Question Number 88951 by M±th+et£s last updated on 14/Apr/20
$$\int_{{a}} ^{{b}} \lceil{x}\rceil\:{dx}=? \\ $$$${a},{b}\in{R}\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\lceil..\rceil\:{is}\:{ceil}\: \\ $$
Answered by mr W last updated on 14/Apr/20
$${let}\:{A}=\lceil{a}\rceil,\:{B}=\lfloor{b}\rfloor,\:{C}=\lceil{b}\rceil,\:{D}=\lfloor{a}\rfloor \\ $$$${I}_{\mathrm{1}} =\int_{{a}} ^{{b}} \lceil{x}\rceil\:{dx} \\ $$$$=\int_{{a}} ^{\lceil{a}\rceil} \lceil{x}\rceil\:{dx}+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \lceil{x}\rceil\:{dx}+\int_{\lfloor{b}\rfloor} ^{{b}} \lceil{x}\rceil\:{dx} \\ $$$$=\int_{{a}} ^{\lceil{a}\rceil} \lceil{a}\rceil\:{dx}+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \left({k}+\mathrm{1}\right)\:{dx}+\int_{\lfloor{b}\rfloor} ^{{b}} \lceil{b}\rceil\:{dx} \\ $$$$=\lceil{a}\rceil\left(\lceil{a}\rceil−{a}\right)+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}\left({k}+\mathrm{1}\right)+\lceil{b}\rceil\left({b}−\lfloor{b}\rfloor\right) \\ $$$$=\lceil{a}\rceil\left(\lceil{a}\rceil−{a}\right)+\underset{{k}=\lceil{a}\rceil+\mathrm{1}} {\overset{\lfloor{b}\rfloor} {\sum}}{k}+\lceil{b}\rceil\left({b}−\lfloor{b}\rfloor\right) \\ $$$$=\lceil{a}\rceil\left(\lceil{a}\rceil−{a}\right)+\frac{\left(\lceil{a}\rceil+\lfloor{b}\rfloor+\mathrm{1}\right)\left(\lfloor{b}\rfloor−\lceil{a}\rceil\right)}{\mathrm{2}}+\lceil{b}\rceil\left({b}−\lfloor{b}\rfloor\right) \\ $$$$={A}\left({A}−{a}\right)+\frac{\left({A}+{B}+\mathrm{1}\right)\left({B}−{A}\right)}{\mathrm{2}}+{C}\left({b}−{B}\right) \\ $$$$=\frac{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{B}−{A}\left(\mathrm{1}+\mathrm{2}{a}\right)+\mathrm{2}\left({b}−{B}\right){C}}{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\frac{\lceil{a}\rceil^{\mathrm{2}} +\lfloor{b}\rfloor^{\mathrm{2}} +\lfloor{b}\rfloor−\left(\mathrm{1}+\mathrm{2}{a}\right)\lceil{a}\rceil+\mathrm{2}\left({b}−\lfloor{b}\rfloor\right)\lceil{b}\rceil}{\mathrm{2}} \\ $$$$ \\ $$$${I}_{\mathrm{2}} =\int_{{a}} ^{{b}} \lfloor{x}\rfloor\:{dx} \\ $$$$=\int_{{a}} ^{\lceil{a}\rceil} \lfloor{x}\rfloor\:{dx}+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \lfloor{x}\rfloor\:{dx}+\int_{\lfloor{b}\rfloor} ^{{b}} \lfloor{x}\rfloor\:{dx} \\ $$$$=\int_{{a}} ^{\lceil{a}\rceil} \lfloor{a}\rfloor\:{dx}+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} {k}\:{dx}+\int_{\lfloor{b}\rfloor} ^{{b}} \lfloor{b}\rfloor\:{dx} \\ $$$$=\lfloor{a}\rfloor\left(\lceil{a}\rceil−{a}\right)+\underset{{k}=\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}} {\sum}}{k}+\lfloor{b}\rfloor\left({b}−\lfloor{b}\rfloor\right) \\ $$$$=\lfloor{a}\rfloor\left(\lceil{a}\rceil−{a}\right)+\frac{\left(\lceil{a}\rceil+\lfloor{b}\rfloor−\mathrm{1}\right)\left(\lfloor{b}\rfloor−\lceil{a}\rceil\right)}{\mathrm{2}}+\lfloor{b}\rfloor\left({b}−\lfloor{b}\rfloor\right) \\ $$$$={D}\left({A}−{a}\right)+\frac{\left({A}+{B}−\mathrm{1}\right)\left({B}−{A}\right)}{\mathrm{2}}+{B}\left({b}−{B}\right) \\ $$$$=\frac{\mathrm{2}{D}\left({A}−{a}\right)+{A}−{B}−{A}^{\mathrm{2}} −{B}^{\mathrm{2}} +\mathrm{2}{Bb}}{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}\lfloor{a}\rfloor\left(\lceil{a}\rceil−{a}\right)+\lceil{a}\rceil+\left(\mathrm{2}{b}−\mathrm{1}\right)\lfloor{b}\rfloor−\lceil{a}\rceil^{\mathrm{2}} −\lfloor{b}\rfloor^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$$${example}: \\ $$$${I}_{\mathrm{2}} =\int_{−\mathrm{1}.\mathrm{5}} ^{\mathrm{3}.\mathrm{7}} \lfloor{x}\rfloor\:{dx} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}\lfloor−\mathrm{1}.\mathrm{5}\rfloor\left(\lceil−\mathrm{1}.\mathrm{5}\rceil+\mathrm{1}.\mathrm{5}\right)+\lceil−\mathrm{1}.\mathrm{5}\rceil+\left(\mathrm{2}×\mathrm{3}.\mathrm{7}−\mathrm{1}\right)\lfloor\mathrm{3}.\mathrm{7}\rfloor−\lceil−\mathrm{1}.\mathrm{5}\rceil^{\mathrm{2}} −\lfloor\mathrm{3}.\mathrm{7}\rfloor^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}\left(−\mathrm{2}\right)\left(−\mathrm{1}+\mathrm{1}.\mathrm{5}\right)−\mathrm{1}+\left(\mathrm{2}×\mathrm{3}.\mathrm{7}−\mathrm{1}\right)×\mathrm{3}−\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{6}.\mathrm{2}}{\mathrm{2}}=\mathrm{3}.\mathrm{1} \\ $$$$ \\ $$$${I}_{\mathrm{2}} =\int_{−\mathrm{2}} ^{\mathrm{4}} \lfloor{x}\rfloor\:{dx} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}\lfloor−\mathrm{2}\rfloor\left(\lceil−\mathrm{2}\rceil+\mathrm{2}\right)+\lceil−\mathrm{2}\rceil+\left(\mathrm{2}×\mathrm{4}−\mathrm{1}\right)\lfloor\mathrm{4}\rfloor−\lceil−\mathrm{2}\rceil^{\mathrm{2}} −\lfloor\mathrm{4}\rfloor^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{−\mathrm{2}+\mathrm{28}−\mathrm{4}−\mathrm{16}}{\mathrm{2}}=\mathrm{3} \\ $$
Commented by M±th+et£s last updated on 14/Apr/20
$${i}\:{am}\:{verry}\:{greatfull}\:{to}\:{you}\:{and}\:{pretiate}\:{what}\: \\ $$$${have}\:{you}\:{done}\:{for}\:{me} \\ $$
Commented by M±th+et£s last updated on 14/Apr/20
$${sir}\:{do}\:{you}\:{have}\:{pdf}\:{or}\:{do}\:{you}\:{have}\:{any}\:{explains} \\ $$$${for}\:{this}\:{subject}\:{because}\:{i}\:{am}\:{a}\:{student}\: \\ $$$${and}\:{i}\:{want}\:{to}\:{learn} \\ $$
Commented by M±th+et£s last updated on 14/Apr/20
$${i}\:{mean}\:{floor}\:{function} \\ $$
Commented by mr W last updated on 14/Apr/20
$${i}\:{don}'{t}\:{have}\:{any}\:{book}.\:{but}\:{it}'{s}\:{just}\:{a} \\ $$$${function}\:{definition},\:{there}\:{is}\:{not} \\ $$$${much}\:{to}\:{read}. \\ $$