91Ó°ÊÓ

Integration by substitution is a powerful technique to solve integrals that are challenging to evaluate in their original form. It transforms a complicated integral into a simpler one. This is done by changing variables. The idea is to choose a substitution variable, usually denoted as \(u\), to simplify the integral.

Here's how it works:

• Select a new variable and set it equal to an expression involving the original variable, for instance, \(u = x^2\).
• Compute the differential of the new variable, \(du = 2x \, dx\).
• Substitute both the new variable \(u\) and its differential \(du\) back into the integral and simplify the expression.
• Don't forget to change the limits of integration if you're handling a definite integral (more on that later).

By using substitution, we break down more complex integrals into simpler ones that are easier to integrate. In our problem, we used \(u = x^2\) to transform the integral of \(x e^{-x^2} \, dx\) into a much simpler integral of \(0.5 e^{-u} \, du\).

A definite integral is an integral that has upper and lower limits. It's represented as \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration.

Definite integrals differ from indefinite integrals because they result in a specific numerical value. They measure the area under the curve of the function \( f(x) \) between the points \( x = a \) and \( x = b \).

In the context of substitution, definite integrals require an adjustment to the limits after completing the substitution. For example, when substituting \( u = x^2 \), the original limits \( x = 0 \) and \( x = 1 \) must be converted into \( u \)-limits. Thus,

• For \( x = 0 \), \( u = 0^2 = 0 \).
• For \( x = 1 \), \( u = 1^2 = 1 \).

After substitution, everything is evaluated in terms of \( u \), including the limits of integration. Therefore, our integral \( \int_{0}^{1} x e^{-x^2} \, dx \) becomes \( 0.5 \, \int_{0}^{1} e^{-u} \, du \).

Exponential functions are a common occurrence in calculus, especially in integrals and derivatives. An exponential function has the form \( e^x \), where \( e \) is a constant approximately equal to 2.71828.

Exponential functions have unique properties that make them easier to work with in calculus:

• The derivative of \( e^x \) is \( e^x \) itself.
• Similarly, the integral of \( e^x \) is also \( e^x \) (plus a constant for indefinite integrals).

For our problem, we faced the integral \( \int e^{-u} \, du \). Integrating this function is straightforward due to its form:

We find the integral is \( -e^{-u} \), evaluated between \( u = 0 \) and \( u = 1 \). This simplification is combined with our substitution and definite integral processes to yield our final result: \( 0.5 \left(1 - \frac{1}{e} \right) \).