Question Number 86993 by Power last updated on 01/Apr/20
Commented by abdomathmax last updated on 01/Apr/20
![I =∫_1 ^5 [10x]dx vhangement 10x =t give I =(1/(10)) ∫_(10) ^(50) [t]dt =(1/(10))Σ_(k=10) ^(49) ∫_k ^(k+1) kdt =(1/(10)) Σ_(k=10) ^(49) k the sequence u_k =k is arithmetic ⇒ Σ_(k=10) ^(49) u_k =((49−10+1)/2)(u_(10) +u_(49) ) =20(10+49) =20(59) =1180 ⇒I =((1180)/(10)) ⇒I =118](https://www.tinkutara.com/question/Q86996.png)
$${I}\:=\int_{\mathrm{1}} ^{\mathrm{5}} \left[\mathrm{10}{x}\right]{dx}\:\:{vhangement}\:\mathrm{10}{x}\:={t}\:{give} \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{10}}\:\int_{\mathrm{10}} ^{\mathrm{50}} \left[{t}\right]{dt}\:=\frac{\mathrm{1}}{\mathrm{10}}\sum_{{k}=\mathrm{10}} ^{\mathrm{49}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{kdt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\:\sum_{{k}=\mathrm{10}} ^{\mathrm{49}} \:{k}\:\:\:\:\:{the}\:{sequence}\:{u}_{{k}} ={k}\:{is}\:{arithmetic}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{10}} ^{\mathrm{49}} \:{u}_{{k}} =\frac{\mathrm{49}−\mathrm{10}+\mathrm{1}}{\mathrm{2}}\left({u}_{\mathrm{10}} \:+{u}_{\mathrm{49}} \right)\:=\mathrm{20}\left(\mathrm{10}+\mathrm{49}\right) \\ $$$$=\mathrm{20}\left(\mathrm{59}\right)\:=\mathrm{1180}\:\Rightarrow{I}\:=\frac{\mathrm{1180}}{\mathrm{10}}\:\Rightarrow{I}\:=\mathrm{118} \\ $$
Answered by TANMAY PANACEA. last updated on 01/Apr/20
![∫_1 ^(1.1) [10x]dx+∫_(1.1) ^(1.2) [10x]dx+∫_(1.2) ^(1.3) [10x]dx+..+∫_(4.9) ^5 [10x] =0.1×10+0.1×11+...+0.1×49 =0.1(10+11+..+49) =(1/(10))×((40)/2)(10+49)=2×59=118](https://www.tinkutara.com/question/Q86997.png)
$$\int_{\mathrm{1}} ^{\mathrm{1}.\mathrm{1}} \left[\mathrm{10}{x}\right]{dx}+\int_{\mathrm{1}.\mathrm{1}} ^{\mathrm{1}.\mathrm{2}} \left[\mathrm{10}{x}\right]{dx}+\int_{\mathrm{1}.\mathrm{2}} ^{\mathrm{1}.\mathrm{3}} \left[\mathrm{10}{x}\right]{dx}+..+\int_{\mathrm{4}.\mathrm{9}} ^{\mathrm{5}} \left[\mathrm{10}{x}\right] \\ $$$$=\mathrm{0}.\mathrm{1}×\mathrm{10}+\mathrm{0}.\mathrm{1}×\mathrm{11}+…+\mathrm{0}.\mathrm{1}×\mathrm{49} \\ $$$$=\mathrm{0}.\mathrm{1}\left(\mathrm{10}+\mathrm{11}+..+\mathrm{49}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}×\frac{\mathrm{40}}{\mathrm{2}}\left(\mathrm{10}+\mathrm{49}\right)=\mathrm{2}×\mathrm{59}=\mathrm{118} \\ $$$$ \\ $$
Commented by Power last updated on 01/Apr/20
$$\mathrm{thanks} \\ $$
Commented by TANMAY PANACEA. last updated on 01/Apr/20
$${most}\:{welcome}\:{sir} \\ $$