Integration Questions

Question Number 79837 by Pratah last updated on 28/Jan/20

Commented by mr W last updated on 28/Jan/20

$${see}\:{Q}\mathrm{79814} \\$$

Commented by Pratah last updated on 28/Jan/20

$$\mathrm{this}\:\mathrm{one}\:\mathrm{is}\:\mathrm{different} \\$$

Commented by mr W last updated on 28/Jan/20

$${if}\:{you}\:{understand}\:{Q}\mathrm{79814},\:{then}\:{you} \\$$$${should}\:{also}\:{be}\:{able}\:{to}\:{solve}\:{this}\:{one}. \\$$

Commented by Pratah last updated on 28/Jan/20

$$\mathrm{no} \\$$

Commented by mr W last updated on 28/Jan/20

$${you}\:{can}'{t}\:{or}\:{you}\:{won}'{t}? \\$$

Commented by abdomathmax last updated on 28/Jan/20

$$\int_{−\mathrm{1}} ^{\mathrm{1}} \left[{x}−\mathrm{1}\right]{dx}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \left[{x}\right]{dx}\:−\int_{−\mathrm{1}} ^{\mathrm{1}} {dx} \\$$$$\left.=\left.\int_{−\mathrm{1}} ^{\mathrm{0}} \left(−\mathrm{1}\right){dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \right){o}\right){dx}\:−\mathrm{2} \\$$$$=\mathrm{1}−\mathrm{2}\:=−\mathrm{1} \\$$

Commented by Pratah last updated on 29/Jan/20

$$\mathrm{no}\:\mathrm{correct} \\$$

Commented by Pratah last updated on 29/Jan/20

$$\mathrm{sir}\:\mathrm{abdomathmax} \\$$

Commented by mr W last updated on 29/Jan/20

$${sir}\:{abdomathmax}: \\$$$${error}\:{in}\:{your}\:{calculation}: \\$$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} \left(−\mathrm{1}\right){dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{0}\right){dx}\:−\mathrm{2} \\$$$$=−\mathrm{1}−\mathrm{2}\:=−\mathrm{3} \\$$$${i}.{e}.\:{you}\:{should}\:{also}\:{get}\:{the}\:{result}\:−\mathrm{3}. \\$$

Commented by mathmax by abdo last updated on 29/Jan/20

$${yes}\:{sir}\:{thank}\:{you}. \\$$

Answered by key of knowledge last updated on 28/Jan/20

$$−\mathrm{1}\leqslant\mathrm{x}<\mathrm{0}\Rightarrow\left[\mathrm{x}−\mathrm{1}\right]=−\mathrm{2}\Rightarrow\int_{−\mathrm{1}} ^{\mathrm{0}} \left(−\mathrm{2}\right)\mathrm{dx}=−\mathrm{2} \\$$$$\mathrm{0}\leqslant\mathrm{x}<\mathrm{1}\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)\mathrm{dx}=−\mathrm{1} \\$$$$\left(−\mathrm{1}\right)+\left(−\mathrm{2}\right)=−\mathrm{3} \\$$

Commented by Pratah last updated on 28/Jan/20

$$\mathrm{answer}:\:−\mathrm{1} \\$$

Commented by mr W last updated on 28/Jan/20

$${sir}\:{Pratah}: \\$$$${to}\:{understand}\:{your}\:{definition},\:{please} \\$$$${answer}: \\$$$$\left[−\mathrm{0}.\mathrm{8}\right]=−\mathrm{1}\:{or}\:\mathrm{0}\:? \\$$

Commented by MJS last updated on 28/Jan/20

wait... he must answer to understand? ...or he must understand to answer? ����

Commented by Pratah last updated on 29/Jan/20

$$\left[−\mathrm{1}+\mathrm{0}.\mathrm{2}\right]=−\mathrm{1}\:\:;\left(\right. \\$$

Commented by mr W last updated on 29/Jan/20

$${pratah}\:{sir}: \\$$$${if}\:\left[−\mathrm{0}.\mathrm{8}\right]=−\mathrm{1} \\$$$${then} \\$$$$\int_{−\mathrm{1}} ^{\:\mathrm{1}} \left[{x}−\mathrm{1}\right]{dx}=−\mathrm{3}\:\neq−\mathrm{1}\:\left({which}\:{you}\:{gave}\right) \\$$

Commented by Pratah last updated on 29/Jan/20

Commented by mr W last updated on 29/Jan/20

$${build}\:{your}\:{own}\:{opinion}\:{if}\:{you}\:{don}'{t} \\$$$${have}\:{one}! \\$$$${that}'{s}\:{why}\:{i}\:{asked}\:{you}\:{the}\:{definition} \\$$$${of}\:\left[−\mathrm{0}.\mathrm{8}\right]\:{at}\:{the}\:{begining}! \\$$$${you}\:{told}\:\left[−\mathrm{0}.\mathrm{8}\right]=−\mathrm{1},\:{but}\:{this}\:{diagram} \\$$$${shows}\:{the}\:{graph}\:\left[{x}−\mathrm{1}\right],\:{and}\:{it}\:{shows} \\$$$${at}\:{x}=\mathrm{0}.\mathrm{5}:\:\left[\mathrm{0}.\mathrm{5}−\mathrm{1}\right]=\left[−\mathrm{0}.\mathrm{5}\right]=\mathrm{0}\neq−\mathrm{1} \\$$$${at}\:{x}=−\mathrm{0}.\mathrm{5}:\:\left[−\mathrm{0}.\mathrm{5}−\mathrm{1}\right]=\left[−\mathrm{1}.\mathrm{5}\right]=−\mathrm{1}\neq−\mathrm{2}. \\$$$${that}\:{means}\:{the}\:{solution}\:{you}\:{gave} \\$$$${uses}\:{a}\:{different}\:{definition}\:{for}\:\left[{x}\right] \\$$$${than}\:{you}! \\$$