Question

evaluate the following integral using the fundamental theorem of calculus.
int_{-9pi/2}^{9pi/2} (cos x - 1) dx
int_{-9pi/2}^{9pi/2} (cos x - 1) dx=square
(type an exact answer.)

Explanation:

Step1: Find the antiderivative

The antiderivative of $\cos x$ is $\sin x$ and of $- 1$ is $-x$. So the antiderivative of $\cos x - 1$ is $F(x)=\sin x - x$.

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 $a =-\frac{9\pi}{2}$ and $b=\frac{9\pi}{2}$. So we calculate $F(\frac{9\pi}{2})-F(-\frac{9\pi}{2})$.
$F(\frac{9\pi}{2})=\sin(\frac{9\pi}{2})-\frac{9\pi}{2}=1 - \frac{9\pi}{2}$
$F(-\frac{9\pi}{2})=\sin(-\frac{9\pi}{2})+\frac{9\pi}{2}=-(- 1)+\frac{9\pi}{2}=1+\frac{9\pi}{2}$
Then $F(\frac{9\pi}{2})-F(-\frac{9\pi}{2})=(1-\frac{9\pi}{2})-(1 + \frac{9\pi}{2})$.

Step3: Simplify the result

$(1-\frac{9\pi}{2})-(1+\frac{9\pi}{2})=1-\frac{9\pi}{2}-1-\frac{9\pi}{2}=-9\pi$

Answer:

$-9\pi$