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Chapter 13: Problem 8

Evaluate the integrals in Exercises \(1-10\) $$ \int_{1}^{\ln 3}\left[t e^{t} \mathbf{i}+e^{t} \mathbf{j}+\ln t \mathbf{k}\right] d t $$

Short Answer

Expert verified
The integral evaluates to \(3(\ln 3 - 1) \mathbf{i} + (3 - e) \mathbf{j} + ((\ln 3)^2 - \ln 3 + 1) \mathbf{k}\)."

Step by step solution

01

Break Down the Integral

The given vector integral is \[\int_{1}^{\ln 3}\left[t e^{t} \mathbf{i}+e^{t} \mathbf{j}+\ln t\mathbf{k}\right] d t\]This can be broken down into three separate integrals for each of the vector components: 1. \(\int_{1}^{\ln 3} t e^t \, dt \mathbf{i} \)2. \(\int_{1}^{\ln 3} e^t \, dt \mathbf{j} \)3. \(\int_{1}^{\ln 3} \ln t \, dt \mathbf{k} \)
02

Evaluate the First Integral

To evaluate \(\int_{1}^{\ln 3} t e^t \, dt\), use integration by parts.Let \(u = t\), \(dv = e^t \, dt\). Then \(du = dt\) and \(v = e^t\).Integration by parts gives:\[\int u \, dv = uv - \int v \, du = t e^t - \int e^t \, dt\]Thus, the integral becomes:\[t e^t - e^t \bigg|_1^{\ln 3}\]Evaluate from 1 to \(\ln 3\):\[\left((\ln 3) e^{\ln 3} - e^{\ln 3}\right) - \left(1\cdot e^1 - e^1\right)\]\[= 3\ln 3 - 3 - 0 = 3(\ln 3 - 1)\]Thus, the first integral evaluates to \(3(\ln 3 - 1) \mathbf{i}\).
03

Evaluate the Second Integral

The second integral is straightforward:\[\int_{1}^{\ln 3} e^t \, dt = e^t \bigg|_1^{\ln 3}\]Evaluate at the limits:\[e^{\ln 3} - e^1 = 3 - e\]Thus, the second integral evaluates to \((3 - e) \mathbf{j}\).
04

Evaluate the Third Integral

The third integral requires integration by parts using\( u = \ln t \) and \( dv = dt \).Then \(du = \frac{1}{t} \, dt\) and \(v = t\).Integration by parts gives:\[\int u \, dv = uv - \int v \, du = t \ln t - \int t \cdot \frac{1}{t} \, dt\]\[= t \ln t - \int dt = t \ln t - t \bigg|_1^{\ln 3}\]Evaluate from 1 to \(\ln 3\):\[\left((\ln 3)(\ln 3) - \ln 3\right) - \left(1 \cdot 0 - 1\right)\]\[= (\ln 3)^2 - \ln 3 + 1\]Thus, the third integral evaluates to \(((\ln 3)^2 - \ln 3 + 1) \mathbf{k}\).
05

Combine the Results

Combine each component's evaluated result:- First integral: \(3(\ln 3 - 1) \mathbf{i}\)- Second integral: \((3 - e) \mathbf{j}\)- Third integral: \(((\ln 3)^2 - \ln 3 + 1) \mathbf{k}\)The final result of the vector integral is:\[3(\ln 3 - 1) \mathbf{i} + (3 - e) \mathbf{j} + ((\ln 3)^2 - \ln 3 + 1) \mathbf{k}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
Integration by parts is a useful technique in calculus used for finding integrals where simpler methods don't work. Its formula is derived from the product rule of differentiation and is \[\int u \, dv = uv - \int v \, du\]This formula helps us to integrate products of functions by assigning one part to be differentiated and the other to be integrated. You often set \(u\) to the polynomial part of the integrand, while \(dv\) is the more complex part, such as an exponential or a trigonometric function.
In this exercise, integration by parts was used for the vector component \( \int_{1}^{\ln 3} t e^t \, dt \). Here, \(u = t\) and \(dv = e^t \, dt\) were chosen because differentiating \(t\) and integrating \(e^t\) are straightforward steps. This combination transforms the original integral into a simpler form that can be solved with basic integration and algebraic manipulation. Understanding how to choose \(u\) and \(dv\) efficiently makes integration by parts a powerful tool for solving challenging integrals.
Vector Integration
Vector integration is the process of integrating functions whose outputs are vectors rather than scalars. In this scenario, each component of the vector needs to be integrated separately, resulting in a new vector made up of the individual integrals of each component.
In our exercise, the given integral is a vector expressed in terms of \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), which are the Cartesian unit vector components. This task involved calculating three separate integrals, one for each of the components associated with \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) respectively.
By breaking down the main integral \[\int_{1}^{\ln 3}\left[t e^{t} \mathbf{i}+e^{t} \mathbf{j}+\ln t\mathbf{k}\right] dt\]it allowed for more manageable pieces: \(\int_{1}^{\ln 3} t e^t \, dt \), \(\int_{1}^{\ln 3} e^t \, dt \), and \(\int_{1}^{\ln 3} \ln t \, dt \). This decomposition leverages our understanding of basic integrals and helps us to translate scalar mathematics into vector mathematics by treating each component separately.
Definite Integration
Definite integration involves calculating the integral of a function between two limits. It differs from indefinite integration, which finds the general form of the antiderivative. Definite integration results in a numerical value (or vector in case of vector integration), representing the area under the curve of the function, between the specified limits.
For the exercise at hand, once each vector component's integral was found through indefinite integration techniques such as integration by parts, definite integration required evaluating each component over the interval \([1, \ln 3]\). The process involves substituting the upper and lower bounds into the antiderivative and subtracting the results, reflecting the fundamental theorem of calculus.
This approach ensures that the final vector result is valid within the specified bounds and accurately represents the cumulative effect of the integrated vector components over the given interval. This step establishes both the magnitude and direction of the integrated vector, aligned with specified limits.

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Most popular questions from this chapter

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