Integrate (4/3)x^3 - (3/4)x^2 - 5 from -1 to 1.

The integral is \(\int_{-1}^{1} \left(\frac{4}{3}x^3 - \frac{3}{4}x^2 - 5\right)dx\).

Integrating term by term gives \(\left[\frac{4}{3}\frac{x^4}{4} - \frac{3}{4}\frac{x^3}{3} - 5x\right]_{-1}^{1} = \left[\frac{x^4}{3} - \frac{x^3}{4} - 5x\right]_{-1}^{1}\).

Evaluating at the limits: \(\left(\frac{1^4}{3} - \frac{1^3}{4} - 5(1)\right) - \left(\frac{(-1)^4}{3} - \frac{(-1)^3}{4} - 5(-1)\right) = \left(\frac{1}{3} - \frac{1}{4} - 5\right) - \left(\frac{1}{3} + \frac{1}{4} + 5\right) = \frac{1}{3} - \frac{1}{4} - 5 - \frac{1}{3} - \frac{1}{4} - 5 = -2\left(\frac{1}{4}\right) - 10 = -\frac{1}{2} - 10 = -10.5\).