Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{4}{5}\).

Step 1: Identify the Substitution

Look for a substitution that simplifies the expression inside the integral. Notice that the expression \((t+2)\) inside the denominator appears to be a good candidate for substitution. Let \(u = t + 2\).

Step 2: Differentiate the Substitute and Express dt

Differentiate the substitution \(u = t + 2\) to find \(du\). This gives us \(du = dt\). Now, express \(dt\) in terms of \(du\). Since they are equal, we have \(dt = du\).

Step 3: Change the Limits of Integration

Substitute the limits of integration in terms of \(u\). For \(t = -1\), \(u = -1 + 2 = 1\). For \(t = 3\), \(u = 3 + 2 = 5\). Thus, the new limits are from \(u = 1\) to \(u = 5\).

Step 4: Rewrite the Integral with Substitution

Substitute \(u = t + 2\) and \(dt = du\) into the integral. The integral becomes \(\int_{1}^{5} \frac{1}{u^2} du\).

Step 5: Integrate

Evaluate the integral \(\int \frac{1}{u^2} du\). The antiderivative of \(\frac{1}{u^2}\) is \(-\frac{1}{u}\). Therefore, \(\int \frac{1}{u^2} du = -\frac{1}{u} + C\).

Step 6: Evaluate the Definite Integral

Substitute the limits of integration into the antiderivative. Calculate \(-\frac{1}{u}\) from \(u = 1\) to \(u = 5\). This results in:\[-\left(\frac{1}{5}\right) + \left(\frac{1}{1}\right) = -\frac{1}{5} + 1 = \frac{4}{5}.\]

Step 7: Conclusion

Thus, the value of the integral \(\int_{-1}^{3} \frac{1}{(t+2)^{2}} dt\) is \(\frac{4}{5}\).

Key Concepts

Substitution Rule

The substitution rule is an essential technique in integral calculus, especially when dealing with definite integrals. It involves replacing a part of an integral with a new variable to simplify the integration process. In simpler terms, when the variable inside the integral is complex, we can substitute it with another variable to make the calculation easier.

To apply substitution:

• Identify a function inside the integral that, when substituted, will simplify the integral. In our exercise, this was \(t+2\).
• Set the substitution variable, usually \(u\), equal to this function. So, we let \(u = t + 2\).
• Differentiate your substitution to find \(du\). In this case, \(du = dt\). Since \(dt = du\), it directly simplifies our integration.

By making the transformation, the complicated \(dt\) is replaced with a simpler \(du\), allowing for more straightforward integration.

Limits of Integration

When performing a substitution in a definite integral, it's not just the integrand that changes, but the limits of integration as well. These limits are specific points where you assess the integral, initially given in terms of the original variable.

Here's how you handle the limits:

• Take each original limit of integration and substitute it into your new variable's equation. If \(u = t + 2\), for the lower limit \(t = -1\), we find \(u = 1\).
• Do the same for the upper limit \(t = 3\), finding \(u = 5\).

Thus, your new limits of integration are from \(u = 1\) to \(u = 5\). This change is critical because it ensures that the integral accurately represents the area under the curve from the original problem.

Antiderivative

Finding an antiderivative is a core step in solving integrals, as it helps us determine the integral's value by finding a function whose derivative is the integrand. The antiderivative essentially "undoes" differentiation.

In this exercise, you're finding the antiderivative of \(\frac{1}{u^2}\). The antiderivative of such a function:

• For any power function \(x^{-n}\), the antiderivative involves increasing the power by one and dividing by the new power.
• Thus, \(\int \frac{1}{u^2} du = -\frac{1}{u} + C\), because raising \(-2\) by one and dividing gives us the expression.

The constant \(C\) is typically included in indefinite integrals as it represents the family of antiderivatives, but for definite integrals like ours, it cancels out.

Integration Techniques

Integration techniques are essential strategies that allow us to tackle a variety of integral problems, each with its specific requirements. These techniques include substitution, integration by parts, partial fractions, and more.

In our case, substitution is highlighted, which is useful when the integrand is a composite function, making it easier to handle by changing variables. Here's why it's crucial:

• It transforms the original integral into a simpler form, allowing straightforward evaluation.
• It's often the first technique tried when encountering complex integrals due to its ability to simplify the structure of the integrand.

Ultimately, mastering a range of integration techniques allows for efficient and accurate solutions to a wide array of calculus problems.