Let \( R \) be the region bounded by the graph of \( y=e^{-a x} \) and \( y=e^{-b x} \) for \( x \geq 0 \), where \( a>b>0 \). Find the area of \( R \). Write the integral. \( \int_{0}(\square) d x \) The area is

The region \( R \) is bounded by the curves \( y = e^{-ax} \) and \( y = e^{-bx} \) for \( x \geq 0 \), where \( a > b > 0 \). Since \( a > b \), for \( x > 0 \) we have \[ e^{-ax} < e^{-bx}. \] Thus, the upper curve is \( y = e^{-bx} \) and the lower curve is \( y = e^{-ax} \). The area of the region \( R \) is given by the integral \[ \int_{0}^{\infty} \left( e^{-bx} - e^{-ax} \right) dx. \] Evaluating the integral, we first integrate term by term: 1. The integral of \( e^{-bx} \) is \[ \int e^{-bx}\, dx = -\frac{1}{b} e^{-bx}. \] 2. The integral of \( e^{-ax} \) is \[ \int e^{-ax}\, dx = -\frac{1}{a} e^{-ax}. \] Thus, the definite integral from \( 0 \) to \( \infty \) is \[ \int_{0}^{\infty} \left( e^{-bx} - e^{-ax} \right) dx = \left[-\frac{1}{b} e^{-bx} + \frac{1}{a} e^{-ax}\right]_{0}^{\infty}. \] Evaluating at the upper limit \( x \to \infty \): \[ \lim_{x \to \infty} \left(-\frac{1}{b} e^{-bx} + \frac{1}{a} e^{-ax}\right) = 0, \] since both \( e^{-bx} \) and \( e^{-ax} \) tend to 0 as \( x \to \infty \). At \( x = 0 \): \[ -\frac{1}{b} e^{-b \cdot 0} + \frac{1}{a} e^{-a \cdot 0} = -\frac{1}{b} + \frac{1}{a}. \] Subtracting the lower limit from the upper limit: \[ \text{Area} = 0 - \left(-\frac{1}{b} + \frac{1}{a}\right) = \frac{1}{b} - \frac{1}{a}. \] Thus, the integral is \[ \int_{0}^{\infty} \left(e^{-bx}-e^{-ax}\right) dx, \] and the area of \( R \) is \[ \frac{1}{b}-\frac{1}{a}. \]