Questions: The integral in this exercise converges. Evaluate the integral without using ∫ from -∞ to -4 θ e^θ d θ ∫ from -∞ to -4 θ e^θ d θ = (Type an exact answer.)

Solution Steps

To evaluate the integral \(\int_{-\infty}^{-4} \theta e^{\theta} d \theta\), we can use integration by parts. Integration by parts is based on the formula \(\int u \, dv = uv - \int v \, du\). We choose \(u = \theta\) and \(dv = e^{\theta} d\theta\). Then, we find \(du\) and \(v\), and apply the integration by parts formula to solve the integral.

We need to evaluate the integral

\[ \int_{-\infty}^{-4} \theta e^{\theta} d\theta. \]

Using integration by parts, we let

\[ u = \theta \quad \text{and} \quad dv = e^{\theta} d\theta. \]

Then, we find

\[ du = d\theta \quad \text{and} \quad v = e^{\theta}. \]

Applying the integration by parts formula \(\int u \, dv = uv - \int v \, du\), we have:

\[ \int \theta e^{\theta} d\theta = \theta e^{\theta} - \int e^{\theta} d\theta. \]

The integral simplifies to:

\[ \int e^{\theta} d\theta = e^{\theta}. \]

Thus, we have:

\[ \int \theta e^{\theta} d\theta = \theta e^{\theta} - e^{\theta} + C = e^{\theta}(\theta - 1) + C. \]

Now we evaluate the definite integral from \(-\infty\) to \(-4\):

\[ \left[ e^{\theta}(\theta - 1) \right]_{-\infty}^{-4}. \]

Calculating at the upper limit \(\theta = -4\):

\[ e^{-4}(-4 - 1) = -5e^{-4}. \]

As \(\theta \to -\infty\), \(e^{\theta}(\theta - 1) \to 0\). Therefore, the definite integral evaluates to:

\[ 0 - (-5e^{-4}) = 5e^{-4}. \]

Final Answer