Math  /  Calculus

QuestionWrite the integral as a sum of integrals without absolute values and evaluate: π/4πcosxdx=\int_{\pi / 4}^{\pi}|\cos x| d x= \square roblem 2. (1 point)

Studdy Solution

STEP 1

1. The function cosx |\cos x| is piecewise and changes sign over the interval [π/4,π][\pi/4, \pi].
2. We need to find the points where cosx\cos x changes sign within the given interval.
3. The integral will be split into intervals where cosx\cos x is either non-negative or non-positive.

STEP 2

1. Determine the intervals where cosx\cos x is non-negative and non-positive.
2. Split the integral into separate integrals over these intervals.
3. Evaluate each integral separately.
4. Sum the results of the integrals.

STEP 3

Determine where cosx\cos x changes sign in the interval [π/4,π][\pi/4, \pi]. The cosine function is non-negative in the first and fourth quadrants and non-positive in the second and third quadrants.
Within [π/4,π][\pi/4, \pi], cosx\cos x changes sign at x=π/2x = \pi/2.

STEP 4

Split the integral π/4πcosxdx\int_{\pi/4}^{\pi} |\cos x| \, dx into two parts based on the sign change:
1. π/4π/2cosxdx\int_{\pi/4}^{\pi/2} \cos x \, dx where cosx\cos x is non-negative.
2. π/2πcosxdx\int_{\pi/2}^{\pi} -\cos x \, dx where cosx\cos x is non-positive.

STEP 5

Evaluate the first integral π/4π/2cosxdx\int_{\pi/4}^{\pi/2} \cos x \, dx:
π/4π/2cosxdx=[sinx]π/4π/2=sin(π/2)sin(π/4)=122\int_{\pi/4}^{\pi/2} \cos x \, dx = [\sin x]_{\pi/4}^{\pi/2} = \sin(\pi/2) - \sin(\pi/4) = 1 - \frac{\sqrt{2}}{2}

STEP 6

Evaluate the second integral π/2πcosxdx\int_{\pi/2}^{\pi} -\cos x \, dx:
π/2πcosxdx=[sinx]π/2π=(sin(π)sin(π/2))=(01)=1\int_{\pi/2}^{\pi} -\cos x \, dx = -[\sin x]_{\pi/2}^{\pi} = -(\sin(\pi) - \sin(\pi/2)) = -(0 - 1) = 1

STEP 7

Sum the results of the integrals:
(122)+1=222\left(1 - \frac{\sqrt{2}}{2}\right) + 1 = 2 - \frac{\sqrt{2}}{2}
The value of the integral is:
222\boxed{2 - \frac{\sqrt{2}}{2}}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord