The area of the region (in square units) boun | vedclass

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(B) The curves are $y=x^3$ and $y=x$. To find the area, we observe that the region is symmetric about the origin because both functions are odd. The a

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$1/2$

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(B) The curves are $y=x^3$ and $y=x$. To find the area, we observe that the region is symmetric about the origin because both functions are odd. The area $A$ is given by $A = \int_{-1}^{1} |x - x^3| dx$. Due to symmetry, $A = 2 \int_{0}^{1} |x - x^3| dx$. In the interval $[0, 1]$, $x \geq x^3$, so $|x - x^3| = x - x^3$. Thus, $A = 2 \int_{0}^{1} (x - x^3) dx$. Evaluating the integral: $A = 2 \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1}$. $A = 2 \left( \frac{1}{2} - \frac{1}{4} \right) = 2 \left( \frac{1}{4} \right) = \frac{1}{2}$ square units.

The area of the region (in square units) bounded by the curves $y=x^3$, $y=x$ and $-1 \leq x \leq 1$ is:

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