1. **State the problem:** We want to find the function $y(t)$ defined by the integral $$y(t) = \int_0^t (0.8^\tau)(t - \tau) \, d\tau$$ 2. **Understand the integral:** This is a convolution-like integral where the integrand is the product of $0.8^\tau$ and $(t - \tau)$. 3. **Rewrite the integral:** $$y(t) = \int_0^t (t - \tau) 0.8^\tau \, d\tau = t \int_0^t 0.8^\tau \, d\tau - \int_0^t \tau 0.8^\tau \, d\tau$$ 4. **Calculate the first integral:** $$\int_0^t 0.8^\tau \, d\tau = \int_0^t e^{\tau \ln(0.8)} \, d\tau = \left[ \frac{e^{\tau \ln(0.8)}}{\ln(0.8)} \right]_0^t = \frac{0.8^t - 1}{\ln(0.8)}$$ 5. **Calculate the second integral:** Use integration by parts with $u=\tau$, $dv=0.8^\tau d\tau$. - $du = d\tau$ - $v = \frac{0.8^\tau}{\ln(0.8)}$ Then, $$\int_0^t \tau 0.8^\tau d\tau = \left. \tau \frac{0.8^\tau}{\ln(0.8)} \right|_0^t - \int_0^t \frac{0.8^\tau}{\ln(0.8)} d\tau = \frac{t 0.8^t}{\ln(0.8)} - \frac{1}{\ln(0.8)} \int_0^t 0.8^\tau d\tau$$ From step 4, substitute the integral: $$= \frac{t 0.8^t}{\ln(0.8)} - \frac{1}{\ln(0.8)} \cdot \frac{0.8^t - 1}{\ln(0.8)} = \frac{t 0.8^t}{\ln(0.8)} - \frac{0.8^t - 1}{(\ln(0.8))^2}$$ 6. **Combine results:** $$y(t) = t \cdot \frac{0.8^t - 1}{\ln(0.8)} - \left( \frac{t 0.8^t}{\ln(0.8)} - \frac{0.8^t - 1}{(\ln(0.8))^2} \right) = \frac{t (0.8^t - 1)}{\ln(0.8)} - \frac{t 0.8^t}{\ln(0.8)} + \frac{0.8^t - 1}{(\ln(0.8))^2}$$ Simplify the $t$ terms: $$\frac{t (0.8^t - 1)}{\ln(0.8)} - \frac{t 0.8^t}{\ln(0.8)} = \frac{t 0.8^t - t - t 0.8^t}{\ln(0.8)} = - \frac{t}{\ln(0.8)}$$ 7. **Final simplified expression:** $$y(t) = - \frac{t}{\ln(0.8)} + \frac{0.8^t - 1}{(\ln(0.8))^2}$$ **Answer:** $$\boxed{y(t) = - \frac{t}{\ln(0.8)} + \frac{0.8^t - 1}{(\ln(0.8))^2}}$$ This expression gives the value of $y(t)$ for any $t \geq 0$.