請問這題微積分怎麼解?

令 f(x) = x^4 + 2x^3 + 6 , g(x) = x^4 + x^3 + 3x^2 - 2x + 6
先求 f 與 g 之交點:
f - g
= x^3 - 3x^2 + 2x
= x( x^2 - 3x + 2 )
= x(x-1)(x-2)
故兩函數交於 x = 0 , 1 , 2

當 0 ≦ x < 1 , f - g = x(x-1)(x-2) ≧ 0 , 故 ∣ f - g ∣ = f - g
當 1 ≦ x < 2 , f - g = x(x-1)(x-2) ≦ 0 , 故 ∣ f - g ∣ = g - f

∫ ∣ f - g ∣ dx , from x = 0 to x = 1
= ∫ ( f - g ) dx
= ∫ ( x^3 - 3x^2 + 2x ) dx
= [ (1/4)x^4 - x^3 + x^2 ] , from x = 0 to x = 1
= 1/4

∫ ∣ f - g ∣ dx , from x = 1 to x = 2
= ∫ ( g - f ) dx
= ∫ ( - x^3 + 3x^2 - 2x ) dx
= [ -(1/4)x^4 + x^3 - x^2 ] , from x = 1 to x = 2
= 0 - (-1/4)
= 1/4

所圍成的區域
= ∫ ∣ f - g ∣ dx , from x = 0 to x = 2
= { ∫ ∣ f - g ∣ dx , from x = 0 to x = 1 } + { ∫ ∣ f - g ∣ dx , from x = 1 to x = 2 }
= 1/4 + 1/4
= 1 / 2 ..... Ans