Join IDNLearn.com and become part of a knowledge-sharing community that thrives on curiosity. Join our interactive Q&A community and get reliable, detailed answers from experienced professionals across a variety of topics.
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
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Thank you for choosing IDNLearn.com for your queries. We’re committed to providing accurate answers, so visit us again soon.
