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This Question was also asked in CBSE Maths Board Exam - 2020 (Question 34 - Set 65/5/1)

Ex 7.8, 4 - Integrate (x2 - x) dx by limit as a sum - Ex 7.8

Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 6 Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 7 Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 8 Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 9 Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 10 Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 11 Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 12 Ex 7.8, 4 - Chapter 7 Class 12 Integrals - Part 13

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Ex 7.8, 4 ∫1_1^4β–’(π‘₯2 βˆ’π‘₯)𝑑π‘₯ Let I = ∫1_1^4β–’(π‘₯2 βˆ’π‘₯)𝑑π‘₯ I = ∫1_1^4β–’γ€– π‘₯2 𝑑π‘₯γ€—βˆ’βˆ«1_1^4β–’γ€– π‘₯ 𝑑π‘₯γ€— Solving I1 and I2 separately Solving I1 ∫1_1^4β–’γ€–π‘₯2 𝑑π‘₯γ€— Putting π‘Ž =1 𝑏 =4 β„Ž=(𝑏 βˆ’ π‘Ž)/𝑛 =(4 βˆ’ 1)/𝑛 =3/𝑛 𝑓(π‘₯)=π‘₯^2 We know that ∫1_π‘Ž^𝑏▒〖π‘₯ 𝑑π‘₯γ€— =(π‘βˆ’π‘Ž) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(π‘Ž)+𝑓(π‘Ž+β„Ž)+𝑓(π‘Ž+2β„Ž)…+𝑓(π‘Ž+(π‘›βˆ’1)β„Ž)) Hence we can write ∫1_1^4β–’γ€–π‘₯2 𝑑π‘₯γ€— =(4βˆ’1) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(1)+𝑓(1+β„Ž)+𝑓(1+2β„Ž)+ …+𝑓(1+(π‘›βˆ’1)β„Ž)) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(1)+𝑓(1+β„Ž)+𝑓(1+2β„Ž)+ …+𝑓(1+(π‘›βˆ’1)β„Ž)) Here, 𝑓(π‘₯)=π‘₯^2 𝑓(1)=(1)^2=1 𝑓(1+β„Ž)=(1+β„Ž)^2 𝑓 (1+2β„Ž)=(1+2β„Ž)^2 … 𝑓(1+(π‘›βˆ’1)β„Ž)=(1+(π‘›βˆ’1)β„Ž)^2 Hence, our equation becomes ∫1_1^4β–’γ€–π‘₯2 𝑑π‘₯γ€— " " =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(1)+𝑓(1+β„Ž)+𝑓(1+2β„Ž)+ …+𝑓(1+(π‘›βˆ’1)β„Ž)) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 ((1)^2+(1+β„Ž)^2+(1+2β„Ž)^2+ …+(1+(π‘›βˆ’1)β„Ž)^2 ) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (β–ˆ(1^2+(1^2+β„Ž^2+2β„Ž)+γ€–(1γ€—^2+ (2β„Ž)^2+4β„Ž)+ …… @ …+(1^2+((π‘›βˆ’1)β„Ž)^2+2(π‘›βˆ’1) β„Ž) )) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 [1^2+1^2+ … +1^2 ] + β„Ž^2+(2β„Ž)^2+ … +(π‘›βˆ’1)β„Ž^2 + [2β„Ž+4β„Ž+ … +2(π‘›βˆ’1)β„Ž] =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (〖𝑛(1)γ€—^2+[β„Ž^2+(2)^2 . β„Ž^2+ … +(π‘›βˆ’1)^2 β„Ž^2 ] +[2β„Ž+2Γ—2β„Ž+ … +(π‘›βˆ’1)Γ—2β„Ž] ) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛(𝑛+𝒉^2 [(1)^2+(2)^2+ …+(π‘›βˆ’1)^2 ] +πŸπ’‰ [1+2+ …+(π‘›βˆ’1)]) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑛+β„Ž^2 [𝑛(𝑛 βˆ’ 1)(2𝑛 βˆ’ 1)/6]+2β„Ž[𝑛(𝑛 βˆ’ 1)/2] ) We know that 1^2+2^2+ …+𝑛^2= (𝑛 (𝑛 + 1)(2𝑛 + 1))/6 1^2+2^2+ ……+(π‘›βˆ’1)^2 = ((𝑛 βˆ’ 1) (𝑛 βˆ’1 + 1)(2(𝑛 βˆ’ 1) + 1))/6 = ((𝑛 βˆ’ 1) 𝑛 (2𝑛 βˆ’ 2 + 1) )/6 = (𝑛 (𝑛 βˆ’ 1) (2𝑛 βˆ’ 1) )/6 We know that 1+2+3+ ……+𝑛= (𝑛 (𝑛 + 1))/2 1+2+3+ ……+(π‘›βˆ’1) = ((𝑛 βˆ’ 1) (𝑛 βˆ’ 1 + 1))/2 = (𝑛 (𝑛 βˆ’ 1) )/2 =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑛+β„Ž^2 [𝑛(𝑛 βˆ’ 1)(2𝑛 βˆ’ 1)]/6+β„Ž[𝑛(𝑛 βˆ’ 1)] ) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑛/𝑛+β„Ž^2 [𝑛(𝑛 βˆ’ 1)(2𝑛 βˆ’ 1)/6𝑛]+β„Ž[𝑛(𝑛 βˆ’ 1)/𝑛]) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (1+β„Ž^2 [(𝑛 βˆ’ 1)(2𝑛 βˆ’ 1)/6]+β„Ž[(𝑛 βˆ’ 1)]) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (1+(3/𝑛)^2 (𝑛 βˆ’ 1)(2𝑛 βˆ’ 1)/6+(3/𝑛)(𝑛 βˆ’ 1)) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (1+9/𝑛^2 . (𝑛 βˆ’ 1)(2𝑛 βˆ’ 1)/6 +3(1 βˆ’ 1/𝑛)) =3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (1+ 9(1 βˆ’ 1/𝑛)(2 βˆ’ 1/𝑛)/6 +3(1 βˆ’ 1/𝑛)) =3(1+ 9(1 βˆ’ 1/∞)(2 βˆ’ 1/∞)/6 +3(1 βˆ’ 1/∞)) =3(1+ 9(1 βˆ’ 0)(2 βˆ’ 0)/6 +3(1 βˆ’0)) =3(1+ (9 Γ— 1 Γ— 2)/6 +3) =3(1+3+3) =3Γ—7 =𝟐𝟏 Solving I2 ∫1_1^4β–’γ€–π‘₯ 𝑑π‘₯γ€— Putting π‘Ž =1 𝑏 =4 β„Ž=(𝑏 βˆ’ π‘Ž)/𝑛 =(4 βˆ’ 1)/𝑛 =3/𝑛 𝑓(π‘₯)=π‘₯ We know that ∫1_π‘Ž^𝑏▒〖π‘₯ 𝑑π‘₯γ€— =(π‘βˆ’π‘Ž) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(π‘Ž)+𝑓(π‘Ž+β„Ž)+𝑓(π‘Ž+2β„Ž)…+𝑓(π‘Ž+(π‘›βˆ’1)β„Ž)) Hence we can write ∫1_1^4β–’γ€–π‘₯ 𝑑π‘₯γ€— =(4βˆ’1) lim┬(nβ†’βˆž) 1/𝑛 (𝑓(1)+𝑓(1+β„Ž)+𝑓(1+2β„Ž)+… +𝑓(1+(π‘›βˆ’1)β„Ž) =3 lim┬(nβ†’βˆž) 1/𝑛 (𝑓(1)+𝑓(1+β„Ž)+𝑓(1+2β„Ž)+… +𝑓(1+(π‘›βˆ’1)β„Ž) Here, 𝑓(π‘₯)=π‘₯ 𝑓(1)=1 𝑓(1+β„Ž)=1+β„Ž 𝑓 (1+2β„Ž)=1+2β„Ž 𝑓(1+(π‘›βˆ’1)β„Ž)=1+(π‘›βˆ’1)β„Ž Hence, our equation becomes ∫_1^4β–’π‘₯ 𝑑π‘₯ =3 lim┬(nβ†’βˆž) 1/𝑛 (𝑓(1)+𝑓(1+β„Ž)+𝑓(1+2β„Ž)+… +𝑓(1+(π‘›βˆ’1)β„Ž) = 3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (1+(1+β„Ž)+(1+2β„Ž)+ …+(1+(π‘›βˆ’1)β„Ž)) = 3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (1+1+ …+1 +β„Ž+2β„Ž+ ……+(π‘›βˆ’1)β„Ž) = 3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 ( 𝑛\ Γ—1+β„Ž (1+2+ ………+(π‘›βˆ’1))) We know that 1+2+3+ ……+𝑛= (𝑛 (𝑛 + 1))/2 1+2+3+ ……+π‘›βˆ’1= ((𝑛 βˆ’ 1) (𝑛 βˆ’ 1 + 1))/2 = (𝑛 (𝑛 βˆ’ 1) )/2 = 3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 ( 𝑛+(β„Ž . 𝑛(𝑛 βˆ’ 1))/2) = 3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) ( 𝑛/𝑛+𝑛(𝑛 βˆ’ 1)β„Ž/2𝑛) = 3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) ( 1+(𝑛 βˆ’ 1)β„Ž/2) = 3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) ( 1+(𝑛 βˆ’ 1)3/(2 . 𝑛)) = 3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) ( 1+(𝑛/𝑛 βˆ’ 1/𝑛) 3/2) [π‘ˆπ‘ π‘–π‘›π‘” β„Ž=3/𝑛] = 3 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) ( 1+(1βˆ’ 1/𝑛) (3 )/2) = 3( 1+(1βˆ’ 1/∞) (3 )/2) = 3( 1+(1βˆ’0) 3/2) = 3(1+ (3 )/2) = 3((5 )/2) = πŸπŸ“/𝟐 Putting the values of I1 and I2 in I ∴ "I = " ∫1_1^4β–’γ€– π‘₯2 𝑑π‘₯γ€—βˆ’βˆ«1_1^4β–’γ€– π‘₯ 𝑑π‘₯γ€— = 21 βˆ’ 15/2 = (42 βˆ’ 15)/2 = πŸπŸ•/𝟐

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.