91Ó°ÊÓ

Integration by substitution is a powerful technique, especially useful when simplifying integrals that seem complex at first glance. Here, the substitution helps transform the integral into a more manageable form. In our problem, we are faced with the integral \( \int_{0}^{\infty} 5x e^{-0.02x^2} \, dx \).
To simplify this, we let \( u = 0.02x^2 \). Then, we find its differential: \( du = 0.04x \, dx \). The goal here is to substitute all expressions of \( x \) with \( u \) so that the integral becomes purely in terms of \( u \).
This substitution turns \( x \, dx \) into \( \frac{1}{0.04} du \), and allows the x-integral to convert to an u-integral, namely \( \int_{0}^{\infty} \frac{125}{2} e^{-u} \, du \). The process of substitution thus simplifies our work, turning a complicated problem into an easier integration task.

Limits are essential when dealing with improper integrals. The integral \( \int_{0}^{\infty} 5x e^{-0.02x^2} \, dx \) is improper because of its infinite limit of integration.
This demands converting it into a finite form using limits. We observe the behavior of the integrand as the variable approaches infinity.
After substitution, we're left with \( \int_{0}^{\infty} \frac{125}{2} e^{-u} \, du \), which must be solved using limits.
To evaluate it, we use the fact that the exponential function \( e^{-u} \) approaches zero as \( u \) tends to infinity. We express this integral using a limit, recognizing how the function behaves at the boundaries of our integration interval.

The antiderivative, sometimes called the indefinite integral, is crucial to solving integrals like ours. In the rewritten form \( \int_{0}^{\infty} \frac{125}{2} e^{-u} \, du \), we seek the antiderivative of \( e^{-u} \).
For this integrand, it is straightforward; the antiderivative is \( -e^{-u} \).
Antiderivatives allow us to transition from the problem of integration to evaluating a function on specific bounds.
After finding the antiderivative, we employ it at the limits of integration, resulting in the evaluation: \( \frac{125}{2} \left[ -e^{-u} \right]_{0}^{\infty} \). Here, the antiderivative helps us determine the overall value of the integral by substituting our boundary values.

Infinite limits come into play in our integral \( \int_{0}^{\infty} 5x e^{-0.02x^2} \, dx \) due to its upper bound. This presents a unique scenario since evaluating does not involve finite numbers.
When using infinite limits, we focus on the behavior of the integrand as the variable within approaches infinity. For \( e^{-u} \), this behavior is well-known: \( e^{-u} \) tends toward zero when \( u \) increases.
This property is critical because it tells us that the impact of the integral's upper limit diminishes, assisting in our use of the limit to evaluate the integral: \( \frac{125}{2} \left(0 - (-1)\right) = \frac{125}{2} \).
The use of limits is what ensures the calculated outcome remains finite and accurate even when dealing with a range that spans up to infinity.