Question

微积分:根号下(8-2y方)在(负根号2，根号2)的定积分

Answer 1

　　解：
　　∫[-√2→√2] √(8-2y²) dy
=√2∫[-√2→√2] √(4-y²) dy
令y=2sinu，则√(4-y²)=2cosu，dy=2cosudu，u：-π/4→π/4
=√2∫[-π/4→π/4] 4cos²u du
利用奇偶对称性
=8√2∫[0→π/4] cos²u du
=4√2∫[0→π/4] (1+cos2u) du
=4√2(u+(1/2)sin2u) |[0→π/4]
=4√2(π/4+1/2)
=√2π+2√2

Answer 2

∫[-√2→√2] √(8-2y²) dy
=√2∫[-√2→√2] √(4-y²) dy
令y=2sinu，则√(4-y²)=2cosu，dy=2cosudu，u：-π/4→π/4
=√2∫[-π/4→π/4] 4cos²u du
利用奇偶对称性
=8√2∫[0→π/4] cos²u du
=4√2∫[0→π/4] (1+cos2u) du
=4√2(u+(1/2)sin2u) |[0→π/4]
=4√2(π/4+1/2)
=√2π+2√2

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