定积分∫[-√2,√2]√(4-x2)dx=

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/30 15:12:53
定积分∫[-√2,√2]√(4-x2)dx=
xQn@UJv: z[hlj#"NTm ġR{-P C=u $2Λv˟κe{8vMa[-(='4Ϩ#xP<~>z@RGyQeS֣THE^(Q?v7q]Z{))PY=SfʔL d+|?e^8ɔ p0ZK/HQVpxxъl{n p`,@C`zK/.ZWb>l9_ȂH]̛7]{LTȲn`k\V_/%YWV۵;<[}o߯ω'L8qL;S/Q- KnQ@ 'Cg1A*7-3 b&٘$ 65o/"cL"

定积分∫[-√2,√2]√(4-x2)dx=
定积分∫[-√2,√2]√(4-x2)dx=

定积分∫[-√2,√2]√(4-x2)dx=

设 x = 2 sint,则 dx = 2cost*dt,积分限变为:[-π/4, π/4]
那么积分可以变换为:
∫√[4 - 4(sint)^2] * 2cost*dt
=∫4(cost)^2 *dt
=2∫2(cost)^2 *dt
=2∫[1 + cos(2t)]*dt
=2∫dt + ∫cos(2t)*d(2t)
=2t + sin2t
=2(π/4 + π/4) + [sin(π/2) - sin(-π/2)]
=π/2 + 2