Problem 36 - Calculus Early Transcendentals

Question

Evaluate the integral. $$ \int_{1}^{2}\left(t^{2} \mathbf{i}+t \sqrt{t-1} \mathbf{j}+t \sin \pi t \mathbf{k}\right) d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{7}{3} \mathbf{i} + \frac{16}{15} \mathbf{j} - \frac{3}{\pi} \mathbf{k} \)."

Step 1: Break Down the Integral

This integral is a vector integral where each component is a separate integral. So, you need to evaluate each component independently. The vector function is given as \( \int_{1}^{2} \left( t^{2} \mathbf{i} + t \sqrt{t-1} \mathbf{j} + t \sin \pi t \mathbf{k} \right) \, dt \). This means you will evaluate three separate integrals: \( \int_{1}^{2} t^2 \, dt \), \( \int_{1}^{2} t \sqrt{t-1} \, dt \), and \( \int_{1}^{2} t \sin \pi t \, dt \).

Step 2: Integrate the First Component

Evaluate the first integral \( \int_{1}^{2} t^2 \, dt \). The antiderivative of \( t^2 \) is \( \frac{t^3}{3} \). Evaluate from 1 to 2: \[ \left[ \frac{t^3}{3} \right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}. \]

Step 3: Integrate the Second Component

Evaluate the second integral \( \int_{1}^{2} t \sqrt{t-1} \, dt \). Use substitution \( u = t-1 \), thus \( du = dt \) and when \( t=1, \; u=0 \) and when \( t=2, \; u=1 \). The integral becomes \( \int_{0}^{1} (u+1) \sqrt{u} \, du \). Expand to \( \int_{0}^{1} (u^{3/2} + u^{1/2}) \, du \). The antiderivatives are \( \frac{u^{5/2}}{5/2} \) and \( \frac{u^{3/2}}{3/2} \). Calculate: \[ \left[ \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} \right]_0^1 = \left( \frac{2}{5} \cdot 1 + \frac{2}{3} \cdot 1 \right) - 0 = \frac{2}{5} + \frac{2}{3} = \frac{6}{15} + \frac{10}{15} = \frac{16}{15}. \]

Step 4: Integrate the Third Component

Evaluate the third integral \( \int_{1}^{2} t \sin(\pi t) \, dt \). Use integration by parts: let \( u = t \), \( dv = \sin(\pi t) \, dt \) then \( du = dt \), \( v = \int \sin(\pi t) \, dt = -\frac{1}{\pi} \cos(\pi t) \). Use integration by parts formula: \( \int u \, dv = uv - \int v \, du \). The integral becomes:\[ \left[ -\frac{t}{\pi} \cos(\pi t) \right]_1^2 + \int_{1}^{2} \frac{1}{\pi} \cos(\pi t) \, dt. \]Evaluating \( \left[ -\frac{t}{\pi} \cos(\pi t) \right]_1^2 \): \[ -\frac{2}{\pi} \cos(2\pi) + \frac{1}{\pi} \cos(\pi) = -\frac{2}{\pi}(1) + \frac{1}{\pi}(-1) = -\frac{2}{\pi} - \frac{1}{\pi} = -\frac{3}{\pi}. \]For \( \int_{1}^{2} \frac{1}{\pi} \cos(\pi t) \, dt \), the antiderivative is \(-\frac{1}{\pi^2} \sin(\pi t)\):\[ \left[ -\frac{1}{\pi^2} \sin(\pi t) \right]_1^2 = -\frac{1}{\pi^2}(\sin 2\pi - \sin \pi) = -0. \]So, the result is \(-\frac{3}{\pi} + 0 = -\frac{3}{\pi}.\)

Step 5: Assemble the Result

Combine all the results to form the final evaluated vector integral: \[ \int_{1}^{2} \left( t^2 \mathbf{i} + t \sqrt{t-1} \mathbf{j} + t \sin \pi t \mathbf{k} \right) dt = \frac{7}{3} \mathbf{i} + \frac{16}{15} \mathbf{j} - \frac{3}{\pi} \mathbf{k}. \]

Key Concepts

Definite Integrals

A definite integral is a powerful tool in calculus used to determine the accumulated quantity, such as total area, volume, or other quantities, over a specific interval. For example, when you calculate the area under a curve from one point to another, you are using the concept of a definite integral. In a vector calculus context, as seen in the given exercise, the definite integral is used to integrate vector functions over a defined interval.

A key characteristic of the definite integral is its limits of integration, represented by the upper and lower bounds. These limits define the interval over which the function is evaluated. To calculate a definite integral, follow these steps:

• Find the antiderivative or indefinite integral of the function.
• Evaluate the antiderivative at the upper limit of the interval.
• Subtract the evaluation of the antiderivative at the lower limit.

This process allows you to find the net area between a function and the x-axis over the interval [a, b]. In vector calculus, each component of a vector function can have its definite integral evaluated independently, leading to a vector result.

Integration by Parts

Integration by parts is a technique used to integrate products of functions by transforming them into simpler integrals. This method comes from the product rule in differentiation and is extremely useful in cases like the third component in the original problem: \( \int_{1}^{2} t \sin(\pi t) \, dt \).

The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]where \( u \) is a function you choose to differentiate, and \( dv \) is a function you choose to integrate. This formula helps to break down complex integrals into simpler parts. Here's how to apply it:

• Choose parts \( u \) and \( dv \) from the original integral.
• Differentiate \( u \) to find \( du \).
• Integrate \( dv \) to find \( v \).
• Substitute into the integration by parts formula.
• Solve the resulting integrals.

In practical terms, by choosing \( u = t \) and \( dv = \sin(\pi t) \, dt \), you can simplify the integral into parts that are easier to evaluate.

Substitution Method

The substitution method, also known as u-substitution, is a strategy for performing integration by changing the variable of integration. This makes an integral easier to solve by converting it into a simpler form. The application of substitution can be seen in the second component of the exercise: \( \int_{1}^{2} t \sqrt{t-1} \, dt \).

Here's how substitution works:

• Identify a substitution \( u = g(t) \) simplifying the integral, with \( du = g'(t) \, dt \).
• Change the limits of integration based on the new variable \( u \).
• Rewrite the integral in terms of \( u \) and \( du \).
• Integrate with respect to \( u \).
• Substitute back to the original variable if needed.

In this problem, substituting \( u = t-1 \) effectively changes the integral into one involving powers of \( u \), which is easier to evaluate and helps in reaching the solution more straightforwardly.

Integral Calculus

Integral calculus involves the concept of accumulation, allowing for the calculation of quantities such as areas, volumes, and other related measurements. It is part of the broader field of calculus, which studies how things change and accumulate.

The main tool of integral calculus is the concept of the integral, both definite and indefinite. Integral calculus provides two fundamental operations:

• Finding the antiderivative, or indefinite integral, of functions.
• Evaluating definite integrals to find accumulated values over intervals.

This branch of mathematics helps to connect rates of change, given by derivatives, with accumulated quantities. In the vector calculus problem provided, integral calculus allows the separate components of a vector field to be analyzed independently through their integrals, which are then combined to give the overall solution.

By developing techniques such as integration by parts and substitution, integral calculus significantly expands the range of functions that can be integrated, making previously complex problems manageable and solvable.