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Answer:
the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;
[tex]\frac{\pi }{2} [e^2 - 1 ][/tex] or 10.036
Step-by-step explanation:
Given the data in the question;
y = [tex]y = e^{(x - 1 )[/tex], y = 0, x = 1, x = 2.
Now, using the integration capabilities of a graphing utility
y = [tex]y = e_2}^{(x - 1 )_[/tex], y = 0
Volume = [tex]\pi \int\limits^2_1 ( e^{x-1)^2} - (0)^2 dx[/tex]
Volume = [tex]\pi \int\limits^2_1 ( e^{x-1)^2 dx[/tex]
Volume = [tex]\pi \int\limits^2_1 e^{2x-2}dx[/tex]
Volume = [tex]\frac{\pi }{e^2} \int\limits^2_1 e^{2x}dx[/tex]
Volume = [tex]\frac{\pi }{e^2} [\frac{e^{2x}}{2}]^2_1[/tex]
Volume = [tex]\frac{\pi }{2e^2} [e^4 - e^2 ][/tex]
Volume = [tex]\frac{\pi }{2} [e^2 - 1 ][/tex] or 10.036
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;
[tex]\frac{\pi }{2} [e^2 - 1 ][/tex] or 10.036