Water is drained from a swimming pool at a rate given by \( R(t)=30 e^{-0.1 t} \mathrm{gal} / \mathrm{hr} \). If the drain is left open indefinitely, how much water drains from the pool? Set up the integral needed to compute the total amount of water that will drain from the pool. 0 The total amount of water that will drain from the pool is gallons. (Simplify your answer.)

We begin by setting up the integral that represents the total volume drained. The drainage rate is given by \[ R(t) = 30 e^{-0.1t} \quad \text{(gal/hr)}. \] The total amount of water drained, \(A\), is the integral of \(R(t)\) from \(t = 0\) to \(t = \infty\): \[ A = \int_{0}^{\infty} 30 e^{-0.1t} \, dt. \] To evaluate the integral, we factor out the constant 30: \[ A = 30 \int_{0}^{\infty} e^{-0.1t} \, dt. \] Recall the integral formula for an exponential decay function: \[ \int_{0}^{\infty} e^{-at} \, dt = \frac{1}{a}, \] where \( a > 0 \). Here, \( a = 0.1 \), so: \[ \int_{0}^{\infty} e^{-0.1t} \, dt = \frac{1}{0.1} = 10. \] Thus, the total volume drained is: \[ A = 30 \times 10 = 300 \quad \text{gallons}. \] The final integral and the simplified amount of water drained are \[ \boxed{\int_{0}^{\infty} 30 e^{-0.1t} \, dt = 300 \text{ gallons}}. \]