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Magazine Chapter 3: Problem 101
Evaluate the following integrals: $$\int_{0}^{1} \mathbf{r}(t) d t,$$ where $$\mathbf{r}(t)=\left\langle\sqrt[3]{t}, \frac{1}{t+1}, e^{-t}\right\rangle$$
Short Answer
Expert verified
\( \left\langle \frac{3}{4}, \ln(2), 1 - \frac{1}{e} \right\rangle \)
Step by step solution
01
Set up the integral component-wise
The vector function \( \mathbf{r}(t) = \langle \sqrt[3]{t}, \frac{1}{t+1}, e^{-t} \rangle \) is made up of three components \( x(t) = \sqrt[3]{t} \), \( y(t) = \frac{1}{t+1} \), and \( z(t) = e^{-t} \). We will evaluate the integral of each component separately over the interval \([0, 1]\).
02
Integrate the first component \(x(t) = \sqrt[3]{t}\)
We need to compute \( \int_{0}^{1} \sqrt[3]{t} \, dt \). \( \sqrt[3]{t} \) can be written as \( t^{1/3} \). Using the power rule for integration, \( \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \), we find: \[ \int t^{1/3} \, dt = \frac{t^{4/3}}{4/3} = \frac{3}{4}t^{4/3} \] Evaluating this from 0 to 1 gives: \[ \frac{3}{4}(1^{4/3} - 0^{4/3}) = \frac{3}{4} \]
03
Integrate the second component \(y(t) = \frac{1}{t+1}\)
The integral \( \int_{0}^{1} \frac{1}{t+1} \, dt \) is solved using the natural logarithm rule: \( \int \frac{1}{u} \, du = \ln|u| + C \). Here, \( u = t+1 \). Thus, \[ \int \frac{1}{t+1} \, dt = \ln|t+1| \] Evaluating from 0 to 1 gives: \[ \ln(1+1) - \ln(0+1) = \ln(2) - \ln(1) = \ln(2) \]
04
Integrate the third component \(z(t) = e^{-t}\)
The integral \( \int_{0}^{1} e^{-t} \, dt \) is computed using the exponential rule: \( \int e^{kt} \, dt = \frac{1}{k} e^{kt} + C \). Here, \( k = -1 \), so: \[ \int e^{-t} \, dt = -e^{-t} \] Evaluating from 0 to 1 gives: \[ -e^{-1} - (-e^{0}) = -\frac{1}{e} + 1 \] Simplifying, we have: \[ 1 - \frac{1}{e} \]
05
Combine the results
We now combine the results from each component to get the final result, \( \mathbf{R} \), for the integral: \[ \int_{0}^{1} \mathbf{r}(t) \, dt = \langle \frac{3}{4}, \ln(2), 1 - \frac{1}{e} \rangle \] This vector represents the integrated values of each of the components of \( \mathbf{r}(t) \) over the interval.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Integration
Vector integration is the process of integrating vector-valued functions over a given interval. In our problem, the vector function \( \mathbf{r}(t) = \langle \sqrt[3]{t}, \frac{1}{t+1}, e^{-t} \rangle \) is considered over the interval from 0 to 1.
Each component of the vector function, \( x(t) = \sqrt[3]{t} \), \( y(t) = \frac{1}{t+1} \), and \( z(t) = e^{-t} \), must be integrated separately. The final result of the vector integration is a vector consisting of the integrals of each of these components.
This technique is used extensively in multivariable calculus, where analyzing the path or space described by a vector-valued function requires examining each part of the motion or position separately by component.
Each component of the vector function, \( x(t) = \sqrt[3]{t} \), \( y(t) = \frac{1}{t+1} \), and \( z(t) = e^{-t} \), must be integrated separately. The final result of the vector integration is a vector consisting of the integrals of each of these components.
This technique is used extensively in multivariable calculus, where analyzing the path or space described by a vector-valued function requires examining each part of the motion or position separately by component.
Power Rule for Integration
The power rule for integration is a fundamental technique used to find antiderivatives. It applies when integrating functions of the form \( t^n \).
According to the power rule, the integral \( \int t^n \, dt \) is calculated as \( \frac{t^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. In our exercise, the first component \( x(t) = \sqrt[3]{t} \) can be rewritten as \( t^{1/3} \).
Applying the power rule, we integrate to find \( \int t^{1/3} \, dt = \frac{t^{4/3}}{4/3} = \frac{3}{4}t^{4/3} \). Evaluating this from 0 to 1 gives a result of \( \frac{3}{4} \), after substituting the limits of integration.
According to the power rule, the integral \( \int t^n \, dt \) is calculated as \( \frac{t^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. In our exercise, the first component \( x(t) = \sqrt[3]{t} \) can be rewritten as \( t^{1/3} \).
Applying the power rule, we integrate to find \( \int t^{1/3} \, dt = \frac{t^{4/3}}{4/3} = \frac{3}{4}t^{4/3} \). Evaluating this from 0 to 1 gives a result of \( \frac{3}{4} \), after substituting the limits of integration.
Exponential Function Integration
Integration of exponential functions involves recognizing the general structure \( e^{kt} \) and applying its corresponding rule. The integral \( \int e^{kt} \, dt = \frac{1}{k} e^{kt} + C \) helps in finding antiderivatives of exponential functions.
In our example, the third component is \( z(t) = e^{-t} \). The parameter is \( k = -1 \), leading to the integration \( \int e^{-t} \, dt = -e^{-t} + C \).
When calculating definite integrals, evaluate the expression at the upper and lower limits, sequentially. For \( e^{-t} \), this results in \( -\frac{1}{e} + 1 \) when evaluated from 0 to 1. This simplifies to \( 1 - \frac{1}{e} \).
In our example, the third component is \( z(t) = e^{-t} \). The parameter is \( k = -1 \), leading to the integration \( \int e^{-t} \, dt = -e^{-t} + C \).
When calculating definite integrals, evaluate the expression at the upper and lower limits, sequentially. For \( e^{-t} \), this results in \( -\frac{1}{e} + 1 \) when evaluated from 0 to 1. This simplifies to \( 1 - \frac{1}{e} \).
Natural Logarithm Integration
Integrating functions that involve fractions similar to \( \frac{1}{t+1} \) relies on the natural logarithm rule. If you have an integral of the form \( \int \frac{1}{u} \, du \), it equals \( \ln |u| + C \).
In the problem, the expression corresponds to \( u = t+1 \). Hence, the integral \( \int \frac{1}{t+1} \, dt \) simplifies to \( \ln|t+1| \).
Evaluating this from 0 to 1 gives \( \ln(2) - \ln(1) \). As \( \ln(1) = 0 \), the result is simply \( \ln(2) \). This integration technique highlights the connection between logarithmic functions and their derivatives, a crucial idea in calculus.
In the problem, the expression corresponds to \( u = t+1 \). Hence, the integral \( \int \frac{1}{t+1} \, dt \) simplifies to \( \ln|t+1| \).
Evaluating this from 0 to 1 gives \( \ln(2) - \ln(1) \). As \( \ln(1) = 0 \), the result is simply \( \ln(2) \). This integration technique highlights the connection between logarithmic functions and their derivatives, a crucial idea in calculus.
