Use integration by parts to evaluate the following integral. \( \int_{-\frac{1}{3}}^{\ln 3} \frac{5 t}{e^{3 t}} d t=\square \)

To evaluate the integral \( \int_{-\frac{1}{3}}^{\ln 3} \frac{5 t}{e^{3t}} \, dt \) using integration by parts, we start by choosing: - \( u = 5t \) (thus, \( du = 5 dt \)) - \( dv = e^{-3t} dt \) (thus, \( v = -\frac{1}{3} e^{-3t} \)) Now, we can apply the integration by parts formula \( \int u \, dv = uv - \int v \, du \): \[ \int \frac{5t}{e^{3t}} dt = -\frac{5t}{3} e^{-3t} - \int -\frac{1}{3} e^{-3t} (5 dt) \] This simplifies to: \[ -\frac{5t}{3} e^{-3t} + \frac{5}{3} \int e^{-3t} dt \] Calculating the last integral: \[ \int e^{-3t} dt = -\frac{1}{3} e^{-3t} \] Thus, substituting back gives: \[ \int \frac{5t}{e^{3t}} dt = -\frac{5t}{3} e^{-3t} - \frac{5}{9} e^{-3t} + C \] Evaluating the definite integral from \( -\frac{1}{3} \) to \( \ln 3 \): \[ \left[-\frac{5t}{3} e^{-3t} - \frac{5}{9} e^{-3t}\right]_{-\frac{1}{3}}^{\ln 3} \] Calculating the upper limit at \( t = \ln 3 \): \[ -\frac{5 \ln 3}{3} e^{-3 \ln 3} - \frac{5}{9} e^{-3 \ln 3} = -\frac{5 \ln 3}{3} \frac{1}{27} - \frac{5}{9} \frac{1}{27} = -\frac{5 \ln 3}{81} - \frac{15}{81} = -\frac{5 \ln 3 + 15}{81} \] Now calculating the lower limit at \( t = -\frac{1}{3} \): \[ -\frac{5 \left(-\frac{1}{3}\right)}{3} e^{1} - \frac{5}{9} e^{1} = \frac{5}{9} e - \frac{5}{9} e = 0 \] Putting it all together, the value of the integral is: \[ 0 - \left(-\frac{5 \ln 3 + 15}{81}\right) = \frac{5 \ln 3 + 15}{81} \] Thus, \[ \int_{-\frac{1}{3}}^{\ln 3} \frac{5t}{e^{3t}} dt = \frac{5 \ln 3 + 15}{81} \] This gives us the final result: \[ \int_{-\frac{1}{3}}^{\ln 3} \frac{5 t}{e^{3 t}} dt = \frac{5 \ln 3 + 15}{81} \]