The area (in square units) bounded by the lin | vedclass

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(D) The area $A$ is given by the integral $\int_{-2}^{4} |y| \, dx = \int_{-2}^{4} |x| \, dx$.Since $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$

Vedclass · Accepted Answer

$10$

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(D) The area $A$ is given by the integral $\int_{-2}^{4} |y| \, dx = \int_{-2}^{4} |x| \, dx$.Since $|x| = -x$ for $x $A = \int_{-2}^{0} (-x) \, dx + \int_{0}^{4} x \, dx$.Evaluating the first part: $\int_{-2}^{0} (-x) \, dx = [-\frac{x^2}{2}]_{-2}^{0} = 0 - (-\frac{(-2)^2}{2}) = 0 - (-2) = 2$.Evaluating the second part: $\int_{0}^{4} x \, dx = [\frac{x^2}{2}]_{0}^{4} = \frac{16}{2} - 0 = 8$.Total area $A = 2 + 8 = 10 	ext{ square units}$.

The area (in square units) bounded by the line $y = x$, the $X$-axis and the lines $x = -2$ and $x = 4$ is...

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