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](Q42823.png)
$$\mathrm{Evaluate}\:: \\ $$$$\int_{−\mathrm{5}} ^{\:\mathrm{5}} \:{x}^{\mathrm{2}} \left[{x}+\frac{\mathrm{1}}{\mathrm{2}}\right]{dx}\:=\:\:? \\ $$$${where}\:\left[.\right]=\:{greatest}\:{integer}\:{function} \\ $$
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} =.....](Q42848.png)
$${changement}\:{x}\:=−\mathrm{5}+{t}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{10}} \left(−\mathrm{5}+{t}\right)^{\mathrm{2}} \left[−\mathrm{5}+{t}\:+\frac{\mathrm{1}}{\mathrm{2}}\right]{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{10}} \left({t}^{\mathrm{2}} −\mathrm{10}{t}\:+\mathrm{25}\right)\left\{−\mathrm{5}\:+\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}\right]\right\}{dt} \\ $$$$=−\mathrm{5}\:\int_{\mathrm{0}} ^{\mathrm{10}} \:\left({t}^{\mathrm{2}} −\mathrm{10}{t}\:+\mathrm{25}\right){dt}\:+\int_{\mathrm{0}} ^{\mathrm{10}} \left({t}^{\mathrm{2}} −\mathrm{10}{t}\:+\mathrm{25}\right)\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}\right]{dt} \\ $$$${but}\:\int_{\mathrm{0}} ^{\mathrm{10}} \left({t}^{\mathrm{2}} −\mathrm{10}{t}\:+\mathrm{25}\right){dt}\:=\left[\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\:−\mathrm{5}{t}^{\mathrm{2}} \:+\mathrm{25}{t}\right]_{\mathrm{0}} ^{\mathrm{10}} \\ $$$$=\frac{\mathrm{10}^{\mathrm{3}} }{\mathrm{3}}\:−\mathrm{500}\:+\mathrm{250}\:=\frac{\mathrm{1000}}{\mathrm{3}}\:−\mathrm{250}\:=\frac{\mathrm{1000}−\mathrm{750}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{250}}{\mathrm{3}}\:\:\:{also}\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{10}} \left({t}^{\mathrm{2}} −\mathrm{10}{t}\:+\mathrm{25}\right)\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}\right]{dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{9}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \left({t}^{\mathrm{2}} −\mathrm{10}{t}\:+\mathrm{25}\right)\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}\right]{dt} \\ $$$${but}\:{we}\:{know}\:{that}\:\left[{x}+{y}\right]=\left[{x}\right]+\left[{y}\right]+\xi\:{with}\xi=\mathrm{0}{or}\:\xi=\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{10}} \left({t}^{\mathrm{2}} −\mathrm{10}{t}\:+\mathrm{25}\right)\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}\right]{dt}= \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{9}} \:\int_{{k}} ^{{k}+\mathrm{1}} \left({t}^{\mathrm{2}} −\mathrm{10}{t}\:+\mathrm{25}\right)\left({k}\:+\xi\right){dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{9}} \left({k}+\xi\right)\:\left[\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\:−\mathrm{5}{t}^{\mathrm{2}} \:+\mathrm{25}{t}\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{9}} \left({k}+\xi\right)\left\{\:\frac{\left({k}+\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{3}}\:−\mathrm{5}\left({k}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{25}\left({k}+\mathrm{1}\right)\right. \\ $$$$\left.−\frac{{k}^{\mathrm{3}} }{\mathrm{3}}\:+\mathrm{5}{k}^{\mathrm{2}} \:−\mathrm{25}{k}\right\}\:=..... \\ $$$$ \\ $$$$ \\ $$
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](Q42824.png)
$$\mathrm{in}\:\mathrm{steps}: \\ $$$$\left[{x}+\frac{\mathrm{1}}{\mathrm{2}}\right]={c} \\ $$$${x}\in\left[−\mathrm{5};\:−\mathrm{4}.\mathrm{5}\left[\:\Rightarrow\:{c}=−\mathrm{5}\right.\right. \\ $$$${x}\in\left[−\mathrm{4}.\mathrm{5};\:−\mathrm{3}.\mathrm{5}\left[\:\Rightarrow\:{c}=−\mathrm{4}\right.\right. \\ $$$${x}\in\:\left[−\mathrm{3}.\mathrm{5};\:−\mathrm{2}.\mathrm{5}\left[\:\Rightarrow\:{c}=−\mathrm{3}\right.\right. \\ $$$${x}\in\:\left[−\mathrm{2}.\mathrm{5};\:−\mathrm{1}.\mathrm{5}\left[\:\Rightarrow\:{c}=−\mathrm{2}\right.\right. \\ $$$${x}\in\:\left[−\mathrm{1}.\mathrm{5};\:−.\mathrm{5}\left[\:\Rightarrow\:{c}=−\mathrm{1}\right.\right. \\ $$$${x}\in\:\left[−.\mathrm{5};\:.\mathrm{5}\left[\:\Rightarrow\:{c}=\mathrm{0}\right.\right. \\ $$$${x}\in\:\left[.\mathrm{5};\:\mathrm{1}.\mathrm{5}\left[\:\Rightarrow\:{c}=\mathrm{1}\right.\right. \\ $$$${x}\in\:\left[\mathrm{1}.\mathrm{5};\:\mathrm{2}.\mathrm{5}\left[\:\Rightarrow\:{c}=\mathrm{2}\right.\right. \\ $$$${x}\in\:\left[\mathrm{2}.\mathrm{5};\:\mathrm{3}.\mathrm{5}\left[\:\Rightarrow\:{c}=\mathrm{3}\right.\right. \\ $$$${x}\in\:\left[\mathrm{3}.\mathrm{5};\:\mathrm{4}.\mathrm{5}\left[\:\Rightarrow\:{c}=\mathrm{4}\right.\right. \\ $$$${x}\in\:\left[\mathrm{4}.\mathrm{5};\:\mathrm{5}\right]\:\Rightarrow\:{c}=\mathrm{5} \\ $$$$\underset{−\mathrm{5}} {\overset{\mathrm{5}} {\int}}{x}^{\mathrm{2}} \left[{x}+\frac{\mathrm{1}}{\mathrm{2}}\right]{dx}=\Sigma{c}\int{x}^{\mathrm{2}} {dx}=\Sigma\frac{{c}}{\mathrm{3}}\left[{x}^{\mathrm{3}} \right]_{{a}} ^{{b}} = \\ $$$$=−\frac{\mathrm{5}}{\mathrm{3}}\left(\left(−\mathrm{4}.\mathrm{5}\right)^{\mathrm{3}} −\left(−\mathrm{5}\right)^{\mathrm{3}} \right)−\frac{\mathrm{4}}{\mathrm{3}}\left(\left(−\mathrm{3}.\mathrm{5}\right)^{\mathrm{3}} −\left(−\mathrm{4}.\mathrm{5}\right)^{\mathrm{3}} \right)−\frac{\mathrm{3}}{\mathrm{3}}\left(\left(−\mathrm{2}.\mathrm{5}\right)^{\mathrm{3}} −\left(−\mathrm{3}.\mathrm{5}\right)^{\mathrm{3}} \right)... \\ $$$$...+\frac{\mathrm{3}}{\mathrm{3}}\left(\mathrm{3}.\mathrm{5}^{\mathrm{3}} −\mathrm{2}.\mathrm{5}^{\mathrm{3}} \right)+\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{4}.\mathrm{5}^{\mathrm{3}} −\mathrm{3}.\mathrm{5}^{\mathrm{3}} \right)+\frac{\mathrm{5}}{\mathrm{3}}\left(\mathrm{5}^{\mathrm{3}} −\mathrm{4}.\mathrm{5}^{\mathrm{3}} \right)= \\ $$$$=−\frac{\mathrm{5}}{\mathrm{3}}\left(\mathrm{5}^{\mathrm{3}} −\mathrm{4}.\mathrm{5}^{\mathrm{3}} \right)+\frac{\mathrm{5}}{\mathrm{3}}\left(\mathrm{5}^{\mathrm{3}} −\mathrm{4}.\mathrm{5}^{\mathrm{3}} \right)...=\mathrm{0} \\ $$
Commented by rahul 19 last updated on 03/Sep/18
Thank you sir ! ��