求定积分∫(1,-1) (|x|+sinx)x^2 dx ?

求步骤,谢谢
2025-01-04 03:13:34
推荐回答(3个)
回答1:

∫(1,-1) (|x|+sinx)x^2 dx
=∫(1,-1) |x|x^2 dx +∫(1,-1) sinxx^2 dx
=2∫(1,0) x^3 dx +0
=2*1/4
=1/2

回答2:

∫(1,-1) (|x|+sinx)x^2 dx
=∫(1,-1) (|x|x^2+sinx*x^2) dx
=2∫(1,0) x^3dx
=1/2

注意:奇函数sinx*x^2在对称区间积分0,偶函数是2倍

回答3:

果断不会做了,惭愧