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Question Number 117006 by TANMAY PANACEA last updated on 08/Oct/20

∫_0 ^(100π) ∣sinx∣ dx

$$\int_{\mathrm{0}} ^{\mathrm{100}\pi} \mid{sinx}\mid\:{dx} \\ $$

Commented by TANMAY PANACEA last updated on 08/Oct/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 08/Oct/20

let  I =∫_0 ^(100π) ∣sinx∣dx  ⇒I =Σ_(k=0) ^(99)  ∫_(kπ) ^((k+1)π)   ∣sinx∣dx  =_(x=kπ +u)   Σ_(k=0) ^(99)   ∫_0 ^π ∣(−1)^k  sinu∣ du =Σ_(k=0) ^(99)  ∫_0 ^π sinu du  =Σ_(k=0) ^(99) [−cosu]_0 ^π   =2 Σ_(k=0) ^(99) (1) =2×100 =200

$$\mathrm{let}\:\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{100}\pi} \mid\mathrm{sinx}\mid\mathrm{dx}\:\:\Rightarrow\mathrm{I}\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{99}} \:\int_{\mathrm{k}\pi} ^{\left(\mathrm{k}+\mathrm{1}\right)\pi} \:\:\mid\mathrm{sinx}\mid\mathrm{dx} \\ $$$$=_{\mathrm{x}=\mathrm{k}\pi\:+\mathrm{u}} \:\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{99}} \:\:\int_{\mathrm{0}} ^{\pi} \mid\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{sinu}\mid\:\mathrm{du}\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{99}} \:\int_{\mathrm{0}} ^{\pi} \mathrm{sinu}\:\mathrm{du} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{99}} \left[−\mathrm{cosu}\right]_{\mathrm{0}} ^{\pi} \:\:=\mathrm{2}\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{99}} \left(\mathrm{1}\right)\:=\mathrm{2}×\mathrm{100}\:=\mathrm{200} \\ $$

Commented by TANMAY PANACEA last updated on 08/Oct/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by Bird last updated on 09/Oct/20

you are welcome sir

$${you}\:{are}\:{welcome}\:{sir} \\ $$

Answered by TANMAY PANACEA last updated on 08/Oct/20

∣sinx∣ graph is alaways +ve and each loop of   sinx area is 2. here 100 loop so area is 2×100=200  ∫_0 ^(100π) ∣sinx∣dx=100∫_0 ^π ∣sinx∣dx=100∣−cosx∣_0 ^π   =−100(−1−1)=200

$$\mid{sinx}\mid\:{graph}\:{is}\:{alaways}\:+{ve}\:{and}\:{each}\:{loop}\:{of}\: \\ $$$${sinx}\:{area}\:{is}\:\mathrm{2}.\:{here}\:\mathrm{100}\:{loop}\:{so}\:{area}\:{is}\:\mathrm{2}×\mathrm{100}=\mathrm{200} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{100}\pi} \mid{sinx}\mid{dx}=\mathrm{100}\int_{\mathrm{0}} ^{\pi} \mid{sinx}\mid{dx}=\mathrm{100}\mid−{cosx}\mid_{\mathrm{0}} ^{\pi} \\ $$$$=−\mathrm{100}\left(−\mathrm{1}−\mathrm{1}\right)=\mathrm{200} \\ $$$$ \\ $$

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