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The integral provided is:

\[
\int_{-1}^{0} (t + 1) \sin(n\pi t) \, dt + \int_{0}^{1} (-t + 1) \sin(n\pi t) \, dt
\]

To solve this, let's address each integral separately and then sum the results.

### First Integral:

\[
\int_{-1}^{0} (t + 1) \sin(n\pi t) \, dt
\]

Let's apply integration by parts where we set:
- $ u = t + 1 $, hence $ du = dt $.
- $ dv = \sin(n\pi t) \, dt $, hence $ v = -\frac{1}{n\pi} \cos(n\pi t) $.

Using the integration by parts formula, $ \int u \, dv = uv - \int v \, du $:

\[
\int_{-1}^{0} (t + 1) \sin(n\pi t) \, dt = \left[ -(t + 1) \frac{1}{n\pi} \cos(n\pi t) \right]_{-1}^{0} + \frac{1}{n\pi} \int_{-1}^{0} \cos(n\pi t) \, dt
\]

Evaluate the boundary terms:

At $ t = 0 $: 
\[
-(0 + 1) \frac{1}{n\pi} \cos(0) = -\frac{1}{n\pi}
\]

At $ t = -1 $:
\[
-(-1 + 1) \frac{1}{n\pi} \cos(-n\pi) = 0
\]

Thus, the first term simplifies to:
\[
-\frac{1}{n\pi} + 0 = -\frac{1}{n\pi}
\]

The integral term is:
\[
\frac{1}{n\pi} \int_{-1}^{0} \cos(n\pi t) \, dt
\]

This integral evaluates to:
\[
\frac{1}{n\pi} \left[\frac{1}{n\pi} \sin(n\pi t) \right]_{-1}^{0}
\]

At $ t = 0 $:
\[
\frac{1}{n\pi} \times \frac{1}{n\pi} \times \sin(0) = 0
\]

At $ t = -1 $:
\[
\frac{1}{n\pi} \times \frac{1}{n\pi} \times \sin(-n\pi) = 0
\]

Thus, the second term contributes 0, and the first integral evaluates to:
\[
-\frac{1}{n\pi}
\]

### Second Integral:

\[
\int_{0}^{1} (-t + 1) \sin(n\pi t) \, dt
\]

Similarly, we use integration by parts:

Set:
- $ u = -t + 1 $, hence $ du = -dt $.
- $ dv = \sin(n\pi t) \, dt $, hence $ v = -\frac{1}{n\pi} \cos(n\pi t) $.

Thus,
\[
\int_{0}^{1} (-t + 1) \sin(n\pi t) \, dt = \left[ -(-t + 1) \frac{1}{n\pi} \cos(n\pi t) \right]_{0}^{1} + \frac{1}{n\pi} \int_{0}^{1} \cos(n\pi t) \, dt
\]

Evaluate the boundary terms:

At $ t = 1 $:
\[
-( -1 + 1 ) \frac{1}{n\pi} \cos(n\pi) = 0
\]

At $ t = 0 $:
\[
-( -0 + 1 ) \frac{1}{n\pi} \cos(0) = -\frac{1}{n\pi}
\]

So the first term simplifies to:
\[
0 - \left( -\frac{1}{n\pi} \right) = \frac{1}{n\pi}
\]

For the integral term:
\[
\frac{1}{n\pi} \left[\frac{1}{n\pi} \sin(n\pi t) \right]_{0}^{1}
\]

At $ t = 1 $:
\[
\frac{1}{n\pi} \times \frac{1}{n\pi} \times \sin(n\pi) = 0
\]

At $ t = 0 $:
\[
\frac{1}{n\pi} \times \frac{1}{n\pi} \times \sin(0) = 0
\]

Thus, the second integral evaluates to:
\[
\frac{1}{n\pi}
\]

### Summing Both Integrals:

Finally, adding the results of the two integrals:
\[
-\frac{1}{n\pi} + \frac{1}{n\pi} = 0
\]

### Conclusion:
The value of the given integral is **0**.

Do you want details on any particular step, or do you have any other questions?

Here are some related questions to explore:
1. How does integration by parts work, and why is it useful in solving integrals?
2. Can the sine function inside the integrals be replaced with another trigonometric function, and how would that affect the solution?
3. What happens if $n$ is not an integer? Would the result still be zero?
4. How would the integration change if the limits were different, such as from $-2$ to $2$?
5. How do you evaluate definite integrals involving complex exponentials instead of trigonometric functions?

**Tip:** When solving integrals, always consider symmetry, especially when dealing with trigonometric functions over symmetric intervals, as it can simplify the computation.

Learn how to solve the integral $\int_{-1}^{0} (t + 1) \sin(n\pi t) \, dt + \int_{0}^{1} (-t + 1) \sin(n\pi t) \, dt$ step-by-step with a detailed explanation. This problem involves integral calculus, integration by parts, and trigonometric integrals, suitable for advanced college-level students.

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