Question

y=(2-x^2)^1.5是什么图形,求在—1到1的定积分

Answer 1

定积分为2.8284

Answer 2

∫(-1,1)(2-x^2)^(3/2)dx=2∫(0,1)(2-x^2)^(3/2) （这是由于被积函数是偶函数得到的）
=2∫(π/2,0)sqrt(2)sint*2(sint)^2*sqrt(2)(-sint)dt (令x=sqrt(2)cost, dx=-sqrt(2)sintdt)
=8 ∫(0,π/2)(sint)^4dt
=8 ∫(0,π/2)[(1-cos2t)/2]^2dt
=2∫(0,π/2)[1-2cos2t+(cos2t)^2]dt
=∫(0,π/2)[3-4os2t+(cos4t)]dt
=3t-2sin2t+1/4*sin4t|(0,π/2)
= 3π/2.

追问

定义域求错了，应该是π/4到π/2

追答

是的，
∫(-1,1)(2-x^2)^(3/2)dx=2∫(0,1)(2-x^2)^(3/2) （这是由于被积函数是偶函数得到的）
=2∫(π/2,π/4)sqrt(2)sint*2(sint)^2*sqrt(2)(-sint)dt (令x=sqrt(2)cost, dx=-sqrt(2)sintdt)
=8 ∫(π/4,π/2)(sint)^4dt
=8 ∫(π/4,π/2)[(1-cos2t)/2]^2dt
=2∫(π/4,π/2)[1-2cos2t+(cos2t)^2]dt
=∫(π/4,π/2)[3-4os2t+(cos4t)]dt
=3t-2sin2t+1/4*sin4t|(π/4,π/2)
= 3π/4+2.

Answer 3

看图片