Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-16, perform the indicated integrations. 1\. \(\int x \sqrt{x+1} d x\)

Step-by-Step Solution

Answer

\( \int x \sqrt{x+1} \, dx = \frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + C \).

Step 1: Choose a Suitable Substitution

Let's use the substitution method and let \( u = x+1 \). Then \( du = dx \), and note that \( x = u-1 \). This substitution will simplify the integral.

Step 2: Replace and Simplify the integral

Replacing the variables in the integral, we have: \( \int x \sqrt{x+1} \, dx = \int (u-1) \sqrt{u} \, du \). Simplify to get: \( \int (u^{3/2} - u^{1/2}) \, du \).

Step 3: Integrate Using Power Rule

Now, integrate each term separately. For \( \int u^{3/2} \, du \), use the power rule \( \frac{u^{5/2}}{5/2} = \frac{2}{5}u^{5/2}\). For \( \int u^{1/2} \, du \), apply the power rule \( \frac{u^{3/2}}{3/2} = \frac{2}{3}u^{3/2}\).

Step 4: Write the Result in Original Variable

The integrated result in terms of \( u \) is \( \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + C \). Substitute back \( u = x+1 \) to express it in terms of \( x \).

Step 5: Final Answer

Substitute \( u = x + 1 \) into the result: The final integral is \( \frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + C \).

Key Concepts

Substitution Method

The substitution method is a handy technique to simplify complex integrals by changing variables. Think of it like swapping a complicated part of the integral with a simpler letter, usually chosen to alleviate complex expressions or exponents. In our exercise, we begin by identifying a suitable substitution. We pick \( u = x+1 \) because it appears under the square root, making it a logical choice for simplifying this specific integral.

Here's why substitution helps:

• Reduces confusing expressions by introducing a new variable \( u \).
• Makes the integral look easier by focusing on terms that are difficult to integrate directly.

After choosing \( u = x+1 \), replacing \( x \, \text{with} \, u-1 \) transforms the integral into forms that are more manageable, separating the initial complexity into straightforward components.

Always remember to account for differential changes too. Here, \( dx = du \), ensuring our integral remains consistent when transitioning between variables.

Power Rule in Integration

The power rule is a go-to tool for integrating polynomials and terms with powers. It allows us to find the integral of terms like \( u^n \) effortlessly. Once our integral was transformed from \( x \) to \( u \), using the substitution method, we apply the power rule to each term.

How it works:

• If you have \( \int u^n \, du \), the power rule says its integral is \( \frac{u^{n+1}}{n+1} \), provided \( n eq -1 \).
• This is because the antiderivative of \( u^n \) forms by increasing the exponent by 1 and dividing by this new exponent.

In our problem, once the expression \( \int (u^{3/2} - u^{1/2}) \, du \) is achieved after substitution, the power rule simplifies it further:

• \( \int u^{3/2} \, du \) becomes \( \frac{2}{5}u^{5/2} \).
• \( \int u^{1/2} \, du \) turns into \( \frac{2}{3}u^{3/2} \).

These transformations demonstrate how the power rule simplifies integration, making seemingly complex expressions much easier to handle.

Indefinite Integrals

Indefinite integrals are all about discovering the family of functions whose derivative leads to the original function, without specifying limits. They represent the antiderivative of a given function. Unlike definite integrals, they do not compute a specific numerical value but rather provide a general formula with a constant \( C \), representing any constant term.

In our example, by the end of the integration process, we arrive at a general form: \[ \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + C \]Here, substituting back \( u = x+1 \) gives:\[ \frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + C \]This expression is an indefinite integral because it describes a set of functions differing only by a constant. By this, whenever you solve an indefinite integral, always include the constant \( C \).

In practice, indefinite integrals are crucial for solving differential equations and modeling scenarios across physics, economics, and other fields where understanding the general behavior of a system is necessary. Knowing both the function and the constant \( C \) ensures complete solutions to these problems.