The area of the region bounded by the curve $y = \frac { x ^ { 3 } } { 16 }$ and the line y = x is 9.

Question

Quizplus · Accepted Answer

To find the area of the region bounded by the curve $y = \frac{x^3}{16}$ and the line $y = x$, we first need to find the points of intersection by setting $\frac{x^3}{16} = x$, which simplifies to $x^3 = 16x$. Solving this equation gives us $x = 0$, $x = 4$, and $x = -4$. However, since we are looking for the area between the curves, we focus on the positive intersection points $x = 0$ and $x = 4$. The area can be calculated by integrating the difference between the two functions from $x = 0$ to $x = 4$:$$\text{Area} = \int_{0}^{4} (x - \frac{x^3}{16}) dx = \left[\frac{1}{2}x^2 - \frac{1}{64}x^4\right]_0^4 = \left(8 - \frac{64}{64}\right) - 0 = 8 - 1 = 7$$Therefore, the area of the region is 7, not 9.

The Area of the Region Bounded by the Curve y=x316y = \frac { x ^ { 3 } } { 16 }y=16x3​

The Area of the Region Bounded by the Curve $y = \frac { x ^ { 3 } } { 16 }$