Answer to Question #342532 in Calculus for kate lim

Question #342532

Find the centroid for the region bounded by y=3−e-X, the x−axis, x=2, and the y−axis.

 


1
Expert's answer
2022-05-19T12:03:20-0400

ANSWER The centroid of the region is "(x_{C},y_{C})" : "x_{C}=\\frac{5+3e^{-2}}{5+e^{-2}}\\cong1.053, y_{C}=\\frac{\\frac{25}{4}+3e^{-2}-\\frac{e^{-4}}{4}}{5+e^{-2}}\\cong1.295"


EXPLANATION

The centroid of the plane region "D=\\left\\{(x,y): 0\\leq x\\leq 2, 0\\leq y\\leq 3-e^{-x}\\right\\}" is "(x_{C},y_{C})" , where "x_{C}=\\frac{M_{x=0}}{A},\\, y_{C}=\\frac{M_{y=0}}{A}" and "A=\\int_{0}^{2}(3-e^{-x} ) dx, \\, M_{x=0}=\\int_{0}^{2}x\\cdot(3-e^{-x} ) dx,\\, M_{y=0}=\\frac{1}{2}\\int_{0}^{2} (3-e^{-x} )^{2} dy."

We calculate these integrals.

1) "A=\\int_{0}^{2}(3-e^{-x} ) dx = \\left [ 3x+e^{-x} \\right ]_{0}^{2}=6+e^{-2} -1=5+e^{ -2}\\cong5.1353" .

2) Let "x=u, \\, (3-e^{-x} )dx=dv" , then "dx=du," "v=3x+e^{-x}" and

"M_{x=0}=\\int_{0}^{2}x\\cdot \\left ( 3-e^{-x} \\right )dx=\\left [ x\\cdot \\left ( 3x+e^{-x} \\right ) \\right ]_{0}^{2}-\\int_{0}^{2}\\left ( 3x+e^{-x} \\right )dx=\\\\=\\left ( 12+2e^{-2} \\right )-\\left [ \\frac{3x^{2}}{2}-e^{-x} \\right ]_{0}^{2}= (12+2e^{-2})-[6-e^{-2}+1]=5+3e^{-2}\\cong5.4060"

3) "M_{y=0}=\\frac{1}{2}\\int_{0}^{2}\\left ( 3-e^{-x} \\right )^{2} dx=\\frac{1}{2}\\int_{0}^{2}\\left ( 9-6e^{-x} +e^{-2x}\\right ) dx=\\frac{1}{2}\\left [ 9x+6e^{-x} -\\frac{e^{-2x}}{2}\\right ]_{0}^{2}=\\frac{1}{2}\\left ( 18+6e^{-2 } -\\frac{e^{-4}}{2}-6+\\frac{1}{2}\\right )=\\frac{25}{4}+3e^{-2}-\\frac{e^{-4}}{4}\\cong6.6514"


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