Along the curve Bold r left parenthesis t right parenthesis equals negative Bold i plus t squared Bold j plus t Bold k​, 0less than or equalstless than or equals​1, evaluate each of the following integrals.
a. ModifyingBelow Integral from nothing to nothing With Upper C left parenthesis x minus y plus z right parenthesis dx     b. ModifyingBelow Integral from nothing to nothing With Upper C left parenthesis x minus y plus z right parenthesis dy     c. ModifyingBelow Integral from nothing to nothing With Upper C left parenthesis x minus y plus z right parenthesis dz


To evaluate each of the integrals along the curve \(\mathbf{r}(t) = -\mathbf{i} + t^2 \mathbf{j} + t \mathbf{k}\), \(0 \leq t \leq 1\), we will follow these steps for each part:

1. Parameterize the integrand.
2. Calculate the differential \(dx\), \(dy\), and \(dz\) as needed.
3. Perform the integration over the given interval.

### Given Curve
\[
\mathbf{r}(t) = -\mathbf{i} + t^2 \mathbf{j} + t \mathbf{k}
\]
Thus,
\[
x(t) = -1, \quad y(t) = t^2, \quad z(t) = t
\]

### Part (a): \(\int_C (x - y + z) \, dx\)

1. **Parameterize the integrand**:
\[
x(t) = -1, \quad y(t) = t^2, \quad z(t) = t
\]
\[
x - y + z = -1 - t^2 + t
\]

2. **Calculate \(dx\)**:
\[
dx = \frac{dx}{dt} dt = 0 dt = 0
\]

Since \(dx = 0\), the integral \(\int_C (x - y + z) \, dx\) is zero:
\[
\int_C (x - y + z) \, dx = \int_0^1 ( -1 - t^2 + t) \cdot 0 \, dt = 0
\]

### Part (b): \(\int_C (x - y + z) \, dy\)

1. **Parameterize the integrand**:
\[
x(t) = -1, \quad y(t) = t^2, \quad z(t) = t
\]
\[
x - y + z = -1 - t^2 + t
\]

2. **Calculate \(dy\)**:
\[
dy = \frac{dy}{dt} dt = 2t dt
\]

3. **Integral**:
\[
\int_C (x - y + z) \, dy = \int_0^1 (-1 - t^2 + t) \cdot 2t \, dt
\]

Evaluate the integral:
\[
\int_0^1 (-1 - t^2 + t) \cdot 2t \, dt = \int_0^1 (-2t - 2t^3 + 2t^2) \, dt
\]

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

Adding them together:
\[
-1 - \frac{1}{2} + \frac{2}{3} = -1.5 + 0.6667 = -0.8333
\]

### Part (c): \(\int_C (x - y + z) \, dz\)

1. **Parameterize the integrand**:
\[
x(t) = -1, \quad y(t) = t^2, \quad z(t) = t
\]
\[
x - y + z = -1 - t^2 + t
\]

2. **Calculate \(dz\)**:
\[
dz = \frac{dz}{dt} dt = 1 dt = dt
\]

3. **Integral**:
\[
\int_C (x - y + z) \, dz = \int_0^1 (-1 - t^2 + t) \, dt
\]

Evaluate the integral:
\[
\int_0^1 (-1 - t^2 + t) \, dt = \int_0^1 -1 \, dt + \int_0^1 -t^2 \, dt + \int_0^1 t \, dt
\]

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

Adding them together:
\[
-1 - \frac{1}{3} + \frac{1}{2} = -1 - 0.3333 + 0.5 = -0.8333
\]

### Summary
- \(\int_C (x - y + z) \, dx = 0\)
- \(\int_C (x - y + z) \, dy = -0.8333\)
- \(\int_C (x - y + z) \, dz = -0.8333\)

Would you like more details or have any questions?

Here are some questions you might ask next:
1. How do you parameterize a curve in three dimensions?
2. What is the significance of each differential \(dx\), \(dy\), and \(dz\) in line integrals?
3. Can you explain the physical meaning of line integrals in vector fields?
4. How do you evaluate line integrals for more complex curves?
5. What are some practical applications of line integrals in engineering?

**Tip**: When performing line integrals, carefully parameterize the integrand and differentials to ensure accurate calculations.

Learn how to evaluate line integrals along a parametrized curve using a step-by-step approach. This problem involves vector calculus and line integrals in three dimensions. Suitable for advanced college-level students and those interested in vector calculus.

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