Problem 41 - Calculus

Question

Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{-3}^{3} \sqrt{7+2 t^{2}}(8 t) d t $$

Step-by-Step Solution

Answer

The definite integral evaluates to 0.

Step 1: Identify the Inner Function

First, identify the inner function in the integrand. Here, the expression inside the square root is the inner function. Thus we let \( u = 7 + 2t^2 \).

Step 2: Differentiate the Inner Function

Differentiate \( u = 7 + 2t^2 \) with respect to \( t \) to find \( du/dt \). This gives \( du = 4t \, dt \).

Step 3: Express \( dt \) in Terms of \( du \)

Solve for \( dt \) in terms of \( du \) and \( t \). We already know \( du = 4t \, dt \), so \( dt = \frac{du}{4t} \).

Step 4: Rearrange the Integrand Using \( u \) Substitution

Rearrange the integrand, \( \sqrt{7 + 2t^2}(8t) dt \), using \( u \). Substitute \( \, \sqrt{u} \,\) for \( \sqrt{7 + 2t^2} \), and use the expression for \( dt \) from the previous step: \( 8t \, \sqrt{u} \frac{du}{4t} = 2 \sqrt{u} \, du \).

Step 5: Change the Limits of Integration

The original limits are in terms of \( t \). When \( t = -3 \), \( u = 7 + 18 = 25 \) and when \( t = 3 \), similarly \( u = 25 \). Thus, the new limits for \( u \) are from 25 to 25, resulting in a zero-width interval.

Step 6: Evaluate the Integral with New Limits

Perform the integration of \( 2 \int_{25}^{25} \sqrt{u} \, du \). Since the limits of integration are the same, the integral evaluates to 0 without needing further calculus steps.

Key Concepts

Substitution Rule

The substitution rule is a technique used in calculus to simplify the process of evaluating an integral. This approach is particularly useful for complex integrands that are difficult to handle directly. By substituting parts of the integrand with a single variable, usually denoted as \( u \), the integration process becomes simpler.

• The goal is to transform the integrand into a more manageable form using a substitution.
• This involves identifying a part of the integrand to replace, which simplifies the original expression.
• Once the substitution is made, we use basic integration techniques to evaluate the integral.

The substitution rule effectively changes variables in the integral, shifting from a more complex function \( f(t) \) to an easier one \( f(u) \). This transformation can significantly reduce the complexity of solving definite integrals.

Inner Function

The concept of the 'inner function' is central to the substitution rule. Identifying the correct inner function is critical for making the substitution process work smoothly. In the given problem, the expression inside the square root, \( 7 + 2t^2 \), functions as the inner function.

• Recognize the part of the integrand that complicates straightforward integration.
• Set this expression equal to a new variable, \( u \).
• This transformation simplifies the rest of the integration process.

By choosing \( u = 7 + 2t^2 \), we isolated the variable part of the integrand. This made the new expression in terms of \( u \) easier to integrate after the original variable \( t \) was expressed in terms of \( u \). This is an essential step in transforming the integral into a simpler form.

Integration Limits

When using substitution in definite integrals, adjusting the limits of integration is an essential step. The original limits, which are in terms of \( t \), need to be transformed to limits in terms of \( u \).

• Calculate the value of \( u \) at each of the original bounds of the integral.
• Substitute the old limits into the expression for \( u \).

For the problem at hand, both \( t = -3 \) and \( t = 3 \) mapped to \( u = 25 \). Therefore, the limits of the new integral became \( 25 \) to \( 25 \). A zero-width interval results because the upper and lower limits are the same. Thus, the integral evaluates to zero immediately, as there's no area to compute.

Calculus Steps

The process of evaluating the integral involves several methodical calculus steps. Following these ensures an accurate solution to the definite integral.

• After identifying the inner function, differentiate it to find \( \frac{du}{dt} \).
• Reexpress \( dt \) in terms of \( du \).
• Substitute into the integrand using the \( u \) substitution.
• Change the integration limits to match the substitution.

In the exercise, once \( u \) was substituted, and limits changed, the integrand became \( 2 \int_{25}^{25} \sqrt{u} \, du \), which simplifies to zero due to equal limits. No further calculus is needed as the integral evaluates to zero without additional computation.