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Area under the curve $y=x^2-4 x$ within the $x$-axis and the line $x=2$, is
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$\frac{16}{3}$ sq.unit
$\int_0^2\left(x^2-4 x\right) d x=\left[\frac{x^3}{3}-\frac{4 x^2}{2}\right]_0^2=\frac{16}{3} s q .$unit.
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