
高数问题?求具体步骤
1个回答
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原式=
21/4∫x^2ydy
=x^2*21/4∫d(y^2/2)
=x^2*21/4*(y^2/2)+C
=x^2*y^2*21/8+C
上限为1,下限为x^2
=x^2*21/8-x^2*x^2*21/8
=(x^2-x^4)*21/8
21/4∫x^2ydy
=x^2*21/4∫d(y^2/2)
=x^2*21/4*(y^2/2)+C
=x^2*y^2*21/8+C
上限为1,下限为x^2
=x^2*21/8-x^2*x^2*21/8
=(x^2-x^4)*21/8
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