The area of the region bounded by the line y | vedclass

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(C) The area $A$ is given by the integral of $|y|$ with respect to $x$ from $x = 2$ to $x = 5$. $A = \int_{2}^{5} |3 - x| \, dx$. Since the line $y =

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(C) The area $A$ is given by the integral of $|y|$ with respect to $x$ from $x = 2$ to $x = 5$. $A = \int_{2}^{5} |3 - x| \, dx$. Since the line $y = 3 - x$ crosses the $X$-axis at $x = 3$,we split the integral: $A = \int_{2}^{3} (3 - x) \, dx + \int_{3}^{5} -(3 - x) \, dx$. $A = \left[ 3x - \frac{x^2}{2} \right]_{2}^{3} + \left[ \frac{x^2}{2} - 3x \right]_{3}^{5}$. $A = \left( (9 - 4.5) - (6 - 2) \right) + \left( (12.5 - 15) - (4.5 - 9) \right)$. $A = (4.5 - 4) + (-2.5 + 4.5) = 0.5 + 2 = 2.5 = \frac{5}{2}$ square units. Thus,the correct option is $C$.

The area of the region bounded by the line $y = 3 - x$,the $X$-axis,and the lines $x = 2$ and $x = 5$ is . . . . . . .

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