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integration from 0 to pi |1/2 + cos x | dx can someone please help me with this

shiya , 5 Years ago
Grade 12th pass
anser 1 Answers
Samyak Jain

Last Activity: 5 Years ago

Examine the graph of 1/2 + cosx from x=0 to \pi, you will find that it is negative for x \small \epsilon (2\small \pi/3 , \small \pi]
\small \therefore \small \int_{0}^{\pi}|1/2 + cosx| = \small \int_{0}^{2\pi /3}(1/2 + cosx)  –  \small \int_{2\pi /3}^{\pi}  (1/2 + cosx)
 
                              = \small \[x/2 + sinx]_{0}^{2\pi /3}  –  \small \[x/2 + sinx]_{2\pi /3}^{\pi}
                              = [(\small \pi/3 + sin2\small \pi/3) – (0 + 0)] – [(\small \pi/2 + 0) – (\small \pi/3 + sin2\small \pi/3)]
                              = (\small \pi/3 + \small \sqrt{3} / 2) – \small \pi/2 + (\small \pi/3 + \small \sqrt{3} / 2)   = 2\small \pi/3 – \small \pi/2 + \small \sqrt{3} 
                              =  \small \pi/6 + \small \sqrt{3}

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