Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \int_{-1}^{1}\left[(1+t)^{3 / 2} \mathbf{i}+(1-t)^{3 / 2} \mathbf{j}\right] d t $$

Step-by-Step Solution

Answer

The result is \(\frac{8\sqrt{2}}{5}(\mathbf{i} + \mathbf{j})\).

Step 1: Identify Components

The function to integrate is a vector function: \((1+t)^{3/2}\mathbf{i} + (1-t)^{3/2}\mathbf{j}\). This means we need to integrate both components separately over the given interval \([-1, 1]\).

Step 2: Integrate the i-component

We first focus on the i-component of the vector, which is \((1+t)^{3/2}\). The integral to solve is: \[ \int_{-1}^{1} (1+t)^{3/2} \, dt \].

Step 3: Solve the Indefinite Integral for i-component

To solve \(\int (1+t)^{3/2} \, dt\), use the substitution method. Set \(u = 1+t\), then \(du = dt\). The integral becomes: \[ \int u^{3/2} \, du = \frac{u^{5/2}}{5/2} + C = \frac{2}{5} u^{5/2} + C \].

Step 4: Apply Limits to the i-component

Substitute back \(u = 1+t\), then evaluate the definite integral: \[ \left. \frac{2}{5} (1+t)^{5/2} \right|_{-1}^{1} = \frac{2}{5} [(1+1)^{5/2} - (1-1)^{5/2}] = \frac{2}{5} [2^{5/2} - 0^{5/2}] = \frac{2}{5} \times 4\sqrt{2} \]. Here, \(2^{5/2} = (\sqrt{4})^5 = 4\sqrt{2} \), so the result is \(\frac{8\sqrt{2}}{5}\).

Step 5: Integrate the j-component

Now consider the j-component, \((1-t)^{3/2}\). The integral is: \[ \int_{-1}^{1} (1-t)^{3/2} \, dt \].

Step 6: Solve the Indefinite Integral for j-component

Same as before, use substitution method \(v = 1-t\), so that \(dv = -dt\). Then the integral becomes: \[ -\int v^{3/2} \, dv = -\frac{v^{5/2}}{5/2} + C = -\frac{2}{5} v^{5/2} + C \].

Step 7: Apply Limits to the j-component

Substitute back \(v = 1-t\), then evaluate the definite integral: \[ \left. -\frac{2}{5} (1-t)^{5/2} \right|_{-1}^{1} = -\frac{2}{5} [(1+1)^{5/2} - (1-(-1))^{5/2}] = -\frac{2}{5} [0^{5/2} - 2^{5/2}] = \frac{8\sqrt{2}}{5} \].

Step 8: Combine Results

Combine the results from evaluating both component integrals:- i-component gives \(\frac{8\sqrt{2}}{5}\mathbf{i}\),- j-component gives \(\frac{8\sqrt{2}}{5}\mathbf{j}\).Thus, the integral evaluates to \(\frac{8\sqrt{2}}{5}\mathbf{i} + \frac{8\sqrt{2}}{5}\mathbf{j} = \frac{8\sqrt{2}}{5} (\mathbf{i} + \mathbf{j}) \).

Key Concepts

Definite Integrals

Definite integrals play a key role in calculus when we need to compute the area under a curve or, as in our exercise, evaluate a vector function over a specific interval. They are defined by the limits placed on the integral, giving a precise value over that range.

For our task, the definite integral is denoted by the integral sign with limits from -1 to 1:

• The lower limit is -1.
• The upper limit is 1.

With definite integrals, the process involves first finding the indefinite integral (the antiderivative) of the function and then evaluating it at the upper and lower limits. This difference gives the area under the function between these bounds.

In our example, once the antiderivative is found, we substitute the bounds and calculate the difference between the two resulting values. This gives us the net area represented by the integral of the vector components over the given interval.

Substitution Method

A powerful technique used in calculus to simplify integrals is the substitution method. By changing the variable of integration, complex expressions become easier to manage. This technique is especially useful when dealing with functions like powers or roots.

Here's how it works in practice:

• Select a substitution that simplifies the integral, such as setting a variable like \(u\) or \(v\) equal to a part of the integrand.
• Rewrite the integral in terms of the new variable.
• Integrate using the new variable.

The exercise here demonstrates this method by using substitutions like \(u = 1+t\) and \(v = 1-t\) for the components of the vector function. Once the substitution is made, the integration process becomes straightforward. After integration, it's necessary to substitute back to the original variable to apply the definite limits. This method streamline the computation and breaks down the problem into simpler parts.

Vector Functions

In our exercise, you're dealing with vector functions, which are functions that yield a vector output from each input. Essentially, instead of outputting a single value, a vector function gives several outputs—one for each component.

Let's break it down:

• A vector function in 2D has components along the \( \mathbf{i} \) and \( \mathbf{j} \) directions, like in our case: \((1+t)^{3/2}\mathbf{i} + (1-t)^{3/2}\mathbf{j}\).
• To integrate a vector function, you deal with each component independently over the same interval.

The integration of vector functions is crucial in fields like physics and engineering where they can describe quantities like velocity and force, which have both magnitude and direction. By integrating each component and then combining them, we derive an overall result vector that represents the behavior of the system over the specified interval. This approach breaks down complex multidimensional problems into manageable one-dimensional tasks.