91Ó°ÊÓ

Integration problems can sometimes be tricky to solve directly. This is where the substitution method, also known as u-substitution, comes into play. It simplifies the integration process by changing variables to make the integral easier to evaluate.
The substitution method involves choosing a part of the integrand to replace with a new variable, typically denoted as \( u \).

• First, identify a function within the integrand whose derivative is also present in the integrand.
• Replace this function with a new variable \( u \).
• Calculate \( du \) by differentiating \( u \) with respect to \( x \), and then rearrange to express \( dx \) in terms of \( du \).

By substituting these expressions back into the integral, the problem often becomes much simpler.
In our example, we set \( u = 9x - 2 \), which simplifies the integral significantly.

The power rule for integration is a straightforward method to find antiderivatives of power functions. It works wonders when you need to integrate terms like \( u^n \), where \( n \) is any real number except \(-1\).
The power rule formula is:

• \( \int u^{\alpha} du = \frac{u^{\alpha+1}}{\alpha+1} + C \)

In our case, after substituting, the integral became \( \int u^{-3} du \). By applying the power rule, you increase the exponent by one, divide by the new exponent, and add an integration constant \( C \).
So, \( \int u^{-3} du \) becomes \( \frac{u^{-2}}{-2} + C \). After multiplying by the factor from our substitution, this results in the partial antiderivative \( - \frac{1}{18} u^{-2} + C \). Use the power rule whenever you have a single power of \( u \) to integrate.

An antiderivative, also known as the primitive or indefinite integral, is the reverse of a derivative. It represents a function whose derivative gives the original function within the integral. When solving integrals, the goal is often to find the antiderivative.
In our given problem, after using substitution and the power rule, we find the antiderivative in terms of \( u \), which was then transformed back into terms of \( x \).

The antiderivative we found was \( - \frac{1}{18} (9x - 2)^{-2} + C \).
The constant \( C \) represents an arbitrary constant because integration can result in multiple equivalent solutions differing by a constant.
This concept of the antiderivative is fundamental in integral calculus, turning boundaries into values and aiding in solving practical problems involving areas, volumes, and accumulated quantities.