Step 1 :First, we recognize that the given integral is an improper integral, so we need to find the limit of the integral as the lower bound approaches negative infinity.
Step 2 :Let's rewrite the integral as a limit: \(\int_{-\infty}^0 z e^{2z} dz = \lim_{a \to -\infty} \int_{a}^0 z e^{2z} dz\)
Step 3 :Now, we can use integration by parts to evaluate the integral. Let \(u = z\) and \(dv = e^{2z} dz\). Then, \(du = dz\) and \(v = \frac{1}{2} e^{2z}\).
Step 4 :Applying integration by parts, we get: \(\int_{a}^0 z e^{2z} dz = \left[ \frac{1}{2} z e^{2z} \right]_a^0 - \int_{a}^0 \frac{1}{2} e^{2z} dz\)
Step 5 :Now, we can evaluate the remaining integral: \(\int_{a}^0 \frac{1}{2} e^{2z} dz = \left[ -\frac{1}{4} e^{2z} \right]_a^0\)
Step 6 :Substituting the bounds, we get: \(\left[ \frac{1}{2} z e^{2z} \right]_a^0 - \left[ -\frac{1}{4} e^{2z} \right]_a^0 = \frac{1}{2}(0) e^{2(0)} - \frac{1}{2}(a) e^{2(a)} - \left(-\frac{1}{4} e^{2(0)} + \frac{1}{4} e^{2(a)}\right)\)
Step 7 :Simplifying, we get: \(\frac{1}{2} - \frac{1}{2} a e^{2a} + \frac{1}{4} - \frac{1}{4} e^{2a}\)
Step 8 :Now, we need to find the limit as \(a \to -\infty\): \(\lim_{a \to -\infty} \left(\frac{1}{2} - \frac{1}{2} a e^{2a} + \frac{1}{4} - \frac{1}{4} e^{2a}\right)\)
Step 9 :As \(a \to -\infty\), \(e^{2a} \to 0\), so the limit becomes: \(\lim_{a \to -\infty} \left(\frac{1}{2} + \frac{1}{4}\right) = \frac{3}{4}\)
Step 10 :Therefore, the value of the integral is \(\boxed{\frac{3}{4}}\)