To determine whether the integral of $xe^{-2x}$ from $-\infty$ to 0 converges, we need to evaluate the following improper integral:

$I = \int_{-\infty}^{0} xe^{-2x} \, dx$

Step 1: Check the behavior as $x \to -\infty$

To analyze whether the integral converges, we should first look at the behavior of the integrand $xe^{-2x}$ as $x \to -\infty$. The exponential term $e^{-2x}$ grows rapidly as $x \to -\infty$, and since $x$ is negative, this will cause the product $xe^{-2x}$ to approach 0. This suggests that the integral might converge, but we need to confirm this by evaluating it.

Step 2: Find the indefinite integral of $xe^{-2x}$

We can solve this integral using integration by parts. Let:

• $u = x$, so $du = dx$,
• $dv = e^{-2x} dx$, so $v = \frac{-1}{2} e^{-2x}$ (since the derivative of $e^{-2x}$ is $-2e^{-2x}$).

Using the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get:

$\int xe^{-2x} \, dx = x \cdot \frac{-1}{2} e^{-2x} - \int \frac{-1}{2} e^{-2x} \, dx$

Simplifying:

$= -\frac{x}{2} e^{-2x} + \frac{1}{4} e^{-2x}$

Thus, the indefinite integral is:

$\int xe^{-2x} \, dx = -\frac{x}{2} e^{-2x} + \frac{1}{4} e^{-2x} + C$

Step 3: Evaluate the definite integral

Now we evaluate the improper integral:

$I = \int_{-\infty}^{0} xe^{-2x} \, dx = \left[ -\frac{x}{2} e^{-2x} + \frac{1}{4} e^{-2x} \right]_{-\infty}^{0}$

At $x = 0$:

$-\frac{0}{2} e^{-2(0)} + \frac{1}{4} e^{-2(0)} = \frac{1}{4}$

As $x \to -\infty$, both terms $-\frac{x}{2} e^{-2x}$ and $\frac{1}{4} e^{-2x}$ approach 0. This is because the exponential term $e^{-2x}$ grows exponentially as $x \to -\infty$, while the factors involving $x$ do not grow fast enough to counteract the rapid decay of the exponential.

Thus, the integral converges and the value is:

$I = \frac{1}{4} - 0 = \frac{1}{4}$

Conclusion

The integral of $xe^{-2x}$ from $-\infty$ to 0 converges, and its value is $\frac{1}{4}$.

Would you like more details on any part of the solution? Here are some related questions you might find interesting:

1. How would the result change if we integrated from 0 to $\infty$ instead of $-\infty$ to 0?
2. What would happen if the exponent in $e^{-2x}$ were different (e.g., $e^{-x}$)?
3. How can we apply integration by parts to solve similar integrals?
4. What are some other techniques to handle improper integrals like this one?
5. How do the limits of integration affect the convergence of the integral?

Tip: When dealing with improper integrals, always check the behavior of the integrand at the boundaries (in this case, $x \to -\infty$) to determine whether the integral will converge.