Wilson Bates
10/11/2023 · High School
Compute the indefinite integral of the following function. \( r(t)=\left\langle t^{4}-7 t, 4 t-7,19\right\rangle \) A. \( \int r(t) d t=\langle \) B. \( \int r(t) d t= \)
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Step-by-step Solution
To compute the indefinite integral of the vector function \( r(t) = \langle t^{4} - 7t, 4t - 7, 19 \rangle \), we will integrate each component of the vector separately.
1. **Integrate the first component**:
\[
\int (t^{4} - 7t) \, dt = \frac{t^{5}}{5} - \frac{7t^{2}}{2} + C_1
\]
2. **Integrate the second component**:
\[
\int (4t - 7) \, dt = 2t^{2} - 7t + C_2
\]
3. **Integrate the third component**:
\[
\int 19 \, dt = 19t + C_3
\]
Now, we can combine these results into a single vector:
\[
\int r(t) \, dt = \left\langle \frac{t^{5}}{5} - \frac{7t^{2}}{2} + C_1, \, 2t^{2} - 7t + C_2, \, 19t + C_3 \right\rangle
\]
Since \( C_1, C_2, \) and \( C_3 \) are arbitrary constants of integration, we can combine them into a single constant \( C \) for the entire vector:
\[
\int r(t) \, dt = \left\langle \frac{t^{5}}{5} - \frac{7t^{2}}{2} + C, \, 2t^{2} - 7t + C, \, 19t + C \right\rangle
\]
Thus, we can express the indefinite integral as:
\[
\int r(t) \, dt = \left\langle \frac{t^{5}}{5} - \frac{7t^{2}}{2} + C, \, 2t^{2} - 7t + C, \, 19t + C \right\rangle
\]
In conclusion, the final answer is:
A. \( \int r(t) dt = \left\langle \frac{t^{5}}{5} - \frac{7t^{2}}{2} + C, \, 2t^{2} - 7t + C, \, 19t + C \right\rangle \)
B. \( \int r(t) dt = \left\langle \frac{t^{5}}{5} - \frac{7t^{2}}{2} + C, \, 2t^{2} - 7t + C, \, 19t + C \right\rangle \)
Quick Answer
A. \( \int r(t) dt = \left\langle \frac{t^{5}}{5} - \frac{7t^{2}}{2} + C, \, 2t^{2} - 7t + C, \, 19t + C \right\rangle \)
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