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

∫_0 ^(100π) ∣sinx∣ dx

0100πsinxdx

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

thank you sir

thankyousir

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

letI=0100πsinxdxI=k=099kπ(k+1)πsinxdx=x=kπ+uk=0990π(1)ksinudu=k=0990πsinudu=k=099[cosu]0π=2k=099(1)=2×100=200

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

thank you sir

thankyousir

Commented by Bird last updated on 09/Oct/20

you are welcome sir

youarewelcomesir

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

sinxgraphisalaways+veandeachloopofsinxareais2.here100loopsoareais2×100=2000100πsinxdx=1000πsinxdx=100cosx0π=100(11)=200

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