int_{-3}^3 |2-x| dx = | vedclass

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(C) The integral represents the area under the curve $y = |2-x|$ from $x = -3$ to $x = 3$. We can split the integral at the point where the expression

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) The integral represents the area under the curve $y = |2-x|$ from $x = -3$ to $x = 3$. We can split the integral at the point where the expression inside the modulus is zero,which is $x = 2$. $\int_{-3}^3 |2-x| dx = \int_{-3}^2 (2-x) dx + \int_{2}^3 (x-2) dx$ Evaluating the first part: $\int_{-3}^2 (2-x) dx = [2x - \frac{x^2}{2}]_{-3}^2 = (4 - 2) - (-6 - \frac{9}{2}) = 2 - (-10.5) = 12.5$. Evaluating the second part: $\int_{2}^3 (x-2) dx = [\frac{x^2}{2} - 2x]_{2}^3 = (4.5 - 6) - (2 - 4) = -1.5 - (-2) = 0.5$. Adding both parts: $12.5 + 0.5 = 13$. Alternatively,using the geometric interpretation from the graph,the area consists of two right-angled triangles: Triangle $1$ (base from $-3$ to $2$,height at $x=-3$ is $|2-(-3)|=5$): Area $= \frac{1}{2} 	imes 5 	imes 5 = 12.5$. Triangle $2$ (base from $2$ to $3$,height at $x=3$ is $|2-3|=1$): Area $= \frac{1}{2} 	imes 1 	imes 1 = 0.5$. Total Area $= 12.5 + 0.5 = 13$.

$\int_{-3}^3 |2-x| dx =$

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