Question Number 56107 by Joel578 last updated on 10/Mar/19
$$\int_{−\mathrm{1}} ^{\mathrm{0}} \:\mid{x}\:\mathrm{sin}\:\left(\pi{x}\right)\mid\:{dx} \\ $$
Commented by maxmathsup by imad last updated on 10/Mar/19
![let I =∫_(−1) ^0 ∣xsin(πx)∣dx changement πx =−t give I =∫_π ^0 ∣((−t)/π)sin(−t)∣((−dt)/π) =(1/π^2 ) ∫_0 ^π t sint dt by parts π^2 I =[−tcost]_0 ^π +∫_0 ^π cost dt =π +[sint]_0 ^π =π ⇒I =(1/π)](https://www.tinkutara.com/question/Q56115.png)
$${let}\:{I}\:=\int_{−\mathrm{1}} ^{\mathrm{0}} \mid{xsin}\left(\pi{x}\right)\mid{dx}\:\:{changement}\:\pi{x}\:=−{t}\:{give}\: \\ $$$${I}\:=\int_{\pi} ^{\mathrm{0}} \mid\frac{−{t}}{\pi}{sin}\left(−{t}\right)\mid\frac{−{dt}}{\pi}\:=\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\pi} {t}\:{sint}\:{dt}\:\:{by}\:{parts} \\ $$$$\pi^{\mathrm{2}} {I}\:=\left[−{tcost}\right]_{\mathrm{0}} ^{\pi} \:+\int_{\mathrm{0}} ^{\pi} \:{cost}\:{dt}\:=\pi\:\:+\left[{sint}\right]_{\mathrm{0}} ^{\pi} \:=\pi\:\Rightarrow{I}\:=\frac{\mathrm{1}}{\pi} \\ $$
Answered by mr W last updated on 10/Mar/19
![I=∫x sin (xπ) dx =−(1/π)∫x d(cos πx) =−(1/π)[x cos (πx)−∫cos (πx)dx] =−(1/π)[x cos (πx)−(1/π)∫cos (πx)d(πx)] =((sin (πx))/π^2 )−((x cos (πx))/π)+C ∫_(−1) ^0 ∣x sin (πx)∣ dx =∫_0 ^1 x sin (πx)dx =[((sin (πx))/π^2 )−((x cos (πx))/π)]_0 ^1 =(1/π)](https://www.tinkutara.com/question/Q56109.png)
$${I}=\int{x}\:\mathrm{sin}\:\left({x}\pi\right)\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\pi}\int{x}\:{d}\left(\mathrm{cos}\:\pi{x}\right) \\ $$$$=−\frac{\mathrm{1}}{\pi}\left[{x}\:\mathrm{cos}\:\left(\pi{x}\right)−\int\mathrm{cos}\:\left(\pi{x}\right){dx}\right] \\ $$$$=−\frac{\mathrm{1}}{\pi}\left[{x}\:\mathrm{cos}\:\left(\pi{x}\right)−\frac{\mathrm{1}}{\pi}\int\mathrm{cos}\:\left(\pi{x}\right){d}\left(\pi{x}\right)\right] \\ $$$$=\frac{\mathrm{sin}\:\left(\pi{x}\right)}{\pi^{\mathrm{2}} }−\frac{{x}\:\mathrm{cos}\:\left(\pi{x}\right)}{\pi}+{C} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{0}} \:\mid{x}\:\mathrm{sin}\:\left(\pi{x}\right)\mid\:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\:\mathrm{sin}\:\left(\pi{x}\right){dx} \\ $$$$=\left[\frac{\mathrm{sin}\:\left(\pi{x}\right)}{\pi^{\mathrm{2}} }−\frac{{x}\:\mathrm{cos}\:\left(\pi{x}\right)}{\pi}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\pi} \\ $$
Answered by MJS last updated on 10/Mar/19
$$\int\mid{x}\mathrm{sin}\:\pi{x}\mid{dx}=\mathrm{sign}\left({x}\mathrm{sin}\:\pi{x}\right)\int{x}\mathrm{sin}\:\pi{x}\:{dx}= \\ $$$$=\mathrm{sign}\left({x}\mathrm{sin}\:\pi{x}\right)\left(\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\mathrm{sin}\:\pi{x}\:−\frac{\mathrm{1}}{\pi}{x}\mathrm{cos}\:\pi{x}\right) \\ $$$$\underset{−\mathrm{1}} {\overset{\mathrm{0}} {\int}}\mid{x}\mathrm{sin}\:\pi{x}\mid{dx}=\frac{\mathrm{1}}{\pi} \\ $$