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Question Number 91217 by mhmd last updated on 28/Apr/20

Commented by Prithwish Sen 1 last updated on 28/Apr/20

∫_0 ^(nπ) ∣cosx∣dx = n∫_0 ^𝛑 ∣cosx∣dx = n[∫_0 ^(𝛑/2) cosxdx+∫_(𝛑/2) ^𝛑 (−cosx)dx] = 2n n=posituve integer. please check

$$\int_{\mathrm{0}} ^{\mathrm{n}\pi} \mid\mathrm{cosx}\mid\mathrm{dx}\:=\:\boldsymbol{\mathrm{n}}\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \mid\mathrm{cosx}\mid\mathrm{dx}\:=\:\mathrm{n}\left[\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{cosxdx}}+\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\boldsymbol{\pi}} \left(−\mathrm{cosx}\right)\mathrm{dx}\right] \\ $$$$=\:\mathrm{2}\boldsymbol{\mathrm{n}}\:\:\:\:\:\:\boldsymbol{\mathrm{n}}=\boldsymbol{\mathrm{posituve}}\:\boldsymbol{\mathrm{integer}}.\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$

Commented by MJS last updated on 28/Apr/20

2n ∫_0 ^(π/2) ∣cos x∣dx=∫_0 ^(π/2) cos x dx=1 ⇒ the area beneath ∣cos x∣ equals 1 for each interval [((kπ)/2); (((k+1)π)/2)]; k∈Z ⇒ the area in each interval [kπ; (k+1)π]; k∈Z equals 2. we have n intervals ⇒ answer is 2n

$$\mathrm{2}{n} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mid\mathrm{cos}\:{x}\mid{dx}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{cos}\:{x}\:{dx}=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{area}\:\mathrm{beneath}\:\mid\mathrm{cos}\:{x}\mid\:\mathrm{equals}\:\mathrm{1}\:\mathrm{for}\:\mathrm{each} \\ $$$$\mathrm{interval}\:\left[\frac{{k}\pi}{\mathrm{2}};\:\frac{\left({k}+\mathrm{1}\right)\pi}{\mathrm{2}}\right];\:{k}\in\mathbb{Z}\:\Rightarrow\:\mathrm{the}\:\mathrm{area}\:\mathrm{in}\:\mathrm{each} \\ $$$$\mathrm{interval}\:\left[{k}\pi;\:\left({k}+\mathrm{1}\right)\pi\right];\:{k}\in\mathbb{Z}\:\mathrm{equals}\:\mathrm{2}.\:\mathrm{we}\:\mathrm{have} \\ $$$${n}\:\mathrm{intervals}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2}{n} \\ $$

Commented by abdomathmax last updated on 28/Apr/20

∫_0 ^(nπ) ∣codx∣dx =Σ_(k=0) ^(n−1) ∫_(kπ) ^((k+1)π) ∣cosx∣dx (ch x=kπ +t =Σ_(k=0) ^(n−1) ∫_0 ^π ∣(−1)^k cost∣dt =Σ_(k=0) ^(n−1) ∫_0 ^π ∣cost∣dt but ∫_0 ^π ∣cost∣dt =∫_0 ^(π/2) cost dt −∫_(π/2) ^π cost dt =1−(−1) =2 ⇒∫_0 ^(nπ) ∣cosx∣dx =2Σ_(k=0) ^(n−1) (1) =2n

$$\int_{\mathrm{0}} ^{{n}\pi} \mid{codx}\mid{dx}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}\pi} ^{\left({k}+\mathrm{1}\right)\pi} \:\:\mid{cosx}\mid{dx}\:\:\left({ch}\:\:{x}={k}\pi\:+{t}\right. \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\int_{\mathrm{0}} ^{\pi} \:\mid\left(−\mathrm{1}\right)^{{k}} \:{cost}\mid{dt}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\pi} \mid{cost}\mid{dt} \\ $$$${but}\:\int_{\mathrm{0}} ^{\pi} \:\mid{cost}\mid{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cost}\:{dt}\:−\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:{cost}\:{dt} \\ $$$$=\mathrm{1}−\left(−\mathrm{1}\right)\:=\mathrm{2}\:\Rightarrow\int_{\mathrm{0}} ^{{n}\pi} \:\mid{cosx}\mid{dx}\:=\mathrm{2}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$=\mathrm{2}{n} \\ $$$$ \\ $$

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