To solve the equation for $\lambda$, we need to evaluate the given integral:

$\int_0^\infty \lambda x e^{-\lambda x} \, dx = 1$

Step 1: Solve the Integral

Let's first solve the integral. The integral is:

$\int_0^\infty \lambda x e^{-\lambda x} \, dx$

To evaluate this, we can use the substitution $u = \lambda x$, so $du = \lambda dx$, and the limits change accordingly:

• When $x = 0$, $u = 0$.
• When $x \to \infty$, $u \to \infty$.

Substituting into the integral:

$\int_0^\infty \lambda x e^{-\lambda x} \, dx = \int_0^\infty u e^{-u} \, du$

The integral $\int_0^\infty u e^{-u} \, du$ is a known result and can be solved using integration by parts or recognized as the Gamma function $\Gamma(n)$ for $n = 2$, which equals $(n-1)!$:

$\int_0^\infty u e^{-u} \, du = 1!$

So,

$\int_0^\infty u e^{-u} \, du = 1$

Step 2: Apply to the Original Equation

Substituting back:

$\lambda \cdot 1 = 1$

Step 3: Solve for $\lambda$

From the equation:

$\lambda = 1$

Thus, the value of $\lambda$ is 1.

Would you like any further details or have any additional questions?

Here are some related questions you might be interested in:

1. How does the Gamma function relate to other integrals?
2. What is the physical interpretation of this type of integral in probability theory?
3. Can you explain the integration by parts method?
4. How does changing the limits of integration affect the result?
5. What happens if we change the exponent in the integrand?

Tip: In many problems involving integrals with an exponential function, recognizing the relationship to the Gamma function can greatly simplify the solution process.