We are asked to evaluate the improper integral:

$\int_{4}^{\infty} x e^{-4x} \, dx$

Step 1: Recognize the integral form

The integral involves the function $x e^{-4x}$, which suggests integration by parts.

Step 2: Apply Integration by Parts

Using the integration by parts formula:

$\int u \, dv = uv - \int v \, du$

Let:

• $u = x$, so $du = dx$
• $dv = e^{-4x} \, dx$, so $v = \int e^{-4x} \, dx = \frac{-1}{4} e^{-4x}$

Now apply the formula:

$\int x e^{-4x} \, dx = x \left( \frac{-1}{4} e^{-4x} \right) - \int \frac{-1}{4} e^{-4x} \, dx$

Step 3: Simplify the terms

This simplifies to:

$= \frac{-x}{4} e^{-4x} + \frac{1}{16} e^{-4x}$

Step 4: Evaluate the definite integral

Now, evaluate the expression from $4$ to $\infty$:

$\int_{4}^{\infty} x e^{-4x} \, dx = \left[ \frac{-x}{4} e^{-4x} + \frac{1}{16} e^{-4x} \right]_{4}^{\infty}$

First, evaluate the limit as $x \to \infty$. Since $e^{-4x}$ decays to zero very quickly as $x$ increases:

$\lim_{x \to \infty} \left( \frac{-x}{4} e^{-4x} + \frac{1}{16} e^{-4x} \right) = 0$

Now, substitute $x = 4$:

$\left( \frac{-4}{4} e^{-16} + \frac{1}{16} e^{-16} \right) = \left( -e^{-16} + \frac{1}{16} e^{-16} \right) = -\frac{15}{16} e^{-16}$

Thus, the value of the integral is:

$\int_{4}^{\infty} x e^{-4x} \, dx = \frac{15}{16} e^{-16}$

Final Answer:

$\frac{15}{16} e^{-16}$

Would you like more details or have any questions?

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Here are 5 related questions to explore further:

1. How do you solve an improper integral when the upper limit is infinity?
2. Can integration by parts be used to solve all exponential functions?
3. What happens if the exponent in the exponential function changes?
4. How do you handle integrals that involve products of polynomials and exponentials?
5. How does the decay rate of $e^{-4x}$ affect the convergence of this integral?

Tip: When evaluating limits at infinity in integrals, remember that exponential decay (like $e^{-4x}$) will often force terms to zero very quickly.