Problem 63 - Calculus III

Question

Evaluate the following integrals. $$ \int_{1}^{4} \mathbf{u}(t) d t, \text { with } \mathbf{u}(t)=\left\langle\frac{\ln (t)}{t}, \frac{1}{\sqrt{t}}, \sin \left(\frac{t \pi}{4}\right)\right\rangle $$

Step-by-Step Solution

Answer

The integral evaluates to \( \left\langle \frac{(\ln 4)^2}{2}, 2, \frac{4}{\pi} \left( 1 + \frac{\sqrt{2}}{2} \right) \right\rangle \).

Step 1: Understand the Integral

The given integral is a vector integral of the form \( \int_{1}^{4} \mathbf{u}(t) \, dt \), where \( \mathbf{u}(t) = \left\langle \frac{\ln(t)}{t}, \frac{1}{\sqrt{t}}, \sin\left(\frac{t\pi}{4}\right) \right\rangle \). This integral represents the integration of each component of the vector.

Step 2: Integrate the First Component

Evaluate \( \int_{1}^{4} \frac{\ln(t)}{t} \, dt \) using integration by parts. Let \( u = \ln(t) \), so \( du = \frac{1}{t} \, dt \), and let \( dv = \frac{1}{t} \, dt \), so \( v = \ln(t) \). Using integration by parts, \( \int u \, dv = uv - \int v \, du \), we have:\[ \int \frac{\ln(t)}{t} \, dt = \left. \frac{(\ln(t))^2}{2} \right|_{1}^{4} = \frac{(\ln(4))^2}{2} - \frac{(\ln(1))^2}{2} = \frac{(\ln(4))^2}{2} \]

Step 3: Integrate the Second Component

Evaluate \( \int_{1}^{4} \frac{1}{\sqrt{t}} \, dt \). Rewrite as \( \int_{1}^{4} t^{-1/2} \, dt \). Antidifferentiate to obtain \( \left. 2t^{1/2} \right|_{1}^{4} = 2\sqrt{4} - 2\sqrt{1} = 4 - 2 = 2 \).

Step 4: Integrate the Third Component

Evaluate \( \int_{1}^{4} \sin\left(\frac{t\pi}{4}\right) \, dt \). Use the substitution \( u = \frac{t\pi}{4} \), so \( du = \frac{\pi}{4} \, dt \) or \( dt = \frac{4}{\pi} \, du \). Change the limits of integration: when \( t = 1 \), \( u = \frac{\pi}{4} \); and when \( t = 4 \), \( u = \pi \). The integral becomes:\[ \frac{4}{\pi} \int_{\frac{\pi}{4}}^{\pi} \sin(u) \, du = -\frac{4}{\pi} \left[\cos(u)\right]_{\frac{\pi}{4}}^{\pi} = -\frac{4}{\pi} \left( -1 - \frac{\sqrt{2}}{2} \right) = \frac{4}{\pi} \left( 1 + \frac{\sqrt{2}}{2} \right) \]

Step 5: Combine the Results

Combine the results of the individual integrals into the vector format. The integral of \( \mathbf{u}(t) \) from 1 to 4 is:\[ \left\langle \frac{(\ln 4)^2}{2}, 2, \frac{4}{\pi} \left( 1 + \frac{\sqrt{2}}{2} \right) \right\rangle \].

Key Concepts

Integration by Parts

Integration by parts is a fundamental technique in calculus that allows us to integrate products of functions easily. It's based on the product rule for differentiation. When you have an integral of a product of functions, \[ \int u \, dv = uv - \int v \, du \]Here's how it works step-by-step:

• Choose your \(u\) and \(dv\):
• Typically, select \(u\) to be a function that becomes simpler when differentiated.
• Choose \(dv\) as the remainder of the integral, something straightforward to integrate.
• Differentiation and Integration:
• Differentiate \(u\) to find \(du\).
• Integrate \(dv\) to find \(v\).
• Substitute and Solve:
• Substitute into the formula: \( uv - \int v \, du \).
• Evaluate both parts of the expression – usually, the integral becomes simpler to solve.
In our exercise, to solve \( \int_{1}^{4} \frac{\ln(t)}{t} \, dt \), we let \( u = \ln(t) \) and \( dv = \frac{1}{t} \, dt \). Using the formula simplifies the integral, resulting in \( \left. \frac{(\ln(t))^2}{2} \right|_{1}^{4} \). It's a powerful method when facing integrals involving logarithmic or polynomial functions in a product form.

Substitution Method

The substitution method, often called "u-substitution," is a vital technique in calculus that simplifies integrals by changing variables. This approach is particularly useful when an integral includes a composition of functions. Here's how to use it:

• Identify the Inner Function:
• Look within the integral for a function and its derivative. Choose \(u\) so that \( du \) replaces a portion of the integrand.
• Change the Variable:
• Replace the identified part of the integral with \(u\), and express \(du\) in terms of the original variable \(dt\).
• Transform the Limits:
• For definite integrals, change the limits corresponding to the limits for \(u\).
• Evaluate the Simplified Integral:
• Integrate with respect to \(u\), and substitute back the original variable if needed.
In the original solution, we used substitution for \( \int_{1}^{4} \sin\left(\frac{t\pi}{4}\right) \, dt \). By substituting \( u = \frac{t\pi}{4} \), the integral simplifies to \( \frac{4}{\pi} \int_{\frac{\pi}{4}}^{\pi} \sin(u) \, du \). This converts the complex trigonometric integral into a more manageable form, making solving much more straightforward.

Definite Integrals

Definite integrals allow us to calculate the exact area under a curve over a specific interval. In vector calculus, we extend this concept to functions with multiple components, integrating each component separately over the interval.

• Understanding Limits:
• The limits of integration represent the interval over which we calculate the area - from the lower limit to the upper limit.
• Calculating the Integral:
• First, find the antiderivative of the function. Then, evaluate it at the upper limit and subtract the evaluation at the lower limit.
• In vector integration, treat each component independently. Integrate each one concerning the same interval limits.
• Interpreting the Result:
• Each integrated component represents the cumulative "effect" over the interval.
In our vector integral \( \int_{1}^{4} \mathbf{u}(t) \, dt \), we evaluated each of the vector's components separately and combined the results for the final vector output: \( \left\langle \frac{(\ln 4)^2}{2}, 2, \frac{4}{\pi} \left( 1 + \frac{\sqrt{2}}{2} \right) \right\rangle \). This shows the process of finding definite integrals in multivariable functions.