𝑦 = −(𝑥 + 1)(𝑥 − 1) find definite integral of y with the lower limit being -1 and upper limit 1

To find the definite integral of \( y = -(x+1)(x-1) \) with the limits \(-1\) to \(1\), we will follow these steps:

1. Simplify the integrand.
2. Integrate the simplified function.
3. Evaluate the integral at the upper and lower limits and find the difference.

### Step 1: Simplify the integrand

First, expand the expression:
\[ y = -(x+1)(x-1) \]
\[ y = -(x^2 - 1) \]
\[ y = -x^2 + 1 \]

So, the function we need to integrate is \( y = -x^2 + 1 \).

### Step 2: Integrate the function

Now, integrate \( -x^2 + 1 \) with respect to \( x \):

\[ \int (-x^2 + 1) \, dx \]

We can integrate term by term:

\[ \int -x^2 \, dx + \int 1 \, dx \]

The antiderivative of \( -x^2 \) is \( -\frac{x^3}{3} \) and the antiderivative of \( 1 \) is \( x \):

\[ \int (-x^2 + 1) \, dx = -\frac{x^3}{3} + x \]

### Step 3: Evaluate the integral at the upper and lower limits

Evaluate the antiderivative from \(-1\) to \(1\):

\[ \left[ -\frac{x^3}{3} + x \right]_{-1}^{1} \]

First, evaluate at the upper limit \( x = 1 \):

\[ \left( -\frac{(1)^3}{3} + (1) \right) = -\frac{1}{3} + 1 = \frac{2}{3} \]

Then, evaluate at the lower limit \( x = -1 \):

\[ \left( -\frac{(-1)^3}{3} + (-1) \right) = -\left( -\frac{1}{3} \right) - 1 = \frac{1}{3} - 1 = -\frac{2}{3} \]

Finally, find the difference between the upper and lower limit evaluations:

\[ \frac{2}{3} - \left( -\frac{2}{3} \right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]

### Conclusion

The definite integral of \( y = -(x+1)(x-1) \) from \(-1\) to \(1\) is:

\[ \int_{-1}^{1} (-(x+1)(x-1)) \, dx = \frac{4}{3} \]

Would you like more details or have any questions about this solution?

Here are 8 related questions for further practice:
1. Find the definite integral of \( y = x^2 - 1 \) from \(-2\) to \(2\).
2. Evaluate \( \int_{0}^{2} (x^2 - 4) \, dx \).
3. Compute the definite integral of \( y = 2x - 3 \) from \(-1\) to \(1\).
4. Determine the area under the curve \( y = x^3 - 4x \) between \( x = -1 \) and \( x = 1 \).
5. Find \( \int_{-1}^{2} (3x^2 - 2x + 1) \, dx \).
6. Evaluate \( \int_{0}^{1} (x^2 + 2x + 1) \, dx \).
7. Compute the definite integral of \( y = \cos(x) \) from \(0\) to \( \pi \).
8. Determine \( \int_{-2}^{2} (4 - x^2) \, dx \).

**Tip:** When integrating polynomials, remember to increase the power by one and divide by the new power, then add the constant of integration if needed for indefinite integrals.

Learn how to find the definite integral of -(x+1)(x-1) from -1 to 1 with a step-by-step solution. This problem involves polynomial integration and the application of the Fundamental Theorem of Calculus. Suitable for college-level students.

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