Question:medium
A curve with equation $y = x^3 - 8x^2 + 16x$ meets the $x$-axis at the origin $O$ and at a point $A$. Then the area of the region, bounded by the curve and the straight-line segment $OA$, is
A curve with equation $y = x^3 - 8x^2 + 16x$ meets the $x$-axis at the origin $O$ and at a point $A$. Then the area of the region, bounded by the curve and the straight-line segment $OA$, is
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If a polynomial has a squared factor like $(x-4)^2$, the curve touches the $x$-axis at that point without crossing it. This means the function doesn't change sign at that root, simplifying the absolute value consideration in the area integral.
Updated On: Jun 24, 2026
- $\frac{61}{3}$
- $\frac{62}{3}$
- $\frac{64}{3}$
- $\frac{65}{3}$
- $\frac{68}{3}$
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The Correct Option is C
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