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f(x)=x^2-x(0→2)f(x)dx+2(0→1)f(x)dx解這種類題目,先要了解(0→2)f(x)dx,(0→1)f(x)dx是定数簡化直感,令a=(0→2)f(x)dx,b=(0→1)f(x)dx,則原式為f(x)=x^2-ax+2b1對一式在(0→1)積分得,b...
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