Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-62, use the Substitution Rule for Definite Integrals to evaluate each definite integral. \(\int_{1}^{3} \frac{x^{2}+1}{\sqrt{x^{3}+3 x}} d x\)

Step-by-Step Solution

Answer

The value of the integral is \(\frac{8}{3}\).

Step 1: Choose a Suitable Substitution

To evaluate the integral using substitution, we start by choosing a substitution that simplifies the integral. In this case, let \[ u = x^3 + 3x. \]This substitution is appropriate because its derivative with respect to x is present in the integral.

Step 2: Find the Derivative and Change of Variables

Differentiate the substitution with respect to x:\[ \frac{du}{dx} = 3x^2 + 3. \]Hence, we can write \[ du = (3x^2 + 3) \, dx = 3(x^2 + 1) \, dx. \]This allows us to express x and dx in terms of u and du. Rewrite the differential as:\[ dx = \frac{1}{3(x^2 + 1)} \, du. \]

Step 3: Substitute and Simplify the Integral

Substitute u and dx in the integral:\[ \int_{1}^{3} \frac{x^2 + 1}{\sqrt{x^3 + 3x}} \, dx = \int_{u(1)}^{u(3)} \frac{1}{\sqrt{u}} \cdot \frac{1}{3(x^2 + 1)} \, du. \]The x terms cancel out, resulting in:\[ \int_{u(1)}^{u(3)} \frac{1}{3\sqrt{u}} \, du. \]

Step 4: Determine the New Limits of Integration

Calculate the new limits of integration:- For x = 1, \( u(1) = 1^3 + 3 \times 1 = 4 \).- For x = 3, \( u(3) = 3^3 + 3 \times 3 = 36 \).This means our integral now evaluates from 4 to 36 in terms of u.

Step 5: Evaluate the Integral

Evaluate the simplified integral:\[ \int_{4}^{36} \frac{1}{3\sqrt{u}} \, du = \frac{1}{3} \int_{4}^{36} u^{-1/2} \, du. \]The antiderivative of \( u^{-1/2} \) is \( 2u^{1/2} \), so we have:\[ \frac{1}{3} \left[ 2u^{1/2} \right]_4^{36}. \]

Step 6: Simplify and Final Evaluation

Continuing with the substitution result:\[ \frac{1}{3} \left[ 2(36)^{1/2} - 2(4)^{1/2} \right]. \]Simplify this expression:- \(2(36)^{1/2} = 2 \times 6 = 12\)- \(2(4)^{1/2} = 2 \times 2 = 4\)Thus, the final calculation is:\[ \frac{1}{3} (12 - 4) = \frac{1}{3} \times 8 = \frac{8}{3}. \]

Key Concepts

Substitution Rule

The substitution rule is a powerful tool for simplifying the process of evaluating definite integrals. Essentially, it transforms a challenging integral into a simpler form that is easier to solve. This is done by substituting a part of the integrand with a new variable. In our example, we substitute the expression \(x^3 + 3x\) with \(u\). This substitution is strategic because:

• It matches a component of the integrand, allowing reduction of complexity.
• The derivative of \(u\) exists within the integrand, facilitating easy replacement of \(dx\) and simplifying the expression.

An important aspect of the substitution rule is expressing \(dx\) in terms of \(du\). This involves differentiating the substitution equation and solving for \(dx\). In this case, with \(u = x^3 + 3x\), we differentiate to get \(du = (3x^2 + 3)dx\) and solve for \(dx\) to complete the transformation.

Change of Variables

Changing variables is an essential concept that complements the substitution rule. It primarily aims to rewrite the integral in terms of a new variable, simplifying its evaluation. When you change variables, you're shifting from the original variable (often \(x\)) to a new one (like \(u\)), chosen to ease the computation.In this context:

• Using \(u = x^3 + 3x\) facilitates a change of perspective from an \(x\)-based description to a \(u\)-focused one.
• The expression remains as a function of \(u\), where tedious polynomials in \(x\) get converted into straightforward terms involving \(u\).

By altering the variables and expressing everything in terms of \(u\), the integral becomes more manageable. This method ties together integrals with their simplifying substitutions, making seemingly complex problems more approachable.

Limits of Integration

When applying the substitution rule for definite integrals, it's crucial to adjust the limits of integration according to the substitution. The limits must change because they originally correspond to values of the initial variable, \(x\). With a change to \(u\), the limits also need to represent \(u\) values.Here's how it's done:

• First, compute the new value of \(u\) at each limit of \(x\). For instance, with \(x = 1\), \(u(1) = 1^3 + 3 \times 1 = 4\).
• Similarly, for \(x = 3\), \(u(3) = 3^3 + 3 \times 3 = 36\).

These transformations yield new, adjusted limits of integration from 4 to 36 in terms of \(u\). Correctly recalculating these limits is vital, as they ensure the integral accurately reflects the function over its intended range.

Antiderivative

The concept of an antiderivative is central to solving integrals. An antiderivative of a function is another function whose derivative equals the original function. In simpler terms, finding an antiderivative is reversing the process of differentiation.In our example problem, once the integral is rewritten using the substitution, we encounter a function of \(u\), specifically \(u^{-1/2}\). The antiderivative of \(u^{-1/2}\) is \(2u^{1/2}\). This is because when you differentiate \(2u^{1/2}\), you obtain \(u^{-1/2}\).Once the antiderivative is found, it's evaluated between the new limits of integration to find the definite integral's value:

• Here, this results in \(\frac{1}{3} \left[ 2u^{1/2} \right]_4^{36}\).
• Finally, substituting the limits \(u = 4\) and \(u = 36\) gives the final result.

Understanding antiderivatives as the inverse of differentiation is key to evaluating integrals effectively, providing the necessary closure for definite integrations.