Math / Calculus

QuestionEvaluate the following integral using the Fundámental Theorem of Calculus. $\int_{-5 \pi / 2}^{5 \pi / 2}(\cos x-4) d x$ $\int_{-5 \pi / 2}^{5 \pi / 2}(\cos x-4) d x=\square$ $\square$ (Type an exact answer.)

Studdy Solution

STEP 1

1. The integral to evaluate is $\int_{-5 \pi / 2}^{5 \pi / 2}(\cos x - 4) \, dx$ .
2. The Fundamental Theorem of Calculus will be used to evaluate the definite integral.
3. The function $\cos x - 4$ is continuous over the interval $[-5\pi/2, 5\pi/2]$ .

STEP 2

1. Find the antiderivative of the integrand $\cos x - 4$ .
2. Apply the Fundamental Theorem of Calculus to evaluate the definite integral.
3. Simplify the result to find the exact answer.

STEP 3

Find the antiderivative of the integrand $\cos x - 4$ .
The antiderivative of $\cos x$ is $\sin x$ , and the antiderivative of a constant $c$ is $cx$ . Therefore, the antiderivative of $\cos x - 4$ is:
$F(x) = \sin x - 4x$

STEP 4

Apply the Fundamental Theorem of Calculus, which states that if $F(x)$ is an antiderivative of $f(x)$ , then:
$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$
Substitute $a = -5\pi/2$ and $b = 5\pi/2$ into the equation:
$\int_{-5\pi/2}^{5\pi/2} (\cos x - 4) \, dx = F(5\pi/2) - F(-5\pi/2)$

STEP 5

Evaluate $F(x)$ at the upper and lower limits:
$F(5\pi/2) = \sin(5\pi/2) - 4(5\pi/2)$ $F(-5\pi/2) = \sin(-5\pi/2) - 4(-5\pi/2)$
Since $\sin(5\pi/2) = \sin(\pi/2) = 1$ and $\sin(-5\pi/2) = \sin(-\pi/2) = -1$ , we have:
$F(5\pi/2) = 1 - 10\pi$ $F(-5\pi/2) = -1 + 10\pi$

STEP 6

Calculate the difference $F(5\pi/2) - F(-5\pi/2)$ :
$F(5\pi/2) - F(-5\pi/2) = (1 - 10\pi) - (-1 + 10\pi)$ $= 1 - 10\pi + 1 - 10\pi$ $= 2 - 20\pi$
The exact value of the integral is:
$\boxed{2 - 20\pi}$

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