91Ó°ÊÓ

The substitution method is a powerful tool used to simplify the evaluation of integrals. It's particularly useful when dealing with complex functions that are difficult to integrate.

In this method, we introduce a new variable to the integrand, which can make the integral easier to solve. This new variable, known as the substitution variable, relates to the original variable through a differentiable function.

For example, in the provided exercise, we set \( u = 5 + x \). This choice is motivated by the expression \( \sqrt{5 + x} \) in the denominator, which becomes \( \sqrt{u} \) under the substitution:

• Recognize parts of the integrand that can be replaced with a new variable.
• Substitute this variable into the integral to simplify the calculations.
• Update the limits if you are evaluating a definite integral.

An effective substitution requires transforming the integral into a more familiar form that's easier to integrate. This often involves rewriting the limits of integration and expressing the differential \( dx \) in terms of \( du \).

Definite integrals represent the area under a curve between two specific points on the x-axis. In our exercise, we evaluate the definite integral of the function \( \frac{x}{\sqrt{5+x}} \) from \( x = -1 \) to \( x = 4 \).

Defining limits of integration helps us know exactly which part of the curve we're evaluating.

When we perform the substitution, the limits change from \(-1\) to \(4\) in the \(x\)-variable, to \(4\) to \(9\) in the \(u\)-variable, because:

• At \( x = -1 \), substituting gives \( u = 4 \).
• At \( x = 4 \), substituting gives \( u = 9 \).

With these new limits in place, we can solve the integral in terms of \(u\). The final step of performing a definite integral is substituting the evaluated limits back into the antiderivative to calculate the exact area.

Antiderivatives, also known as indefinite integrals, are functions that represent the original function from which a derivative was taken. They are the cornerstone of integration, allowing us to "undo" differentiation.

In our exercise, after substituting \( u = 5 + x \), we needed the antiderivatives to simplify and solve the integrals:

• The first integral, \( \int u^{1/2} \, du \), has the antiderivative \( \frac{2}{3} u^{3/2} \).
• The second integral, \( \int u^{-1/2} \, du \), has the antiderivative \( 2 u^{1/2} \), but multiplied by \(-5\).

Finding these antiderivatives is crucial. The Fundamental Theorem of Calculus allows us to then evaluate these antiderivatives over the given limits to calculate the exact value of the definite integral. This process of using the antiderivative to evaluate definite integrals highlights the interconnected roles of differentiation and integration.