91Ó°ÊÓ

U-substitution is a very handy tool used in integration to simplify the function you need to integrate. In essence, it's about finding a new variable, commonly denoted as "u," to replace a more cumbersome expression in the integral.

The goal is to transform your integral into a form that's easier to work with. In our exercise, we started by letting \( u = 5 - x \). Once you define "u," you'll need to differentiate it with respect to \( x \) to express \( dx \) in terms of \( du \). For example, from \( u = 5 - x \), differentiating gives \( du = -dx \) or \( dx = -du \). This helps you rewrite the entire integral using "u."

• First, identify the part of the integral to substitute.
• Set \( u \) as that expression.
• Find \( du \) in terms of \( dx \) or vice versa.
• Transform the integral into terms of \( u \).

This way, you simplify the integration process by dealing with a simpler expression.

Differentiation is intimately tied to u-substitution because it's used to find \( du \) from \( u \). Differentiation describes how a function changes as its input changes. It gives us rates at which function values are changing.

In our integral, after setting \( u = 5 - x \), we differentiate to find \( du \). The differentiation of \( u = 5 - x \) gives us \( -1 \) since the derivative of \( 5 \) is \( 0 \) and the derivative of \( -x \) is \( -1 \). Thus, we have \( du = -dx \), or rearranged, \( dx = -du \).

• Differentiate the expression for \( u \) with respect to \( x \).
• Use the derivative to express \( dx \) in terms of \( du \).

By using differentiation in this way, we can smoothly transition from an integral in \( x \) to one in \( u \).

Integration is the process of finding the original function from its derivative. It's essentially the reverse of differentiation. Once you've performed u-substitution and differentiation, integration becomes easier.

In our example, we've transformed the integral into two parts: \(-\int \frac{6}{u} \, du + \int 1 \, du\). Each part can be integrated separately:

• The integral \( \int \frac{6}{u} \, du \) results in \( 6 \ln |u| \).
• \( \int 1 \, du \) simply equals \( u \).

By integrating each piece, we're accumulating the signed area under each part of the curve described by the functions. This leads to a combined result of \( 6 \ln |u| - u \).

Substitution and separation of terms simplify complex integrals into manageable calculations.

Every indefinite integral includes an arbitrary integration constant, commonly denoted as \( C \). Without any boundary conditions, it's impossible to determine the specific value of \( C \).

In our problem, after integrating and getting \( -6 \ln |u| + u \), you must add \( C \) because indefinite integrals represent families of functions, all differing only by a constant amount. This unknown constant allows for potential initial conditions or specific real-world values to update or specify the solution more precisely.

• Always include \( C \) when writing the solution to an indefinite integral.
• Remember, \( C \) reflects the freedom of having infinite functions that share the same derivative.

In our final mathematical expression \(-6 \ln |5-x| + (5-x) + C\), \( C \) accounts for any potential vertical shifts of the curve.