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How do you find the indefinite integral of \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle ?\)

Short Answer

Expert verified
Now, suppose we have the following vector function: $$ \mathbf{r}(t) = \langle t^2, e^t, \sin(t)\rangle $$ Using the step-by-step solution, find the indefinite integral of the given vector function.

Step by step solution

01

Identify the Components of the Vector Function

Given the vector function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\), identify the components as \(f(t)\), \(g(t)\), and \(h(t)\).
02

Integrate Each Component Separately

Integrate each component with respect to the variable \(t\). This means finding the antiderivative \(F(t)\) for \(f(t)\), \(G(t)\) for \(g(t)\), and \(H(t)\) for \(h(t)\). These antiderivative functions will be used to construct the resulting vector function. $$ F(t) = \int f(t) dt \\ G(t) = \int g(t) dt \\ H(t) = \int h(t) dt $$
03

Construct the Resulting Vector Function

Combine the antiderivative functions \(F(t)\), \(G(t)\), and \(H(t)\) that were found in Step 2 to create a new vector function, which is the indefinite integral of the given function. Don't forget to include the constant of integration \(C\) for each component. $$ \mathbf{R}(t) = \langle F(t)+C_1, G(t)+C_2, H(t)+C_3\rangle $$
04

Result

The indefinite integral of the given vector function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is: $$ \mathbf{R}(t) = \langle \int f(t) dt + C_1, \int g(t) dt + C_2, \int h(t) dt + C_3\rangle $$

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Let \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\). a. Assume that \(\lim \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3}\right\rangle,\) which means that \(\lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0 .\) Prove that \(\lim _{t \rightarrow a} f(t)=L_{1}, \quad \lim _{t \rightarrow a} g(t)=L_{2}, \quad\) and \(\quad \lim _{t \rightarrow a} h(t)=L_{3}\). b. Assume that \(\lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2},\) and \(\lim _{t \rightarrow a} h(t)=L_{3} .\) Prove that \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3}\right\rangle\) which means that \(\lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0\).

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