1. **State the problem:** Calculate the definite integral from 1 to -1 of the function $f(y) = (y^2 - 1) - (1 - y^2)$. 2. **Simplify the integrand:** $$ (y^2 - 1) - (1 - y^2) = y^2 - 1 - 1 + y^2 = 2y^2 - 2 $$ 3. **Rewrite the integral:** $$ \int_1^{-1} (2y^2 - 2) \, dy $$ 4. **Note the bounds are reversed:** The integral from 1 to -1 is the negative of the integral from -1 to 1, so $$ \int_1^{-1} (2y^2 - 2) \, dy = - \int_{-1}^1 (2y^2 - 2) \, dy $$ 5. **Integrate the function:** Use the power rule for integration: $$ \int (2y^2 - 2) \, dy = 2 \int y^2 \, dy - 2 \int 1 \, dy = 2 \cdot \frac{y^3}{3} - 2y = \frac{2y^3}{3} - 2y $$ 6. **Evaluate the definite integral from -1 to 1:** $$ \left[ \frac{2y^3}{3} - 2y \right]_{-1}^1 = \left( \frac{2(1)^3}{3} - 2(1) \right) - \left( \frac{2(-1)^3}{3} - 2(-1) \right) = \left( \frac{2}{3} - 2 \right) - \left( -\frac{2}{3} + 2 \right) $$ $$ = \left( \frac{2}{3} - 2 \right) - \left( -\frac{2}{3} + 2 \right) = \frac{2}{3} - 2 + \frac{2}{3} - 2 = \frac{4}{3} - 4 = -\frac{8}{3} $$ 7. **Apply the negative sign from step 4:** $$ - \int_{-1}^1 (2y^2 - 2) \, dy = - \left(-\frac{8}{3} \right) = \frac{8}{3} $$ **Final answer:** $$ \int_1^{-1} ((y^2 - 1) - (1 - y^2)) \, dy = \frac{8}{3} $$