Get expert insights and community-driven knowledge on IDNLearn.com. Our experts are available to provide in-depth and trustworthy answers to any questions you may have.
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.
Learn more about area of infinite curve here
https://brainly.com/question/3139455
#SPJ4
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Your questions deserve reliable answers. Thanks for visiting IDNLearn.com, and see you again soon for more helpful information.
