3-3a^2+2=0


2010-12-29 14:31:12 補充：
"Swt " should be "Set"