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Sagot :
The area of the resulting surface of this infinite curve is π units.
In this question,
The infinite curve is y = e^−2x, x ≥ 0.
The curve is rotated about x-axis.
Since x ≥ 0, the limits will be 0 to ∞.
Then the area of the resulting surface is,
[tex]A= 2\pi \lim_{b\to \infty} (\int\limits^\infty_0{e^{-2x} } \, dx )[/tex]
Now substitute,
u = -2x
⇒ du = -2dx
⇒ dx = [tex]-\frac{1}{2} du[/tex]
Then,
[tex]\int\limits{-\frac{1}{2}e^{u} } \, du =-\frac{1}{2}\int\limits{e^{u} } \, du[/tex]
Now substitute u and du, we get
⇒ [tex]-\frac{1}{2} \int\limits {e^{-2x} }(-2) \, dx[/tex]
⇒ [tex]-\frac{-2}{2} \int\limits {e^{-2x} } \, dx[/tex]
⇒ [tex](1) \int\limits {e^{-2x} } \, dx[/tex]
⇒ [tex]\int\limits {e^{-2x} } \, dx[/tex]
Thus the area of the resulting surface is
[tex]A= 2\pi \int\limits^\infty_0{e^{-2x} } \, dx[/tex]
⇒ [tex]A= 2\pi [{e^{-2x}(\frac{1}{-2} ) } \,]\limits^\infty_0[/tex]
⇒ [tex]A= \frac{2\pi}{-2} [{e^{-2(\infty)}-e^{-2(0)} } \,]\\[/tex]
⇒ [tex]A= -\pi [0-1} \,]\\[/tex]
⇒ [tex]A= \pi[/tex]
Hence we can conclude that the area of the resulting surface is π units.
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