Question

Find the exact area of the surface obtained by rotating the curve about the x-axis.

y=sqrt(1+e^x) for x gt-eq 0 and lt-eq 1

Answer 1

\[y = \sqrt{1 + e^x}, 0 \le x \le 1\]

Answer 2

I know the formula is\[S = \int\limits_{a}^{b} 2\pi f(x) \sqrt{1 + f'(x)^2} dx\]

Answer 3

and \[y' = \frac{ e^x }{ 2\sqrt{1 + e^x} }\]

Answer 4

But this gives an enormous integral that I can't get started solving.\[S = 2\pi \int\limits_{0}^{1} \sqrt{1+ e^x} \sqrt{1 + (\frac{ e^x }{ 2\sqrt{1 + e^x} })^2}\]

Answer 5

you can do it just consentrate

Answer 6

ok I was able to simplify it to this: \[2\pi \int\limits_{0}^{1} \frac{ e^x + 2 }{ 2 } dx\]