设连续函数f(x)f(x)f(x)满足f(x)=x2−∫02f(x)dxf(x) = x^2 - \int_{0}^{2}f(x)dxf(x)=x2−∫02f(x)dx，则∫02f(x)dx=\int_{0}^{2}f(x)dx =∫02f(x)dx=（ ）.
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