∫(上限1,下限-1)(X^3*cosx+2)dx 等于多少?
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∫(上限1,下限-1)(X^3*cosx+2)dx 等于多少?
X的三次方乘以cosx加2括号dx.
X的三次方乘以cosx加2括号dx.
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嗯.我来给你解答吧:
原式=x^3*sinx|(1,-1)-∫(1,-1)(3x^2sinx)dx
=sin1-(-sin(-1))-[(-3x^2cosx)|(1,-1)-∫(1,-1)(-6xcosx)dx
=-[-3cos1+3cos(-1)]+(-6xsinx)|(1,-1)-∫(1,-1)(-6sinx)dx
=[-6sin1-6(sin-1)]+(-6cosx)|(1,-1)
=-6cos1+6cos(-1)
=0
原式=x^3*sinx|(1,-1)-∫(1,-1)(3x^2sinx)dx
=sin1-(-sin(-1))-[(-3x^2cosx)|(1,-1)-∫(1,-1)(-6xcosx)dx
=-[-3cos1+3cos(-1)]+(-6xsinx)|(1,-1)-∫(1,-1)(-6sinx)dx
=[-6sin1-6(sin-1)]+(-6cosx)|(1,-1)
=-6cos1+6cos(-1)
=0
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