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Question Number 168677 by infinityaction last updated on 15/Apr/22

Commented by infinityaction last updated on 15/Apr/22

please evaluate this problem

$${please}\:{evaluate}\:{this}\:{problem} \\ $$

Answered by mnjuly1970 last updated on 16/Apr/22

Ω= ∫_0 ^( 2020) x.∣sin(πx)∣dx = 2020∫_0 ^( 2020) ∣sin(πx)∣dx−Ω 2Ω= 2020{∫_0 ^( 1) sin(πx) −∫_1 ^( 2) sin(πx)+∫_2 ^( 3) sin(πx)dx − ∫_3 ^( 4) sin(πx)dx+...−∫_(2019) ^( 2020) sin(πx)dx = (2020) (2020)∫_0 ^( 1) sin(πx)dx 2 Ω= 2020^( 2) ((2/π)) ∴ (π∫_0 ^( 2020) x.∣sin(πx)∣dx)^( (1/2)) = 2020

$$\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{2020}} {x}.\mid{sin}\left(\pi{x}\right)\mid{dx} \\ $$$$\:\:\:\:\:\:\:\:=\:\mathrm{2020}\int_{\mathrm{0}} ^{\:\mathrm{2020}} \mid{sin}\left(\pi{x}\right)\mid{dx}−\Omega \\ $$$$\:\:\:\:\:\mathrm{2}\Omega=\:\mathrm{2020}\left\{\int_{\mathrm{0}} ^{\:\mathrm{1}} {sin}\left(\pi{x}\right)\:\:−\int_{\mathrm{1}} ^{\:\mathrm{2}} {sin}\left(\pi{x}\right)+\int_{\mathrm{2}} ^{\:\mathrm{3}} {sin}\left(\pi{x}\right){dx}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\int_{\mathrm{3}} ^{\:\mathrm{4}} {sin}\left(\pi{x}\right){dx}+...−\int_{\mathrm{2019}} ^{\:\mathrm{2020}} {sin}\left(\pi{x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\:\left(\mathrm{2020}\right)\:\left(\mathrm{2020}\right)\int_{\mathrm{0}} ^{\:\mathrm{1}} {sin}\left(\pi{x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}\:\Omega=\:\mathrm{2020}^{\:\mathrm{2}} \:\left(\frac{\mathrm{2}}{\pi}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\left(\pi\int_{\mathrm{0}} ^{\:\mathrm{2020}} {x}.\mid{sin}\left(\pi{x}\right)\mid{dx}\right)^{\:\frac{\mathrm{1}}{\mathrm{2}}} =\:\mathrm{2020} \\ $$$$ \\ $$

Commented by infinityaction last updated on 16/Apr/22

thank you sir

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

Commented by mnjuly1970 last updated on 16/Apr/22

grateful....

$${grateful}.... \\ $$

Answered by mr W last updated on 16/Apr/22

for n=even: ∫_n ^(n+1) x sin πx dx =−(1/π)∫_n ^(n+1) x d(cos πx) =−(1/π){[x cos πx]_n ^(n+1) −∫_n ^(n+1) cos πx dx} =−(1/π){−(2n+1)−(1/π)[sin πx]_n ^(n+1) } =((2n+1)/π) π∫_n ^(n+1) x sin πx dx=2n+1 π∫_0 ^(2020) x ∣sin πx∣ dx=1+3+5+...+4039=2020^2 (√(π∫_0 ^(2020) x ∣sin πx∣ dx))=2020

$${for}\:{n}={even}: \\ $$$$\int_{{n}} ^{{n}+\mathrm{1}} {x}\:\mathrm{sin}\:\pi{x}\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\pi}\int_{{n}} ^{{n}+\mathrm{1}} {x}\:{d}\left(\mathrm{cos}\:\pi{x}\right) \\ $$$$=−\frac{\mathrm{1}}{\pi}\left\{\left[{x}\:\mathrm{cos}\:\pi{x}\right]_{{n}} ^{{n}+\mathrm{1}} −\int_{{n}} ^{{n}+\mathrm{1}} \:\mathrm{cos}\:\pi{x}\:{dx}\right\} \\ $$$$=−\frac{\mathrm{1}}{\pi}\left\{−\left(\mathrm{2}{n}+\mathrm{1}\right)−\frac{\mathrm{1}}{\pi}\left[\mathrm{sin}\:\pi{x}\right]_{{n}} ^{{n}+\mathrm{1}} \right\} \\ $$$$=\frac{\mathrm{2}{n}+\mathrm{1}}{\pi} \\ $$$$\pi\int_{{n}} ^{{n}+\mathrm{1}} {x}\:\mathrm{sin}\:\pi{x}\:{dx}=\mathrm{2}{n}+\mathrm{1} \\ $$$$\pi\int_{\mathrm{0}} ^{\mathrm{2020}} {x}\:\mid\mathrm{sin}\:\pi{x}\mid\:{dx}=\mathrm{1}+\mathrm{3}+\mathrm{5}+...+\mathrm{4039}=\mathrm{2020}^{\mathrm{2}} \\ $$$$\sqrt{\pi\int_{\mathrm{0}} ^{\mathrm{2020}} {x}\:\mid\mathrm{sin}\:\pi{x}\mid\:{dx}}=\mathrm{2020} \\ $$

Commented by infinityaction last updated on 16/Apr/22

but sir answer is 2020

$${but}\:{sir}\:{answer}\:{is}\:\mathrm{2020} \\ $$

Commented by mr W last updated on 16/Apr/22

yes. i′ve fixed.

$${yes}.\:{i}'{ve}\:{fixed}. \\ $$

Commented by infinityaction last updated on 16/Apr/22

thank you sir

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

Commented by amin96 last updated on 16/Apr/22

bravo

$${bravo} \\ $$$$ \\ $$

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