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Question Number 88951 by M±th+et£s last updated on 14/Apr/20

∫_a ^b ⌈x⌉ dx=? a,b∈R ⌈..⌉ is ceil

$$\int_{{a}} ^{{b}} \lceil{x}\rceil\:{dx}=? \\ $$$${a},{b}\in{R}\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\lceil..\rceil\:{is}\:{ceil}\: \\ $$

Answered by mr W last updated on 14/Apr/20

let A=⌈a⌉, B=⌊b⌋, C=⌈b⌉, D=⌊a⌋ I_1 =∫_a ^b ⌈x⌉ dx =∫_a ^(⌈a⌉) ⌈x⌉ dx+Σ_(k=⌈a⌉) ^(⌊b⌋−1) ∫_k ^(k+1) ⌈x⌉ dx+∫_(⌊b⌋) ^b ⌈x⌉ dx =∫_a ^(⌈a⌉) ⌈a⌉ dx+Σ_(k=⌈a⌉) ^(⌊b⌋−1) ∫_k ^(k+1) (k+1) dx+∫_(⌊b⌋) ^b ⌈b⌉ dx =⌈a⌉(⌈a⌉−a)+Σ_(k=⌈a⌉) ^(⌊b⌋−1) (k+1)+⌈b⌉(b−⌊b⌋) =⌈a⌉(⌈a⌉−a)+Σ_(k=⌈a⌉+1) ^(⌊b⌋) k+⌈b⌉(b−⌊b⌋) =⌈a⌉(⌈a⌉−a)+(((⌈a⌉+⌊b⌋+1)(⌊b⌋−⌈a⌉))/2)+⌈b⌉(b−⌊b⌋) =A(A−a)+(((A+B+1)(B−A))/2)+C(b−B) =((A^2 +B^2 +B−A(1+2a)+2(b−B)C)/2) I_1 =((⌈a⌉^2 +⌊b⌋^2 +⌊b⌋−(1+2a)⌈a⌉+2(b−⌊b⌋)⌈b⌉)/2) I_2 =∫_a ^b ⌊x⌋ dx =∫_a ^(⌈a⌉) ⌊x⌋ dx+Σ_(k=⌈a⌉) ^(⌊b⌋−1) ∫_k ^(k+1) ⌊x⌋ dx+∫_(⌊b⌋) ^b ⌊x⌋ dx =∫_a ^(⌈a⌉) ⌊a⌋ dx+Σ_(k=⌈a⌉) ^(⌊b⌋−1) ∫_k ^(k+1) k dx+∫_(⌊b⌋) ^b ⌊b⌋ dx =⌊a⌋(⌈a⌉−a)+Σ_(k=⌈a⌉) ^(⌊b⌋−1) k+⌊b⌋(b−⌊b⌋) =⌊a⌋(⌈a⌉−a)+(((⌈a⌉+⌊b⌋−1)(⌊b⌋−⌈a⌉))/2)+⌊b⌋(b−⌊b⌋) =D(A−a)+(((A+B−1)(B−A))/2)+B(b−B) =((2D(A−a)+A−B−A^2 −B^2 +2Bb)/2) I_2 =((2⌊a⌋(⌈a⌉−a)+⌈a⌉+(2b−1)⌊b⌋−⌈a⌉^2 −⌊b⌋^2 )/2) example: I_2 =∫_(−1.5) ^(3.7) ⌊x⌋ dx I_2 =((2⌊−1.5⌋(⌈−1.5⌉+1.5)+⌈−1.5⌉+(2×3.7−1)⌊3.7⌋−⌈−1.5⌉^2 −⌊3.7⌋^2 )/2) I_2 =((2(−2)(−1+1.5)−1+(2×3.7−1)×3−(−1)^2 −3^2 )/2) I_2 =((6.2)/2)=3.1 I_2 =∫_(−2) ^4 ⌊x⌋ dx I_2 =((2⌊−2⌋(⌈−2⌉+2)+⌈−2⌉+(2×4−1)⌊4⌋−⌈−2⌉^2 −⌊4⌋^2 )/2) I_2 =((−2+28−4−16)/2)=3

$${let}\:{A}=\lceil{a}\rceil,\:{B}=\lfloor{b}\rfloor,\:{C}=\lceil{b}\rceil,\:{D}=\lfloor{a}\rfloor \\ $$$${I}_{\mathrm{1}} =\int_{{a}} ^{{b}} \lceil{x}\rceil\:{dx} \\ $$$$=\int_{{a}} ^{\lceil{a}\rceil} \lceil{x}\rceil\:{dx}+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \lceil{x}\rceil\:{dx}+\int_{\lfloor{b}\rfloor} ^{{b}} \lceil{x}\rceil\:{dx} \\ $$$$=\int_{{a}} ^{\lceil{a}\rceil} \lceil{a}\rceil\:{dx}+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \left({k}+\mathrm{1}\right)\:{dx}+\int_{\lfloor{b}\rfloor} ^{{b}} \lceil{b}\rceil\:{dx} \\ $$$$=\lceil{a}\rceil\left(\lceil{a}\rceil−{a}\right)+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}\left({k}+\mathrm{1}\right)+\lceil{b}\rceil\left({b}−\lfloor{b}\rfloor\right) \\ $$$$=\lceil{a}\rceil\left(\lceil{a}\rceil−{a}\right)+\underset{{k}=\lceil{a}\rceil+\mathrm{1}} {\overset{\lfloor{b}\rfloor} {\sum}}{k}+\lceil{b}\rceil\left({b}−\lfloor{b}\rfloor\right) \\ $$$$=\lceil{a}\rceil\left(\lceil{a}\rceil−{a}\right)+\frac{\left(\lceil{a}\rceil+\lfloor{b}\rfloor+\mathrm{1}\right)\left(\lfloor{b}\rfloor−\lceil{a}\rceil\right)}{\mathrm{2}}+\lceil{b}\rceil\left({b}−\lfloor{b}\rfloor\right) \\ $$$$={A}\left({A}−{a}\right)+\frac{\left({A}+{B}+\mathrm{1}\right)\left({B}−{A}\right)}{\mathrm{2}}+{C}\left({b}−{B}\right) \\ $$$$=\frac{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{B}−{A}\left(\mathrm{1}+\mathrm{2}{a}\right)+\mathrm{2}\left({b}−{B}\right){C}}{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\frac{\lceil{a}\rceil^{\mathrm{2}} +\lfloor{b}\rfloor^{\mathrm{2}} +\lfloor{b}\rfloor−\left(\mathrm{1}+\mathrm{2}{a}\right)\lceil{a}\rceil+\mathrm{2}\left({b}−\lfloor{b}\rfloor\right)\lceil{b}\rceil}{\mathrm{2}} \\ $$$$ \\ $$$${I}_{\mathrm{2}} =\int_{{a}} ^{{b}} \lfloor{x}\rfloor\:{dx} \\ $$$$=\int_{{a}} ^{\lceil{a}\rceil} \lfloor{x}\rfloor\:{dx}+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \lfloor{x}\rfloor\:{dx}+\int_{\lfloor{b}\rfloor} ^{{b}} \lfloor{x}\rfloor\:{dx} \\ $$$$=\int_{{a}} ^{\lceil{a}\rceil} \lfloor{a}\rfloor\:{dx}+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} {k}\:{dx}+\int_{\lfloor{b}\rfloor} ^{{b}} \lfloor{b}\rfloor\:{dx} \\ $$$$=\lfloor{a}\rfloor\left(\lceil{a}\rceil−{a}\right)+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}{k}+\lfloor{b}\rfloor\left({b}−\lfloor{b}\rfloor\right) \\ $$$$=\lfloor{a}\rfloor\left(\lceil{a}\rceil−{a}\right)+\frac{\left(\lceil{a}\rceil+\lfloor{b}\rfloor−\mathrm{1}\right)\left(\lfloor{b}\rfloor−\lceil{a}\rceil\right)}{\mathrm{2}}+\lfloor{b}\rfloor\left({b}−\lfloor{b}\rfloor\right) \\ $$$$={D}\left({A}−{a}\right)+\frac{\left({A}+{B}−\mathrm{1}\right)\left({B}−{A}\right)}{\mathrm{2}}+{B}\left({b}−{B}\right) \\ $$$$=\frac{\mathrm{2}{D}\left({A}−{a}\right)+{A}−{B}−{A}^{\mathrm{2}} −{B}^{\mathrm{2}} +\mathrm{2}{Bb}}{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}\lfloor{a}\rfloor\left(\lceil{a}\rceil−{a}\right)+\lceil{a}\rceil+\left(\mathrm{2}{b}−\mathrm{1}\right)\lfloor{b}\rfloor−\lceil{a}\rceil^{\mathrm{2}} −\lfloor{b}\rfloor^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$$${example}: \\ $$$${I}_{\mathrm{2}} =\int_{−\mathrm{1}.\mathrm{5}} ^{\mathrm{3}.\mathrm{7}} \lfloor{x}\rfloor\:{dx} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}\lfloor−\mathrm{1}.\mathrm{5}\rfloor\left(\lceil−\mathrm{1}.\mathrm{5}\rceil+\mathrm{1}.\mathrm{5}\right)+\lceil−\mathrm{1}.\mathrm{5}\rceil+\left(\mathrm{2}×\mathrm{3}.\mathrm{7}−\mathrm{1}\right)\lfloor\mathrm{3}.\mathrm{7}\rfloor−\lceil−\mathrm{1}.\mathrm{5}\rceil^{\mathrm{2}} −\lfloor\mathrm{3}.\mathrm{7}\rfloor^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}\left(−\mathrm{2}\right)\left(−\mathrm{1}+\mathrm{1}.\mathrm{5}\right)−\mathrm{1}+\left(\mathrm{2}×\mathrm{3}.\mathrm{7}−\mathrm{1}\right)×\mathrm{3}−\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{6}.\mathrm{2}}{\mathrm{2}}=\mathrm{3}.\mathrm{1} \\ $$$$ \\ $$$${I}_{\mathrm{2}} =\int_{−\mathrm{2}} ^{\mathrm{4}} \lfloor{x}\rfloor\:{dx} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}\lfloor−\mathrm{2}\rfloor\left(\lceil−\mathrm{2}\rceil+\mathrm{2}\right)+\lceil−\mathrm{2}\rceil+\left(\mathrm{2}×\mathrm{4}−\mathrm{1}\right)\lfloor\mathrm{4}\rfloor−\lceil−\mathrm{2}\rceil^{\mathrm{2}} −\lfloor\mathrm{4}\rfloor^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{−\mathrm{2}+\mathrm{28}−\mathrm{4}−\mathrm{16}}{\mathrm{2}}=\mathrm{3} \\ $$

Commented by M±th+et£s last updated on 14/Apr/20

i am verry greatfull to you and pretiate what have you done for me

$${i}\:{am}\:{verry}\:{greatfull}\:{to}\:{you}\:{and}\:{pretiate}\:{what}\: \\ $$$${have}\:{you}\:{done}\:{for}\:{me} \\ $$

Commented by M±th+et£s last updated on 14/Apr/20

sir do you have pdf or do you have any explains for this subject because i am a student and i want to learn

$${sir}\:{do}\:{you}\:{have}\:{pdf}\:{or}\:{do}\:{you}\:{have}\:{any}\:{explains} \\ $$$${for}\:{this}\:{subject}\:{because}\:{i}\:{am}\:{a}\:{student}\: \\ $$$${and}\:{i}\:{want}\:{to}\:{learn} \\ $$

Commented by M±th+et£s last updated on 14/Apr/20

i mean floor function

$${i}\:{mean}\:{floor}\:{function} \\ $$

Commented by mr W last updated on 14/Apr/20

i don′t have any book. but it′s just a function definition, there is not much to read.

$${i}\:{don}'{t}\:{have}\:{any}\:{book}.\:{but}\:{it}'{s}\:{just}\:{a} \\ $$$${function}\:{definition},\:{there}\:{is}\:{not} \\ $$$${much}\:{to}\:{read}. \\ $$

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