Question

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. $\frac{61}{3}$
• B. $\frac{62}{3}$
• C. $\frac{64}{3}$
• D. $\frac{65}{3}$
• E. $\frac{68}{3}$

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To find the area bounded by a curve and the $x$-axis, we first find the points where the curve intersects the $x$-axis (the roots) and then integrate the absolute value of the function between those points.

Step 2: Detailed Explanation:
1. Find the intersection points with the $x$-axis (set $y=0$):
\[ x^3 - 8x^2 + 16x = 0 \]
\[ x(x^2 - 8x + 16) = 0 \]
\[ x(x - 4)^2 = 0 \]
The roots are $x = 0$ (origin $O$) and $x = 4$ (point $A$). The line segment $OA$ lies on the $x$-axis from $x=0$ to $x=4$.
2. Calculate the area by integration:
\[ \text{Area} = \int_{0}^{4} (x^3 - 8x^2 + 16x) dx \]
\[ \text{Area} = \left[ \frac{x^4}{4} - \frac{8x^3}{3} + \frac{16x^2}{2} \right]_{0}^{4} \]
\[ \text{Area} = \left[ \frac{x^4}{4} - \frac{8x^3}{3} + 8x^2 \right]_{0}^{4} \]
3. Substitute the limits:
\[ \text{Area} = \left( \frac{4^4}{4} - \frac{8(4^3)}{3} + 8(4^2) \right) - (0) \]
\[ \text{Area} = \left( 64 - \frac{512}{3} + 128 \right) \]
\[ \text{Area} = 192 - \frac{512}{3} \]
\[ \text{Area} = \frac{576 - 512}{3} = \frac{64}{3} \]

Step 3: Final Answer:
The area of the region is $\frac{64}{3}$ square units.