This Question was also asked in CBSE Maths Board Exam - 2020 (Question 34 - Set 65/5/1)
Chapter 7 Class 12 Integrals
Ex 7.1, 18 Important
Ex 7.1, 20
Ex 7.2, 20 Important
Ex 7.2, 26 Important
Ex 7.2, 35
Ex 7.2, 36 Important
Ex 7.3, 6 Important
Ex 7.3, 13 Important
Ex 7.3, 18 Important
Ex 7.3, 22 Important
Ex 7.3, 24 (MCQ) Important
Example 9 (i)
Example 10 (i)
Ex 7.4, 8 Important
Ex 7.4, 15 Important
Ex 7.4, 21 Important
Ex 7.4, 22
Ex 7.4, 25 (MCQ) Important
Example 15 Important
Ex 7.5, 9 Important
Ex 7.5, 11 Important
Ex 7.5, 17
Ex 7.5, 18 Important
Ex 7.5, 21 Important
Example 20 Important
Example 22 Important
Ex 7.6, 13 Important
Ex 7.6, 14 Important
Ex 7.6, 18 Important
Ex 7.6, 19
Ex 7.6, 24 (MCQ) Important
Ex 7.7, 5 Important
Ex 7.7, 10
Ex 7.7, 11 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams You are here
Question 6 Important Deleted for CBSE Board 2024 Exams
Example 25 (i)
Ex 7.8, 15
Ex 7.8, 16 Important
Ex 7.8, 20 Important
Ex 7.8, 22 (MCQ)
Ex 7.9, 4
Ex 7.9, 7 Important
Ex 7.9, 8
Ex 7.9, 9 (MCQ) Important
Example 28 Important
Example 32 Important
Example 34 Important
Ex 7.10,8 Important
Ex 7.10, 18 Important
Example 38 Important
Example 39 Important
Example 42 Important
Misc 18 Important
Misc 8 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Misc 23 Important
Misc 29 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Misc 38 (MCQ) Important
Question 4 (MCQ) Important Deleted for CBSE Board 2024 Exams
Integration Formula Sheet - Chapter 41 Class 41 Formulas Important
Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Question 4 ∫1_1^4▒(𝑥2 −𝑥)𝑑𝑥 Let I = ∫1_1^4▒(𝑥2 −𝑥)𝑑𝑥 I = ∫1_1^4▒〖 𝑥2 𝑑𝑥〗−∫1_1^4▒〖 𝑥 𝑑𝑥〗 Solving I1 and I2 separately Solving I1 ∫1_1^4▒〖𝑥2 𝑑𝑥〗 Putting 𝑎 =1 𝑏 =4 ℎ=(𝑏 − 𝑎)/𝑛 =(4 − 1)/𝑛 =3/𝑛 𝑓(𝑥)=𝑥^2 We know that ∫1_𝑎^𝑏▒〖𝑥 𝑑𝑥〗 =(𝑏−𝑎) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(𝑎)+𝑓(𝑎+ℎ)+𝑓(𝑎+2ℎ)…+𝑓(𝑎+(𝑛−1)ℎ)) Hence we can write ∫1_1^4▒〖𝑥2 𝑑𝑥〗 =(4−1) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(1)+𝑓(1+ℎ)+𝑓(1+2ℎ)+ …+𝑓(1+(𝑛−1)ℎ)) =3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(1)+𝑓(1+ℎ)+𝑓(1+2ℎ)+ …+𝑓(1+(𝑛−1)ℎ)) Here, 𝑓(𝑥)=𝑥^2 𝑓(1)=(1)^2=1 𝑓(1+ℎ)=(1+ℎ)^2 𝑓 (1+2ℎ)=(1+2ℎ)^2 … 𝑓(1+(𝑛−1)ℎ)=(1+(𝑛−1)ℎ)^2 Hence, our equation becomes ∫1_1^4▒〖𝑥2 𝑑𝑥〗 " " =3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(1)+𝑓(1+ℎ)+𝑓(1+2ℎ)+ …+𝑓(1+(𝑛−1)ℎ)) =3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 ((1)^2+(1+ℎ)^2+(1+2ℎ)^2+ …+(1+(𝑛−1)ℎ)^2 ) =3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (█(1^2+(1^2+ℎ^2+2ℎ)+〖(1〗^2+ (2ℎ)^2+4ℎ)+ …… @ …+(1^2+((𝑛−1)ℎ)^2+2(𝑛−1) ℎ) )) =3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 [1^2+1^2+ … +1^2 ] + ℎ^2+(2ℎ)^2+ … +(𝑛−1)ℎ^2 + [2ℎ+4ℎ+ … +2(𝑛−1)ℎ] =3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (〖𝑛(1)〗^2+[ℎ^2+(2)^2 . ℎ^2+ … +(𝑛−1)^2 ℎ^2 ] +[2ℎ+2×2ℎ+ … +(𝑛−1)×2ℎ] ) =3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛(𝑛+𝒉^2 [(1)^2+(2)^2+ …+(𝑛−1)^2 ] +𝟐𝒉 [1+2+ …+(𝑛−1)]) =3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑛+ℎ^2 [𝑛(𝑛 − 1)(2𝑛 − 1)/6]+2ℎ[𝑛(𝑛 − 1)/2] ) We know that 1^2+2^2+ …+𝑛^2= (𝑛 (𝑛 + 1)(2𝑛 + 1))/6 1^2+2^2+ ……+(𝑛−1)^2 = ((𝑛 − 1) (𝑛 −1 + 1)(2(𝑛 − 1) + 1))/6 = ((𝑛 − 1) 𝑛 (2𝑛 − 2 + 1) )/6 = (𝑛 (𝑛 − 1) (2𝑛 − 1) )/6 We know that 1+2+3+ ……+𝑛= (𝑛 (𝑛 + 1))/2 1+2+3+ ……+(𝑛−1) = ((𝑛 − 1) (𝑛 − 1 + 1))/2 = (𝑛 (𝑛 − 1) )/2 =3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑛+ℎ^2 [𝑛(𝑛 − 1)(2𝑛 − 1)]/6+ℎ[𝑛(𝑛 − 1)] ) =3 (𝑙𝑖𝑚)┬(𝑛→∞) (𝑛/𝑛+ℎ^2 [𝑛(𝑛 − 1)(2𝑛 − 1)/6𝑛]+ℎ[𝑛(𝑛 − 1)/𝑛]) =3 (𝑙𝑖𝑚)┬(𝑛→∞) (1+ℎ^2 [(𝑛 − 1)(2𝑛 − 1)/6]+ℎ[(𝑛 − 1)]) =3 (𝑙𝑖𝑚)┬(𝑛→∞) (1+(3/𝑛)^2 (𝑛 − 1)(2𝑛 − 1)/6+(3/𝑛)(𝑛 − 1)) =3 (𝑙𝑖𝑚)┬(𝑛→∞) (1+9/𝑛^2 . (𝑛 − 1)(2𝑛 − 1)/6 +3(1 − 1/𝑛)) =3 (𝑙𝑖𝑚)┬(𝑛→∞) (1+ 9(1 − 1/𝑛)(2 − 1/𝑛)/6 +3(1 − 1/𝑛)) =3(1+ 9(1 − 1/∞)(2 − 1/∞)/6 +3(1 − 1/∞)) =3(1+ 9(1 − 0)(2 − 0)/6 +3(1 −0)) =3(1+ (9 × 1 × 2)/6 +3) =3(1+3+3) =3×7 =𝟐𝟏 Solving I2 ∫1_1^4▒〖𝑥 𝑑𝑥〗 Putting 𝑎 =1 𝑏 =4 ℎ=(𝑏 − 𝑎)/𝑛 =(4 − 1)/𝑛 =3/𝑛 𝑓(𝑥)=𝑥 We know that ∫1_𝑎^𝑏▒〖𝑥 𝑑𝑥〗 =(𝑏−𝑎) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(𝑎)+𝑓(𝑎+ℎ)+𝑓(𝑎+2ℎ)…+𝑓(𝑎+(𝑛−1)ℎ)) Hence we can write ∫1_1^4▒〖𝑥 𝑑𝑥〗 =(4−1) lim┬(n→∞) 1/𝑛 (𝑓(1)+𝑓(1+ℎ)+𝑓(1+2ℎ)+… +𝑓(1+(𝑛−1)ℎ) =3 lim┬(n→∞) 1/𝑛 (𝑓(1)+𝑓(1+ℎ)+𝑓(1+2ℎ)+… +𝑓(1+(𝑛−1)ℎ) Here, 𝑓(𝑥)=𝑥 𝑓(1)=1 𝑓(1+ℎ)=1+ℎ 𝑓 (1+2ℎ)=1+2ℎ 𝑓(1+(𝑛−1)ℎ)=1+(𝑛−1)ℎ Hence, our equation becomes ∫_1^4▒𝑥 𝑑𝑥 =3 lim┬(n→∞) 1/𝑛 (𝑓(1)+𝑓(1+ℎ)+𝑓(1+2ℎ)+… +𝑓(1+(𝑛−1)ℎ) = 3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (1+(1+ℎ)+(1+2ℎ)+ …+(1+(𝑛−1)ℎ)) = 3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (1+1+ …+1 +ℎ+2ℎ+ ……+(𝑛−1)ℎ) = 3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 ( 𝑛\ ×1+ℎ (1+2+ ………+(𝑛−1))) We know that 1+2+3+ ……+𝑛= (𝑛 (𝑛 + 1))/2 1+2+3+ ……+𝑛−1= ((𝑛 − 1) (𝑛 − 1 + 1))/2 = (𝑛 (𝑛 − 1) )/2 = 3 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 ( 𝑛+(ℎ . 𝑛(𝑛 − 1))/2) = 3 (𝑙𝑖𝑚)┬(𝑛→∞) ( 𝑛/𝑛+𝑛(𝑛 − 1)ℎ/2𝑛) = 3 (𝑙𝑖𝑚)┬(𝑛→∞) ( 1+(𝑛 − 1)ℎ/2) = 3 (𝑙𝑖𝑚)┬(𝑛→∞) ( 1+(𝑛 − 1)3/(2 . 𝑛)) = 3 (𝑙𝑖𝑚)┬(𝑛→∞) ( 1+(𝑛/𝑛 − 1/𝑛) 3/2) [𝑈𝑠𝑖𝑛𝑔 ℎ=3/𝑛] = 3 (𝑙𝑖𝑚)┬(𝑛→∞) ( 1+(1− 1/𝑛) (3 )/2) = 3( 1+(1− 1/∞) (3 )/2) = 3( 1+(1−0) 3/2) = 3(1+ (3 )/2) = 3((5 )/2) = 𝟏𝟓/𝟐 Putting the values of I1 and I2 in I ∴ "I = " ∫1_1^4▒〖 𝑥2 𝑑𝑥〗−∫1_1^4▒〖 𝑥 𝑑𝑥〗 = 21 − 15/2 = (42 − 15)/2 = 𝟐𝟕/𝟐