Practicing Success
Practicing Success
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The area enclosed between the curves $y = x$ and $y=2x-x^2$ (in square units), is |
$\frac{1}{2}$ $\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{4}$ |
$\frac{1}{6}$ |
Clearly, Shaded Area = $\int\limits_0^1\{(2x-x^2)-x\}dx$ ⇒ Shaded Area = $\int\limits_0^1(x-x^2)dx=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac{1}{6}$
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