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Find-0-2-1-x-dx-




Question Number 208409 by hardmath last updated on 15/Jun/24
Find:   ∫_0 ^( 2)  ∣1 − x∣ dx = ?
$$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\mid\mathrm{1}\:−\:\mathrm{x}\mid\:\mathrm{dx}\:=\:? \\ $$
Answered by mr W last updated on 15/Jun/24
=2∫_1 ^2 (x−1)dx  =2∫_0 ^1 tdt  =1
$$=\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{2}} \left({x}−\mathrm{1}\right){dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {tdt} \\ $$$$=\mathrm{1} \\ $$
Commented by hardmath last updated on 15/Jun/24
dear professor,  ∫_0 ^( 1) (x−1)dx + ∫_1 ^( 2) (1−x)dx = ... 1 ?
$$\mathrm{dear}\:\mathrm{professor}, \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{x}−\mathrm{1}\right)\mathrm{dx}\:+\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}\:=\:…\:\mathrm{1}\:? \\ $$
Commented by mr W last updated on 15/Jun/24
yes!
$${yes}! \\ $$
Answered by lepuissantcedricjunior last updated on 15/Jun/24
pour x∈[0;1]  ∣1−x∣=1−x  pour x∈[1;2] ∣1−x∣=−(1−x)=x−1  alors   ∫_0 ^2 ∣1−x∣dx=∫_0 ^1 (1−x)dx+∫_1 ^2 (x−1)dx                          =[x^1^2^2^2^2^2^2^2^2^(22222222222222)          −(1/2)x^2^1^1^1^1^(1....^1 )      ]_0 ^1 +[(1/2)x^2^1^3^(45^6^  )    −x^1^2^3^4^(45^4^4^   )     ]_1 ^2      =1−(1/2)+2−2−(1/2)+1=0      =1
$$\boldsymbol{{pour}}\:\boldsymbol{{x}}\in\left[\mathrm{0};\mathrm{1}\right]\:\:\mid\mathrm{1}−\boldsymbol{{x}}\mid=\mathrm{1}−\boldsymbol{{x}} \\ $$$$\boldsymbol{{pour}}\:\boldsymbol{{x}}\in\left[\mathrm{1};\mathrm{2}\right]\:\mid\mathrm{1}−\boldsymbol{{x}}\mid=−\left(\mathrm{1}−\boldsymbol{{x}}\right)=\boldsymbol{{x}}−\mathrm{1} \\ $$$$\boldsymbol{{alors}}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \mid\mathrm{1}−\boldsymbol{{x}}\mid\boldsymbol{{dx}}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\boldsymbol{{x}}\right)\boldsymbol{{dx}}+\int_{\mathrm{1}} ^{\mathrm{2}} \left(\boldsymbol{{x}}−\mathrm{1}\right)\boldsymbol{{dx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\boldsymbol{{x}}^{\mathrm{1}^{\mathrm{2}^{\mathrm{2}^{\mathrm{2}^{\mathrm{2}^{\mathrm{2}^{\mathrm{2}^{\mathrm{2}^{\mathrm{2}^{\mathrm{22222222222222}} } } } } } } } } } −\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}^{\mathrm{1}^{\mathrm{1}^{\mathrm{1}^{\mathrm{1}^{\mathrm{1}….^{\mathrm{1}} } } } } } } \right]_{\mathrm{0}} ^{\mathrm{1}} +\left[\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}^{\mathrm{1}^{\mathrm{3}^{\mathrm{45}^{\mathrm{6}^{} } } } } } −\boldsymbol{{x}}^{\mathrm{1}^{\mathrm{2}^{\mathrm{3}^{\mathrm{4}^{\mathrm{45}^{\mathrm{4}^{\mathrm{4}^{} } } } } } } } \right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\:\:\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}−\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:=\mathrm{1} \\ $$

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