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.