Qual é ∫012x dx\int_{0}^{1} 2x \, dx∫012xdx?
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∫012x dx=[x2]01=1−0=1\int_{0}^{1} 2x \, dx = [x^{2}]_{0}^{1} = 1 - 0 = 1∫012xdx=[x2]01=1−0=1