Evaluate the integral. \( \int_{0}^{4}\left(x^{2}-5\right) d x=\square \) (Simplify your answer.)

To evaluate the integral \( \int_{0}^{4} \left(x^{2} - 5\right) dx \), we can start by finding the antiderivative of the integrand. 1. Find the antiderivative: \[ \int (x^{2} - 5) \, dx = \frac{x^{3}}{3} - 5x + C \] 2. Evaluate the definite integral from 0 to 4: \[ \left[ \frac{x^{3}}{3} - 5x \right]_{0}^{4} \] 3. Calculate at the upper limit (when \( x = 4 \)): \[ \frac{4^{3}}{3} - 5(4) = \frac{64}{3} - 20 \] 4. Convert 20 to a fraction to combine: \[ 20 = \frac{60}{3} \] \[ \frac{64}{3} - \frac{60}{3} = \frac{4}{3} \] 5. Now calculate at the lower limit (when \( x = 0 \)): \[ \left( \frac{0^{3}}{3} - 5(0) \right) = 0 \] 6. Subtract the lower limit evaluation from the upper limit evaluation: \[ \frac{4}{3} - 0 = \frac{4}{3} \] Thus, the value of the integral is: \[ \int_{0}^{4} \left(x^{2} - 5\right) dx = \frac{4}{3} \] Final answer: \( \frac{4}{3} \)