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Question Number 42823 by rahul 19 last updated on 03/Sep/18
Evaluate : ∫_(−5) ^( 5) x^2 [x+(1/2)]dx = ? where [.]= greatest integer function
Evaluate:∫−55x2[x+12]dx=?where[.]=greatestintegerfunction
Commented by prof Abdo imad last updated on 03/Sep/18
changement x =−5+t give I = ∫_0 ^(10) (−5+t)^2 [−5+t +(1/2)]dt =∫_0 ^(10) (t^2 −10t +25){−5 +[t+(1/2)]}dt =−5 ∫_0 ^(10) (t^2 −10t +25)dt +∫_0 ^(10) (t^2 −10t +25)[t+(1/2)]dt but ∫_0 ^(10) (t^2 −10t +25)dt =[(t^3 /3) −5t^2 +25t]_0 ^(10) =((10^3 )/3) −500 +250 =((1000)/3) −250 =((1000−750)/3) =((250)/3) also we have ∫_0 ^(10) (t^2 −10t +25)[t+(1/2)]dt =Σ_(k=0) ^9 ∫_k ^(k+1) (t^2 −10t +25)[t+(1/2)]dt but we know that [x+y]=[x]+[y]+ξ withξ=0or ξ=1 ∫_0 ^(10) (t^2 −10t +25)[t+(1/2)]dt= =Σ_(k=0) ^9 ∫_k ^(k+1) (t^2 −10t +25)(k +ξ)dt =Σ_(k=0) ^9 (k+ξ) [(t^3 /3) −5t^2 +25t]_k ^(k+1) =Σ_(k=0) ^9 (k+ξ){ (((k+1)^3 )/3) −5(k+1)^2 +25(k+1) −(k^3 /3) +5k^2 −25k} =.....
changementx=−5+tgiveI=∫010(−5+t)2[−5+t+12]dt=∫010(t2−10t+25){−5+[t+12]}dt=−5∫010(t2−10t+25)dt+∫010(t2−10t+25)[t+12]dtbut∫010(t2−10t+25)dt=[t33−5t2+25t]010=1033−500+250=10003−250=1000−7503=2503alsowehave∫010(t2−10t+25)[t+12]dt=∑k=09∫kk+1(t2−10t+25)[t+12]dtbutweknowthat[x+y]=[x]+[y]+ξwithξ=0orξ=1∫010(t2−10t+25)[t+12]dt==∑k=09∫kk+1(t2−10t+25)(k+ξ)dt=∑k=09(k+ξ)[t33−5t2+25t]kk+1=∑k=09(k+ξ){(k+1)33−5(k+1)2+25(k+1)−k33+5k2−25k}=…..
Answered by MJS last updated on 03/Sep/18
in steps: [x+(1/2)]=c x∈[−5; −4.5[ ⇒ c=−5 x∈[−4.5; −3.5[ ⇒ c=−4 x∈ [−3.5; −2.5[ ⇒ c=−3 x∈ [−2.5; −1.5[ ⇒ c=−2 x∈ [−1.5; −.5[ ⇒ c=−1 x∈ [−.5; .5[ ⇒ c=0 x∈ [.5; 1.5[ ⇒ c=1 x∈ [1.5; 2.5[ ⇒ c=2 x∈ [2.5; 3.5[ ⇒ c=3 x∈ [3.5; 4.5[ ⇒ c=4 x∈ [4.5; 5] ⇒ c=5 ∫_(−5) ^5 x^2 [x+(1/2)]dx=Σc∫x^2 dx=Σ(c/3)[x^3 ]_a ^b = =−(5/3)((−4.5)^3 −(−5)^3 )−(4/3)((−3.5)^3 −(−4.5)^3 )−(3/3)((−2.5)^3 −(−3.5)^3 )... ...+(3/3)(3.5^3 −2.5^3 )+(4/3)(4.5^3 −3.5^3 )+(5/3)(5^3 −4.5^3 )= =−(5/3)(5^3 −4.5^3 )+(5/3)(5^3 −4.5^3 )...=0
insteps:[x+12]=cx∈[−5;−4.5[⇒c=−5x∈[−4.5;−3.5[⇒c=−4x∈[−3.5;−2.5[⇒c=−3x∈[−2.5;−1.5[⇒c=−2x∈[−1.5;−.5[⇒c=−1x∈[−.5;.5[⇒c=0x∈[.5;1.5[⇒c=1x∈[1.5;2.5[⇒c=2x∈[2.5;3.5[⇒c=3x∈[3.5;4.5[⇒c=4x∈[4.5;5]⇒c=5∫5−5x2[x+12]dx=Σc∫x2dx=Σc3[x3]ab==−53((−4.5)3−(−5)3)−43((−3.5)3−(−4.5)3)−33((−2.5)3−(−3.5)3)……+33(3.53−2.53)+43(4.53−3.53)+53(53−4.53)==−53(53−4.53)+53(53−4.53)…=0
Commented by rahul 19 last updated on 03/Sep/18
Thank you sir ! ��

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