Integration Full Chapter Explained - Integration Class 12 - Everything you need

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

Slide17.JPG

Slide18.JPG
Slide19.JPG Slide20.JPG Slide21.JPG Slide22.JPG Slide23.JPG Slide24.JPG Slide25.JPG Slide26.JPG Slide27.JPG Slide28.JPG Slide29.JPG

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

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

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.