∫(-3~3) [√(9 - x²) - x³] dx
= 2∫(0~3) √(9 - x²) - 0,第一個偶函數(shù),第二個奇函數(shù)
令x = 3sinz,dx = 3cosz dz
= 2∫(0~π/2) (3cosz)² dz
= 18∫(0~π/2) (1 + cos2z)/2 dz
= 9(z + 1/2 • sin2z) |(0~π/2)
= 9π/2
= 4.5π= 2∫(0~3) √(9 - x²) - 0，第一個偶函數(shù)，第二個奇函數(shù)令x = 3sinz，dx = 3cosz dz沒看懂f(x) = √(9 - x²)若f(-x) = f(x)則f(x)是偶函數(shù)，并且∫(-a~a) f(x) dx = 2∫(0~a) f(x) dx相反，奇函數(shù)就是f(-x) = - f(x)，∫(-a~a) f(x) dx = 0之后的做法就是第二換元法，需用三角函數(shù)代換觀察√[9 - (3sinz)²] = √(9 - 9sin²z) = √(9cos²z) = 3cosz剛好消掉根號，所以作代換x = 3sinz上下限會改變的，當(dāng)x = 0時0 = 3sinz => z = 0當(dāng)x = 3時3 = 3sinz => sinz = 1 => z = π/2于是由(0~3)變?yōu)?0~π/2)∫(-3~3) [√(9 - x²) - x³] dx-x^3到第二步為什么為0是∫(-3~3) x³ dx = 0，另一個變?yōu)?∫(0~3) √(9 - x²) dx