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(C) The given curve is $y = 3 - \left|x - \frac{1}{2}\right| - |x + 1|$.We define the function in different intervals:$y = \begin{cases} 3 + (x + 1) +

Vedclass · Accepted Answer

$\frac{27}{8}$

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(C) The given curve is $y = 3 - \left|x - \frac{1}{2}ight| - |x + 1|$.We define the function in different intervals:$y = \begin{cases} 3 + (x + 1) + (x - \frac{1}{2}) = 2x + \frac{7}{2}, & x Correct piecewise definition:For $x For $-1 \leq x For $x \geq 1/2$: $y = 3 - (x - 1/2) - (x + 1) = 3 - x + 1/2 - x - 1 = 5/2 - 2x$. Setting $y=0$,$x = 5/4$.The region is a trapezoid with parallel sides of length $3/2$ (height) and base on the $x$-axis from $x = -7/4$ to $x = 5/4$.The length of the base is $5/4 - (-7/4) = 12/4 = 3$.The height of the trapezoid is $3/2$.Area $= \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$.Here,the parallel sides are the base on the $x$-axis (length $3$) and the top horizontal segment from $x = -1$ to $x = 1/2$ (length $1/2 - (-1) = 3/2$).Area $= \frac{1}{2} 	imes (3 + 3/2) 	imes (3/2) = \frac{1}{2} 	imes (9/2) 	imes (3/2) = \frac{27}{8}$ sq. units.

The area of the bounded region enclosed by the curve $y=3-\left|x-\frac{1}{2}\right|-|x+1|$ and the $x-$axis is

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