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

Slide16.JPG
Slide17.JPG Slide18.JPG Slide19.JPG Slide20.JPG Slide21.JPG Slide22.JPG Slide23.JPG Slide24.JPG Slide25.JPG

  1. Class 12
  2. Important Question for exams Class 12
Ask Download

Transcript

Ex 7.8, 4 ﷐1﷮4﷮﷐𝑥2 −𝑥﷯𝑑𝑥﷯ Let I1 = ﷐1﷮4﷮﷐𝑥2 −𝑥﷯𝑑𝑥﷯ I1 = ﷐1﷮4﷮ 𝑥2 𝑑𝑥﷯−﷐1﷮4﷮ 𝑥 𝑑𝑥﷯ Solving I2 I2 = ﷐1﷮4﷮ 𝑥2 𝑑𝑥﷯ Putting 𝑎 =1 𝑏 =4 ℎ= ﷐𝑏 − 𝑎﷮𝑛﷯ = ﷐4 − 1﷮𝑛﷯ = ﷐3﷮𝑛﷯ & 𝑓﷐𝑥﷯=﷐𝑥﷮2﷯ Hence we can write it as ﷐1﷮4﷮ 𝑥2 𝑑𝑥﷯ =﷐4−1﷯﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑓﷐1﷯+𝑓﷐1+ℎ﷯+𝑓﷐1+2ℎ﷯+ …+𝑓﷐1+﷐𝑛−1﷯ℎ﷯﷯ 𝑓﷐𝑥﷯=﷐𝑥﷮2﷯ 𝑓﷐−1﷯=﷐﷐1﷯﷮2﷯=1 𝑓﷐1+ℎ﷯=﷐﷐1+ℎ﷯﷮2﷯ =﷐﷐1﷯﷮2﷯+﷐ℎ﷮2﷯+2﷐1﷯ ℎ 𝑓 ﷐1+2ℎ﷯=﷐﷐1+2ℎ﷯﷮2﷯=﷐﷐1﷯﷮2﷯+﷐﷐2ℎ﷯﷮2﷯+2﷐1﷯ 2ℎ 𝑓﷐−1+﷐𝑛−1﷯ℎ﷯=﷐﷐1+﷐𝑛−1﷯ ℎ﷯﷮2﷯ =﷐﷐1﷯﷮2﷯+﷐﷐﷐𝑛−1﷯ℎ﷯﷮2﷯+2﷐1﷯﷐𝑛−1﷯ℎ Thus, ﷐1﷮4﷮ 𝑥2 𝑑𝑥﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ (﷐﷐1﷯﷮2﷯+﷐﷐1﷯﷮2﷯+﷐﷐ℎ﷯﷮2﷯+2﷐1﷯﷐ℎ﷯+ ﷐﷐1﷯﷮2﷯+﷐﷐2ℎ﷯﷮2﷯+2﷐1﷯﷐2ℎ﷯+ …… ﷐﷐1﷯﷮2﷯+﷐﷐﷐𝑛−1﷯ℎ﷯﷮2﷯+2﷐1﷯﷐﷐𝑛−1﷯ℎ﷯) =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ (﷐﷐1﷯﷮2﷯+﷐﷐1﷯﷮2﷯+﷐﷐1﷯﷮2﷯+ ……+﷐﷐1﷯﷮2﷯+ ﷐﷐ℎ﷯﷮2﷯+﷐﷐2ℎ﷯﷮2﷯+ ……﷐﷐﷐𝑛−1﷯ℎ﷯﷮2﷯+ 2﷐1﷯﷐ℎ﷯+2﷐1﷯﷐2ℎ﷯+ ……+2﷐1﷯﷐﷐𝑛−1﷯ℎ﷯) =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ (﷐𝑛 .﷐1﷯﷮2﷯+﷐﷐1﷯﷮2﷯﷐ℎ﷮2﷯+﷐﷐2﷯﷮2﷯﷐ℎ﷮2﷯+ ……+﷐﷐𝑛−1﷯﷮2﷯﷐ℎ﷮2﷯+ 2ℎ+2﷐2ℎ﷯+ ……﷐𝑛−1﷯﷐2ℎ﷯) =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ [𝑛+﷐ℎ﷮2﷯﷐﷐1﷮2﷯+﷐2﷮2﷯+ ……﷐﷐𝑛−1﷯﷮2﷯﷯ +2ℎ﷐1+2+ ……+﷐𝑛−1﷯﷯ ] =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑛+﷐ℎ﷮2﷯﷐﷐𝑛﷐𝑛 − 1﷯﷐2𝑛 − 1﷯﷮6﷯﷯+2ℎ﷐﷐𝑛﷐𝑛 − 1﷯﷮2﷯﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑛+﷐﷐﷐3﷮𝑛﷯﷯﷮2﷯﷐﷐𝑛﷐𝑛 − 1﷯﷐2𝑛 − 1﷯﷮6﷯﷯+2﷐﷐3﷮𝑛﷯﷯﷐﷐𝑛(𝑛 − 1)﷮2﷯﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑛+﷐9﷮﷐𝑛﷮2﷯﷯﷐﷐𝑛﷐𝑛 − 1﷯﷐2𝑛 − 1﷯﷮6﷯﷯+3﷐﷐𝑛(𝑛 − 1)﷮𝑛﷯﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑛+﷐3﷮2﷯﷐﷐𝑛﷐𝑛 − 1﷯﷐2𝑛 − 1﷯﷮﷐𝑛﷮2﷯﷯﷯+3﷐𝑛−1﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑛+﷐3﷮2﷯﷐﷐𝑛﷮𝑛﷯﷐1 − ﷐1﷮𝑛﷯﷯﷐2𝑛−1﷯﷯+3﷐𝑛−1﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐﷐𝑛﷮𝑛﷯+﷐3﷮2﷯﷐﷐𝑛﷮𝑛﷯﷐1 − ﷐1﷮𝑛﷯﷯﷐﷐2𝑛 − 1﷮𝑛﷯﷯﷯+3﷐﷐𝑛−1﷮𝑛﷯﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1+﷐3﷮2﷯﷐1﷐1 − ﷐1﷮𝑛﷯﷯﷐2−﷐1﷮𝑛﷯﷯+3﷐1 − ﷐1﷮𝑛﷯﷯﷯﷯ =3﷐﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷯+﷐3﷮2﷯﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1 − ﷐1﷮𝑛﷯﷯﷐2−﷐1﷮𝑛﷯﷯+﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ 3﷐1 − ﷐1﷮𝑛﷯﷯﷯ =3﷐1+﷐3﷮2﷯﷐1 − ﷐1﷮𕔴uc1﷯﷯﷐2−﷐1﷮𕔴uc1﷯﷯+3﷐1 − ﷐1﷮𕔴uc1﷯﷯﷯ =3﷐1+﷐3﷮2﷯﷐1 −0﷯﷐2−0﷯+3﷐1 −0﷯﷯ =3﷐1+﷐3﷮2﷯﷐1﷯﷐2﷯+3﷐1﷯﷯ =3﷐7﷯ =21 Solving I3 I3 = ﷐1﷮4﷮ 𝑥 𝑑𝑥﷯ Putting 𝑎 =1 𝑏 =4 ℎ= ﷐𝑏 − 𝑎﷮𝑛﷯ = ﷐4 − 1﷮𝑛﷯ = ﷐3﷮𝑛﷯ & 𝑓﷐𝑥﷯=𝑥 Hence we can write it as ﷐1﷮4﷮ 𝑥 𝑑𝑥﷯ =﷐4−1﷯﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑓﷐1﷯+𝑓﷐1+ℎ﷯+𝑓﷐1+2ℎ﷯+ …+𝑓﷐1+﷐𝑛−1﷯ℎ﷯﷯ 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 Thus, ﷐1﷮4﷮ 𝑥 𝑑𝑥﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐1+﷐1+ℎ﷯+﷐1+2ℎ﷯+ ……+﷐1+﷐𝑛−1﷯ℎ﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐1+1+1+ ……+1+ℎ+2ℎ+ ……+﷐𝑛−1﷯ℎ﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐𝑛+ℎ﷐1+2+ ……+(𝑛−1)﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑛+﷐ℎ﷐𝑛﷯﷐𝑛 − 1﷯﷮2﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐﷐𝑛﷮𝑛﷯+﷐ℎ﷐𝑛﷯﷐𝑛 − 1﷯﷮2𝑛﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1+﷐ℎ ﷐𝑛 − 1﷯﷮2﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1+﷐3﷮𝑛﷯ ﷐﷐𝑛 − 1﷮2﷯﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1+﷐3﷮2﷯ ﷐﷐𝑛 − 1﷮𝑛﷯﷯﷯ =3﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1+﷐3﷮2﷯ ﷐1−﷐1﷮𝑛﷯﷯﷯ =3﷐1+﷐3﷮2﷯ ﷐1−﷐1﷮𕔴uc1﷯﷯﷯ =3﷐1+﷐3﷮2﷯ ﷐1−0﷯﷯ =3﷐1+﷐3﷮2﷯﷯ =3﷐﷐5﷮2﷯﷯ =﷐15﷮2﷯ Putting the values of I2 and I3 in I1 ∴ I1 = ﷐1﷮4﷮ 𝑥2 𝑑𝑥﷯−﷐1﷮4﷮ 𝑥 𝑑𝑥﷯ = 21 − ﷐15﷮2﷯ = ﷐42 − 15﷮2﷯ = ﷐𝟐𝟕﷮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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.