The area of the region bounded by y=x^3, the | vedclass

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(A) The area $A$ of the region bounded by the curve $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$ is given by $A = \int_{a}^{b} |f(x)| \, dx$.

Vedclass · Accepted Answer

$68$

Vedclass · Answer

(A) The area $A$ of the region bounded by the curve $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$ is given by $A = \int_{a}^{b} |f(x)| \, dx$. Here, $f(x) = x^3$, $a = -2$, and $b = 4$. The function $x^3$ is negative for $x  0$. Thus, the integral is split at $x = 0$: $A = \int_{-2}^{0} |x^3| \, dx + \int_{0}^{4} |x^3| \, dx = \int_{-2}^{0} (-x^3) \, dx + \int_{0}^{4} x^3 \, dx$. Evaluating the first integral: $\int_{-2}^{0} (-x^3) \, dx = [-\frac{x^4}{4}]_{-2}^{0} = 0 - (-\frac{(-2)^4}{4}) = \frac{16}{4} = 4$. Evaluating the second integral: $\int_{0}^{4} x^3 \, dx = [\frac{x^4}{4}]_{0}^{4} = \frac{4^4}{4} - 0 = 4^3 = 64$. Total area $A = 4 + 64 = 68$ square units.

The area of the region bounded by $y=x^3$, the $x$-axis, $x=-2$, and $x=4$ is

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