Find the line integral of Bold Upper F equals left parenthesis 3 x squared minus 5 x right parenthesis Bold i plus 5 z Bold j plus Bold k from​ (0,0,0) to​ (1,1,1) over each of the following paths in the accompanying figure.
a. Upper C 1​: Bold r left parenthesis t right parenthesis equals t Bold i plus t Bold j plus t Bold k comma 0 less than or equals t less than or equals 1
b. Upper C 2​: Bold r left parenthesis t right parenthesis equals t Bold i plus t squared Bold j plus t Superscript 4 Baseline Bold k comma 0 less than or equals t less than or equals 1
c. Upper C 3 union Upper C 4​: the path consisting of the line segment from​ (0,0,0) to​ (1,1,0) followed by the segment from​ (1,1,0) to​ (1,1,1)

To find the line integral of the vector field \(\mathbf{F} = (3x^2 - 5x) \mathbf{i} + 5z \mathbf{j} + \mathbf{k}\) along the given paths, we will use the following formula for the line integral:

\[
\int_C \mathbf{F} \cdot d\mathbf{r}
\]

### Path \(C_1\): \(\mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} + t \mathbf{k}\), \(0 \leq t \leq 1\)

1. **Calculate \(d\mathbf{r}\)**:
\[
d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (\mathbf{i} + \mathbf{j} + \mathbf{k}) dt
\]

2. **Parameterize \(\mathbf{F}\)**:
\[
\mathbf{F}(x, y, z) = \mathbf{F}(t, t, t) = (3t^2 - 5t) \mathbf{i} + 5t \mathbf{j} + \mathbf{k}
\]

3. **Dot product \(\mathbf{F} \cdot d\mathbf{r}\)**:
\[
\mathbf{F} \cdot d\mathbf{r} = (3t^2 - 5t) dt + 5t dt + dt = (3t^2 - 5t + 5t + 1) dt = (3t^2 + 1) dt
\]

4. **Integral**:
\[
\int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 (3t^2 + 1) dt = \left[ t^3 + t \right]_0^1 = 1^3 + 1 - (0^3 + 0) = 2
\]

### Path \(C_2\): \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^4 \mathbf{k}\), \(0 \leq t \leq 1\)

1. **Calculate \(d\mathbf{r}\)**:
\[
d\mathbf{r} = \left( \mathbf{i} + 2t \mathbf{j} + 4t^3 \mathbf{k} \right) dt
\]

2. **Parameterize \(\mathbf{F}\)**:
\[
\mathbf{F}(x, y, z) = \mathbf{F}(t, t^2, t^4) = (3t^2 - 5t) \mathbf{i} + 5t^4 \mathbf{j} + \mathbf{k}
\]

3. **Dot product \(\mathbf{F} \cdot d\mathbf{r}\)**:
\[
\mathbf{F} \cdot d\mathbf{r} = (3t^2 - 5t) dt + 5t^4 \cdot 2t dt + 1 \cdot 4t^3 dt = (3t^2 - 5t + 10t^5 + 4t^3) dt
\]

4. **Integral**:
\[
\int_{C_2} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 (3t^2 - 5t + 10t^5 + 4t^3) dt
\]

Breaking it into parts:
\[
\int_0^1 3t^2 dt = \left[ t^3 \right]_0^1 = 1
\]
\[
\int_0^1 -5t dt = \left[ -\frac{5}{2}t^2 \right]_0^1 = -\frac{5}{2}
\]
\[
\int_0^1 10t^5 dt = \left[ \frac{10}{6}t^6 \right]_0^1 = \frac{5}{3}
\]
\[
\int_0^1 4t^3 dt = \left[ t^4 \right]_0^1 = 1
\]

Adding them together:
\[
1 - \frac{5}{2} + \frac{5}{3} + 1 = 1 + 1 - \frac{5}{2} + \frac{5}{3} = 2 - \frac{15}{6} + \frac{10}{6} = 2 - \frac{5}{6} = \frac{7}{6}
\]

### Path \(C_3 \cup C_4\)

#### Path \(C_3\): from \((0,0,0)\) to \((1,1,0)\)

1. **Parameterize \(\mathbf{r}(t)\)**:
\[
\mathbf{r}(t) = t \mathbf{i} + t \mathbf{j}, \quad 0 \leq t \leq 1
\]

2. **Calculate \(d\mathbf{r}\)**:
\[
d\mathbf{r} = (\mathbf{i} + \mathbf{j}) dt
\]

3. **Parameterize \(\mathbf{F}\)**:
\[
\mathbf{F}(t, t, 0) = (3t^2 - 5t) \mathbf{i} + 0 \mathbf{j} + \mathbf{k}
\]

4. **Dot product \(\mathbf{F} \cdot d\mathbf{r}\)**:
\[
\mathbf{F} \cdot d\mathbf{r} = (3t^2 - 5t) dt
\]

5. **Integral**:
\[
\int_{C_3} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 (3t^2 - 5t) dt = 1 - \frac{5}{2} = -\frac{3}{2}
\]

#### Path \(C_4\): from \((1,1,0)\) to \((1,1,1)\)

1. **Parameterize \(\mathbf{r}(t)\)**:
\[
\mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t \mathbf{k}, \quad 0 \leq t \leq 1
\]

2. **Calculate \(d\mathbf{r}\)**:
\[
d\mathbf{r} = \mathbf{k} dt
\]

3. **Parameterize \(\mathbf{F}\)**:
\[
\mathbf{F}(1, 1, t) = (3 \cdot 1^2 - 5 \cdot 1) \mathbf{i} + 5t \mathbf{j} + \mathbf{k} = (-2) \mathbf{i} + 5t \mathbf{j} + \mathbf{k}
\]

4. **Dot product \(\mathbf{F} \cdot d\mathbf{r}\)**:
\[
\mathbf{F} \cdot d\mathbf{r} = dt
\]

5. **Integral**:
\[
\int_{C_4} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 dt = 1
\]

#### Combined \(C_3 \cup C_4\):

\[
\int_{C_3 \cup C_4} \mathbf{F} \cdot d\mathbf{r} = \int_{C_3} \mathbf{F} \cdot d\mathbf{r} + \int_{C_4} \mathbf{F} \cdot d\mathbf{r} = -\frac{3}{2} + 1 = -\frac{1}{2}
\]

### Summary
- \(\int_{C_1} \mathbf{F} \cdot d\mathbf{r} = 2\)
- \(\int_{C_2} \mathbf{F} \cdot d\mathbf{r} = \frac{7}{6}\)
- \(\int_{C_3 \cup C_4} \mathbf{F} \cdot d\mathbf{r} = -\frac{1}{2}\)

Would you like more details or have any questions?

Here are some questions you might ask next:
1. How is a line integral calculated for vector fields?
2. Can you explain more about parameterizing paths?
3. What are some applications of line integrals in physics?
4. How do you interpret the result of a line integral?
5. Can you provide another example of a line integral with a different vector field?

**Tip**: Always carefully choose your parameterization to simplify the calculation of line integrals.

Explore how to calculate line integrals of vector fields along various paths with detailed solutions. This example covers paths \(C_1\), \(C_2\), and \(C_3 \cup C_4\) using the vector field \(\mathbf{F} = (3x^2 - 5x) \mathbf{i} + 5z \mathbf{j} + \mathbf{k}\).

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