Problem 37 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the definite integral. $$ \int_{0}^{4} \frac{1}{\sqrt{2 x+1}} d x $$

Step-by-Step Solution

Answer

The value of the given definite integral is 2.

Step 1: Setting up the substitution

Initiate by choosing a substitution that simplifies the integrand. Let \(u = 2x + 1\). Then, \(du = 2dx\). Now, isolate \(dx\) to get \(dx = du/2\).

Step 2: Changing the integral limits

As the original limits were in terms of x, replace them with corresponding values in terms of u. When \(x=0\), \(u=2*0 + 1 = 1\). When \(x=4\), \(u=2*4 + 1 = 9\). So the integral now ranges from 1 to 9.

Step 3: Substituting

Substitute \(u = 2x + 1\) and \(dx = du/2\) into the integral. It would become \(\int_{1}^{9} \frac{1}{\sqrt{u}} * {du/2}\).

Step 4: Simplifying the integral

The integral above simplifies to \(1/2 \int_{1}^{9} u^{-1/2} du = 1/2 * [2u^{1/2}]_{1}^{9}\). The factor of 1/2 is pulled out of integral because it is a constant.

Step 5: Evaluating the Integral

To finalize, evaluate the definite integral at the endpoints. This yields: \(1/2 * [2\sqrt{u}]_{1}^{9} = [ \sqrt{u} ]_{1}^{9} = \sqrt{9} - \sqrt{1} = 3 - 1 = 2\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used in the evaluation of integrals, particularly when dealing with complex integrands. By choosing an appropriate substitution, you can simplify the integrand into a more manageable form. In this exercise, the substitution chosen was \( u = 2x + 1 \). The primary goal of this step is to transform the integral in terms of \( u \) instead of \( x \).

Here's how you can approach it:

• Identify a part of the integrand that, when substituted, will simplify the expression. In this case, \( 2x + 1 \) is a good choice since it appears inside the square root.
• Express \( dx \) in terms of \( du \) by differentiating \( u \) with respect to \( x \). So, if \( du = 2dx \), then \( dx = \frac{du}{2} \).
• Substitute \( u \) and \( dx \) back into the integral.

This method not only simplifies the integral but also eases the process of evaluating it later.

Integration Limits

When using the substitution method, it's crucial to change the integration limits to match the new variable. The limits originally given are in terms of \( x \), but they must be converted to \( u \) for the substitution.

This requires substituting the values of \( x \) into the substitution equation \( u = 2x + 1 \).

• For the lower limit: When \( x = 0 \), substitute into \( u = 2 \cdot 0 + 1 = 1 \).
• For the upper limit: When \( x = 4 \), substitute into \( u = 2 \cdot 4 + 1 = 9 \).

Thus, the integral should now be evaluated from \( u = 1 \) to \( u = 9 \). Changing the limits ensures the definite integral accurately reflects the variable transformation you've made.

Evaluating Integrals

Evaluating integrals, especially after substitution, involves carrying out the integration process on the transformed integral and then using the limits to find the definite solution.

Once the new limits are applied, substitute and simplify the integral. For this exercise:

• The integral becomes \( \int_{1}^{9} \frac{1}{\sqrt{u}} \cdot \frac{du}{2} \), which simplifies to \( \frac{1}{2} \int_{1}^{9} u^{-1/2} du \).
• Integrate \( u^{-1/2} \), which results in \( 2u^{1/2} \).
• Consider the constant factor \( \frac{1}{2} \), which multiplies the entire definite integral as \( \frac{1}{2} \cdot [2u^{1/2}]_{1}^{9} \).

The definite integral is then evaluated by computing the result at the upper limit and subtracting the result at the lower limit.

Square Root Functions

Square root functions often appear in integrals, requiring careful handling because of their non-linear nature. In this exercise, the integrand includes the term \( \frac{1}{\sqrt{2x+1}} \), which involves a square root.

Understanding and simplifying square root expressions is essential:

• When substituted, the expression turns into \( \frac{1}{\sqrt{u}} \), which is a common form suitable for integration.
• Square root terms are transformed into exponent terms using the transformation \( \sqrt{u} = u^{1/2} \). This makes it easier to apply integration rules.

With square root functions, using substitutions can significantly simplify the integral, thus making the evaluation process more straightforward. This method allows for expressions to be handled in their simplest form, enabling easier manipulation and calculation.