Evaluate the following integral using the Fundamental Theorem of Calculus. \( \int_{-9 \pi / 2}^{9 \pi / 2}(\cos x-2) \mathrm{dx} \) \( \int_{-9 \pi / 2}^{9 \pi / 2}(\cos x-2) \mathrm{dx}=\square \) \( ( \) Type an exact answer.)

To evaluate the integral \( \int_{-9 \pi / 2}^{9 \pi / 2}(\cos x-2) \mathrm{dx} \), we can use the Fundamental Theorem of Calculus. First, we will find the antiderivative of \( \cos x - 2 \): \[ \int (\cos x - 2) \, dx = \sin x - 2x + C \] Now we will evaluate the definite integral from \( -9 \pi / 2 \) to \( 9 \pi / 2 \): \[ \left[ \sin x - 2x \right]_{-9 \pi / 2}^{9 \pi / 2} \] Calculating the upper limit: \[ \sin(9 \pi / 2) - 2(9 \pi / 2) = \sin(4\pi + \pi/2) - 9\pi = 1 - 9\pi \] Calculating the lower limit: \[ \sin(-9 \pi / 2) - 2(-9 \pi / 2) = \sin(-4\pi - \pi/2) + 9\pi = -1 + 9\pi \] Now, putting it all together: \[ \left[ \sin(9 \pi / 2) - 2(9 \pi / 2) \right] - \left[ \sin(-9 \pi / 2) - 2(-9 \pi / 2) \right] = (1 - 9\pi) - (-1 + 9\pi) \] This simplifies to: \[ 1 - 9\pi + 1 - 9\pi = 2 - 18\pi \] Thus, the value of the integral is: \[ \int_{-9 \pi / 2}^{9 \pi / 2}(\cos x-2) \mathrm{dx} = 2 - 18\pi \] So the final answer is: \[ \boxed{2 - 18\pi} \]