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 the curve and the x-axis (line segment OA), we need to find the roots where it intersects the x-axis, which act as our integration limits.
Step 2: Key Formula or Approach:
Set \(y = 0\) to find the x-intercepts.
The area is \(\int_{x_1}^{x_2} y \, dx\).
Step 3: Detailed Explanation:
Find the roots:
\[ x^3 - 8x^2 + 16x = 0 \] Factor out \(x\):
\[ x(x^2 - 8x + 16) = 0 \] Factor the quadratic:
\[ x(x - 4)^2 = 0 \] The roots are \(x = 0\) (Origin O) and \(x = 4\) (Point A).
The region lies between \(x = 0\) and \(x = 4\). Because \((x-4)^2\) is always non-negative, the function is positive on \((0, 4)\).
Compute the area:
\[ \text{Area} = \int_{0}^{4} (x^3 - 8x^2 + 16x) dx \] Integrate term by term:
\[ = \left[ \frac{x^4}{4} - \frac{8x^3}{3} + \frac{16x^2}{2} \right]_{0}^{4} \] \[ = \left[ \frac{x^4}{4} - \frac{8x^3}{3} + 8x^2 \right]_{0}^{4} \] Evaluate at upper limit (lower limit gives 0):
\[ = \frac{4^4}{4} - \frac{8(4^3)}{3} + 8(4^2) \] \[ = \frac{256}{4} - \frac{8(64)}{3} + 8(16) \] \[ = 64 - \frac{512}{3} + 128 \] \[ = 192 - \frac{512}{3} \] Find a common denominator:
\[ = \frac{576}{3} - \frac{512}{3} = \frac{64}{3} \] Step 4: Final Answer:
The area is \(\frac{64}{3}\).