evaluate the following integral using the fundamental theorem of calculus.\n int_{-9pi/2}^{9pi/2} (cos x…

evaluate the following integral using the fundamental theorem of calculus.\n int_{-9pi/2}^{9pi/2} (cos x - 2) dx \n int_{-9pi/2}^{9pi/2} (cos x - 2) dx=square \n(type an exact answer.)

evaluate the following integral using the fundamental theorem of calculus.\n int_{-9pi/2}^{9pi/2} (cos x - 2) dx \n int_{-9pi/2}^{9pi/2} (cos x - 2) dx=square \n(type an exact answer.)

Answer

Explanation:

Step1: Find the antiderivative

The antiderivative of $\cos x$ is $\sin x$ and the antiderivative of $- 2$ is $-2x$. So the antiderivative of $\cos x - 2$ is $F(x)=\sin x-2x$.

Step2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that $\int_{a}^{b}f(x)dx=F(b)-F(a)$. Here $a =-\frac{9\pi}{2}$, $b=\frac{9\pi}{2}$ and $F(x)=\sin x - 2x$. [ \begin{align*} F\left(\frac{9\pi}{2}\right)-F\left(-\frac{9\pi}{2}\right)&=\left(\sin\frac{9\pi}{2}-2\times\frac{9\pi}{2}\right)-\left(\sin\left(-\frac{9\pi}{2}\right)-2\times\left(-\frac{9\pi}{2}\right)\right)\ &=(1 - 9\pi)-(- 1+9\pi)\ &=1 - 9\pi + 1-9\pi\ &=2 - 18\pi \end{align*} ]

Answer:

$2 - 18\pi$