Question

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Answer 1

The volume of the solid that results when the region enclosed by the given curves y = e⁻²ˣ, y = 0, x = 0, and x = 1 is revolved around the x-axis and the y-axis is 0.77 and 0.7135.

What is the revolution of the curve?

Revolving the region bordered by y = f(x) and the x-axis on the interval [a, b] around the x-axis generates the volume (V) of a solid.

The volume is given as

$\rm Volume = \int _a^b \pi y^2 \ dx$

The volume of the solid that results when the region enclosed by the given curves y = e⁻²ˣ, y = 0, x = 0, and x = 1 is revolved around the x-axis.

$\rm Volume = \int _0^1 \pi (e^{-2x})^2 \ dx\\\\\\Volume = \int _0^1 \pi (e^{-4x}) dx\\\\\\Volume = \pi [\dfrac{e^{-4x}}{-4}]_0^1\\\\\\Volume = - \dfrac{\pi}{4} [e^{-4} - e^0]\\\\\\Volume = -\dfrac{\pi}{4} [-0.98168]\\\\\\Volume = 0.77$

The volume of the solid that results when the region enclosed by the given curves y = e⁻²ˣ, y = 0, x = 0, and x = 1 is revolved around the y-axis.

$\rm Volume = \pi \left [\int_{0}^{0.135} dy + \int_{0.135}^{1}\left ( \dfrac{\ln x}{-2} \right )^{2}dy \right ]\\\\Volume =0.71375$

More about the revolution of the curve link is given below.

brainly.com/question/14640419

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