Question

求0到1√4-x^2的定积分

Answer 1

设x=2sina dx=2cosada
x从0--1, a从0--π/6
(0-1)∫(√4-x²)dx
=∫[√4(1-sin²a)]*2cosada
=∫2cosa*2cosa da
=4∫cos²a da
=2∫[1+cos(2a)]da
=2∫1da+∫cos(2a)d(2a)
=2a+sin2a
a从0--π/6 2*π/6+sin2π/6=(π/3)+(√3)/2

Answer 2

换元法令x=2sint
x∈[0,1] t∈[0,π/6]
∫[0,1] √(4-x²) dx=∫[0,π/6] 2cost d(2sint)
=∫[0,π/6] 4cos²t dt
=∫[0,π/6] 2cos2t+2 dt
=sin2t+2t | [0,π/6]
=√3/2 + π/3
希望对你有帮助

Answer 3

被积函数是什么？你给的是积分区间吧