Problem 92 - Applied Calculus

Question

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int(x-1) \sqrt{x+2} d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{2}{5}(x+2)^{5/2} - 2(x+2)^{3/2} + C \).

Step 1: Choose the substitution

For the integral \( \int(x-1) \sqrt{x+2} \, dx \), we choose the substitution \( u = x + 2 \). This is because \( x + 2 \) is under the square root, which makes the integration more straightforward. Then, differentiate to find \( du \) such that \( du = dx \).

Step 2: Express x in terms of u

Since \( u = x + 2 \), we can express \( x \) in terms of \( u \) by rearranging the equation as \( x = u - 2 \).

Step 3: Rewrite the integral in terms of u

Substitute \( x = u - 2 \) and \( dx = du \) into the integral: \( \int (x-1) \sqrt{x+2} \, dx = \int ((u-2)-1) \sqrt{u} \, du = \int (u-3) \sqrt{u} \, du \).

Step 4: Simplify the integral

Split the integral: \( \int (u-3) \sqrt{u} \, du = \int u^{1/2} u \, du - 3\int u^{1/2} \, du = \int u^{3/2} \, du - 3\int u^{1/2} \, du \).

Step 5: Integrate each term

Use the power rule for integration on each term separately: \( \int u^{3/2} \, du = \frac{2}{5}u^{5/2} \) and \( \int u^{1/2} \, du = \frac{2}{3}u^{3/2} \).

Step 6: Combine integrated terms

Combine the results from integration: \( \frac{2}{5}u^{5/2} - 3\left(\frac{2}{3}u^{3/2}\right) = \frac{2}{5}u^{5/2} - 2u^{3/2} \).

Step 7: Substitute back for x

Replace \( u \) with \( x + 2 \) to get the integral in terms of \( x \): \( \frac{2}{5}(x+2)^{5/2} - 2(x+2)^{3/2} + C \).

Key Concepts

Integration Techniques

Integration techniques help us find the integral of a function, especially when the direct application of basic integration formulas is not feasible. One common approach is known as substitution, an essential technique for handling more complex integrals.

The substitution method utilizes a change of variables to simplify the integral. This technique is particularly useful when dealing with composite functions or when the integrand involves expressions raised to a power or nested within another function. Here's how substitution generally works:

• You identify a part of the integrand to substitute with a new variable, often denoted as \(u\).
• Differentiating this new variable gives you \(du\), which replaces \(dx\) in the integral.
• The substitution transforms the integral into a simpler form, often making it resemble a basic integral that we can manage more easily.

By implementing these steps effectively, substitution streamlines the integration process. The choice of \(u\) is crucial. It's best to pick a component of the function inside an outer operation (like a square root or power) that, when replaced, simplifies the equation significantly.

Indefinite Integrals

Indefinite integrals are a fundamental concept in calculus representing a family of functions. When we integrate a function without upper and lower limits of integration, we speak of an indefinite integral. This form of integration results in a function known as the antiderivative.

An indefinite integral is expressed as \(\int f(x) \, dx = F(x) + C\), where

• \(f(x)\) is the integrand, the function we are integrating,
• \(F(x)\) is any antiderivative of \(f(x)\), and
• \(C\) is the constant of integration, representing an infinite number of vertical shifts of \(F(x)\) on a graph.

The constant of integration is crucial because it reflects the fact that there isn't just one antiderivative but rather a whole family of them. When performing indefinite integration, it’s important to correctly apply integration rules or techniques, like substitution, to find \(F(x)\). After integrating, always remember to add \(C\) to acknowledge that the solution includes all possible shifts of the basic antiderivative function.

Calculus Substitution

Calculus substitution, or even more specifically, the substitution method in integration, transforms challenging integrals into more manageable ones. It's like rephrasing a complicated problem into an easier version that uses familiar techniques. This process is not only efficient but also a powerful way to solve integrals that can't be approached directly.

The key to using substitution lies in identifying part of the integrand as a separate term. This term, often expressed as \(u\), captures a composition within the integrand. By rewriting parts of the integrand and replacing \(dx\) with \(du\), we simplify the problem significantly.

• First, choose \(u\) such that it simplifies the integral, making it clear which substitution achieves this effectively.
• Differentiate \(u\) to find \(du\); this replaces \(dx\) and helps to convert all terms of the integral in terms of \(u\).
• Once you simplify and integrate in \(u\), the final step is substituting back the original variable \(x\) to express the function in its original variable terms.

Proper execution of these steps can convert a daunting integral into a much simpler form, thus unlocking solutions with finesse and clarity.