Find the volume of the solid generated by revolving the region \( R \) bounded by \( y=e^{-2 x}, y=0, x=0 \) and \( x=\ln 5 \) about the \( x \)-axis.

To find the volume of the solid generated by revolving the region \( R \) bounded by \( y = e^{-2x} \), \( y = 0 \), \( x = 0 \), and \( x = \ln 5 \) about the \( x \)-axis, we can use the disk method. The formula for the volume \( V \) when a region is revolved around the \( x \)-axis is given by: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] In this case, \( f(x) = e^{-2x} \), and our bounds are \( a = 0 \) and \( b = \ln 5 \). Therefore, we have: \[ V = \pi \int_{0}^{\ln 5} (e^{-2x})^2 \, dx \] This simplifies to: \[ V = \pi \int_{0}^{\ln 5} e^{-4x} \, dx \] Now, we can evaluate the integral. The integral of \( e^{-4x} \) is: \[ \int e^{-4x} \, dx = -\frac{1}{4} e^{-4x} + C \] Next, we evaluate it from \( 0 \) to \( \ln 5 \): \[ V = \pi \left[ -\frac{1}{4} e^{-4x} \right]_{0}^{\ln 5} \] Calculating this gives: \[ V = \pi \left( -\frac{1}{4} e^{-4 \ln 5} - \left( -\frac{1}{4} e^0 \right) \right) \] Remembering that \( e^{\ln a} = a \), we can compute \( e^{-4 \ln 5} = (e^{\ln 5})^{-4} = 5^{-4} = \frac{1}{625} \): \[ V = \pi \left( -\frac{1}{4} \cdot \frac{1}{625} + \frac{1}{4} \right) \] Now this simplifies to: \[ V = \pi \left( \frac{1}{4} - \frac{1}{2500} \right) \] We need a common denominator to combine the fractions. The common denominator of \( 4 \) and \( 2500 \) is \( 2500 \): \[ \frac{1}{4} = \frac{625}{2500} \] Thus: \[ V = \pi \left( \frac{625}{2500} - \frac{1}{2500} \right) = \pi \left( \frac{624}{2500} \right) \] Therefore, the volume \( V \) of the solid generated by revolving the region \( R \) about the \( x \)-axis is: \[ \boxed{\frac{624\pi}{2500}} \]