求2∫(0,π)(6X눀+4-sin눀X)dx详细过程

2025-06-26 23:27:12
推荐回答(1个)
回答1:

2∫(0->π)[6x^2+4-(sinx)^2 ]dx
=∫(0->π)[12x^2+8-2(sinx)^2 ]dx
=∫(0->π)(12x^2+7+cos2x ) dx
= [4x^3+7x+ sin(2x)/2](0->π)
= 4π^3+7π