91Ó°ÊÓ

U-substitution is a powerful technique in calculus used to simplify and solve integrals. The idea is to transform a complex integral into a simpler one by substituting a part of the integral with a new variable, often denoted as \( u \). In our example, the integral \( \int_{0}^{1}(2x-1)^{3} \, dx \) becomes more straightforward when we substitute \( u = 2x - 1 \).
This transforms the integral into expressions in terms of \( u \), which are often easier to compute.

• The substitution expresses \( x \) in terms of \( u \): if \( u = 2x - 1 \), then \( x = \frac{u + 1}{2} \).
• Find \( dx \) in terms of \( du \): differentiate \( u = 2x - 1 \) to get \( du = 2 \, dx \), so \( dx = \frac{du}{2} \).

By replacing \( x \) and \( dx \) with their equivalents in terms of \( u \), we simplify the process, thus breaking down the complicated polynomial expression into an integral involving a simpler expression \( u^3 \). This transformation is fundamental in successfully evaluating the integral.

When applying the u-substitution method to definite integrals, changing the limits of integration is crucial. After substituting \( u = 2x - 1 \), the original limits based on \( x \) are transformed accordingly. This ensures the integral evaluates correctly over the same interval, but in terms of \( u \).Here is how you adjust the limits of integration:

• For the lower limit, where \( x = 0 \), substitute into \( u = 2x - 1 \), resulting in \( u = -1 \).
• For the upper limit, where \( x = 1 \), substitute into \( u = 2x - 1 \), giving \( u = 1 \).

This adjustment transforms the limits from \( x = 0 \) to \( x = 1 \) into new limits \( u = -1 \) to \( u = 1 \). Ensuring the limits are consistent with the variable substitution helps maintain the integrity of the evaluation, allowing the integral to be solved accurately.

An indefinite integral represents a family of functions and includes an arbitrary constant \( C \). While calculating indefinite integrals via u-substitution, limits aren't involved, unlike definite integrals. You focus instead on finding a general form of the solution function.For our example, starting with the integral \( \int (2x-1)^3 \, dx \), we use the substitution \( u = 2x - 1 \). This changes the integral to \( \int u^3 \frac{du}{2} \).

• Integrate with respect to \( u \) to get \( \int u^3 \frac{du}{2} = \frac{u^4}{8} + C \).
• Substitute back \( u = 2x - 1 \) to return to the original variable: \( \frac{(2x-1)^4}{8} + C \).

This expression represents the indefinite integral, capturing all possible antiderivatives of the function \((2x-1)^3 \), with \( C \) accounting for any constant shifts in the integrated function, emphasizing the indefinite nature.