Answer
$$S = \pi(e^2 + 2e -6) $$
Work Step by Step
1. Write the formula for the area of the surface obtained by rotating a curve about the $x$-axis.
$$S = \int_{\alpha}^{\beta} 2\pi y \sqrt {(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$
2. Calculate: $\frac{dx}{dt}$ and $\frac{dy}{dt}$:
$\frac{dx}{dt} = \frac{d(e^t - t)}{dt} = e^t - 1$
$\frac{dy}{dt} = \frac{d(4e^{t/2})}{dt} = \frac 1 2(4)(e^{t/2}) = 2e^{t/2} $
3. Substitute the given values into the formula:
$S = \int_{0}^{1} 2\pi (e^t - t) \sqrt {(e^t - 1)^2 + (2e^{t/2})^2} \space dt$
$S = \int_{0}^{1} 2\pi (e^t - t) \sqrt {(e^{2t} -2e^t + 1) + (4e^{t})} \space dt$
$S = \int_{0}^{1} 2\pi (e^t - t) \sqrt {e^{2t} + 2e^t + 1} \space dt$
$S = \int_{0}^{1} 2\pi (e^t - t) \sqrt {(e^{t} + 1)}^2 \space dt$
$S = \int_{0}^{1} 2\pi (e^t - t) (e^{t} + 1)\space dt$
$S = \pi \int_{0}^{1} 2(e^{2t} + e^{t} -te^t - t)\space dt$
$S = \pi \int_{0}^{1} 2e^{2t} + 2e^{t} -2(te^t) - 2t\space dt$
$S = \pi[e^{2t} + 2e^t -2(t-1)^*e^t - t^2]_0^1$
$S = \pi[e^{2(1)} + 2e^1 -2(1-1)e^1 - (1)^2] - \pi[e^{2(0)} + 2e^0 -2(0-1)e^0 - (0)^2] $
$S = \pi(e^2 + 2e - 1 - (1 + 2 +2)) = \pi(e^2 + 2e -6) $
* Integral by parts:
$$\int te^t \space dt$$
$u= t$
$\frac {du}{dt} = 1$
$\frac{dv}{dt} = e^t$
$v = e^t$
$$\int te^t \space dt = uv - \int v\frac{du}{dt} dt$$
$$\int te^t \space dt = te^t - \int e^t (1) = te^t - e^t = e^t(t-1)$$
