The area of the region bounded by the curves | vedclass

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(B) The area $A$ is given by the integral of the absolute difference between the curves: $A = \int_{0}^{2} |x^3 - x^2| \, dx$. First, find the interse

Vedclass · Accepted Answer

$\frac{3}{2}$

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(B) The area $A$ is given by the integral of the absolute difference between the curves: $A = \int_{0}^{2} |x^3 - x^2| \, dx$. First, find the intersection points: $x^3 = x^2 \implies x^2(x-1) = 0$, so $x=0$ and $x=1$. In the interval $[0, 1]$, $x^2 \ge x^3$, so $|x^3 - x^2| = x^2 - x^3$. In the interval $[1, 2]$, $x^3 \ge x^2$, so $|x^3 - x^2| = x^3 - x^2$. Thus, $A = \int_{0}^{1} (x^2 - x^3) \, dx + \int_{1}^{2} (x^3 - x^2) \, dx$. Evaluating the first integral: $[\frac{x^3}{3} - \frac{x^4}{4}]_{0}^{1} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$. Evaluating the second integral: $[\frac{x^4}{4} - \frac{x^3}{3}]_{1}^{2} = (\frac{16}{4} - \frac{8}{3}) - (\frac{1}{4} - \frac{1}{3}) = (4 - \frac{8}{3}) - (-\frac{1}{12}) = \frac{4}{3} + \frac{1}{12} = \frac{16+1}{12} = \frac{17}{12}$. Total area $A = \frac{1}{12} + \frac{17}{12} = \frac{18}{12} = \frac{3}{2}$.

The area of the region bounded by the curves $y=x^3$, $y=x^2$ and the lines $x=0$ and $x=2$ is

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