The graphs of f(x) = x^2 and g(x) = cx^3 (whe | vedclass

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(C) The area $A$ between the two curves $f(x) = x^2$ and $g(x) = cx^3$ on the interval $[0, 1/c]$ is given by the integral of the difference between t

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$1/2$

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(C) The area $A$ between the two curves $f(x) = x^2$ and $g(x) = cx^3$ on the interval $[0, 1/c]$ is given by the integral of the difference between the upper and lower functions. From the graph,in the interval $[0, 1/c]$,$x^2 \ge cx^3$. Thus,the area $A$ is: $A = \int_{0}^{1/c} (x^2 - cx^3) dx$ $= \left[ \frac{x^3}{3} - \frac{cx^4}{4} ight]_{0}^{1/c}$ $= \left( \frac{(1/c)^3}{3} - \frac{c(1/c)^4}{4} ight) - 0$ $= \frac{1}{3c^3} - \frac{1}{4c^3}$ $= \frac{4 - 3}{12c^3} = \frac{1}{12c^3}$ Given that the area $A = 2/3$,we have: $\frac{1}{12c^3} = \frac{2}{3}$ $12c^3 = \frac{3}{2} 	imes 1 = 1.5$ (Wait,let's re-calculate: $\frac{1}{12c^3} = \frac{2}{3} \implies 12c^3 = \frac{3}{2} = 1.5$ is incorrect. Let's solve $\frac{1}{12c^3} = \frac{2}{3}$ correctly: $12c^3 	imes 2 = 3 \implies 24c^3 = 3 \implies c^3 = 3/24 = 1/8$. Therefore,$c = \sqrt[3]{1/8} = 1/2$.

The graphs of $f(x) = x^2$ and $g(x) = cx^3$ (where $c > 0$) intersect at the points $(0, 0)$ and $\left( \frac{1}{c}, \frac{1}{c^2} \right)$. If the area of the region lying between these graphs over the interval $[0, 1/c]$ is equal to $2/3$,then the value of $c$ is:

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