The area of the region bounded by the curves | vedclass

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(D) The area $A$ is given by the integral $\int_{3}^{5} |x - 4| \, dx$.Since $|x - 4| = -(x - 4)$ for $x < 4$ and $|x - 4| = (x - 4)$ for $x \ge 4$, w

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$1 	ext{ sq. unit}$

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(D) The area $A$ is given by the integral $\int_{3}^{5} |x - 4| \, dx$.Since $|x - 4| = -(x - 4)$ for $x $A = \int_{3}^{4} -(x - 4) \, dx + \int_{4}^{5} (x - 4) \, dx$.Evaluating the first part: $\int_{3}^{4} (-x + 4) \, dx = [-\frac{x^2}{2} + 4x]_{3}^{4} = (-8 + 16) - (-4.5 + 12) = 8 - 7.5 = 0.5$.Evaluating the second part: $\int_{4}^{5} (x - 4) \, dx = [\frac{x^2}{2} - 4x]_{4}^{5} = (12.5 - 20) - (8 - 16) = -7.5 - (-8) = 0.5$.Total area $A = 0.5 + 0.5 = 1 	ext{ sq. unit}$.

The area of the region bounded by the curves $y = |x - 4|$, $x = 3$, $x = 5$, and the $X$-axis is

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