           1. Chapter 7 Class 12 Integrals
2. Serial order wise
3. Ex 7.8

Transcript

Ex 7.8, 4 1 4 2 Let I1 = 1 4 2 I1 = 1 4 2 1 4 Solving I2 I2 = 1 4 2 Putting =1 =4 = = 4 1 = 3 & = 2 Hence we can write it as 1 4 2 = 4 1 uc1 1 1 + 1+ + 1+2 + + 1+ 1 = 2 1 = 1 2 =1 1+ = 1+ 2 = 1 2 + 2 +2 1 1+2 = 1+2 2 = 1 2 + 2 2 +2 1 2 1+ 1 = 1+ 1 2 = 1 2 + 1 2 +2 1 1 Thus, 1 4 2 =3 uc1 1 ( 1 2 + 1 2 + 2 +2 1 + 1 2 + 2 2 +2 1 2 + 1 2 + 1 2 +2 1 1 ) =3 uc1 1 ( 1 2 + 1 2 + 1 2 + + 1 2 + 2 + 2 2 + 1 2 + 2 1 +2 1 2 + +2 1 1 ) =3 uc1 1 ( . 1 2 + 1 2 2 + 2 2 2 + + 1 2 2 + 2 +2 2 + 1 2 ) =3 uc1 1 [ + 2 1 2 + 2 2 + 1 2 +2 1+2+ + 1 ] =3 uc1 1 + 2 1 2 1 6 +2 1 2 =3 uc1 1 + 3 2 1 2 1 6 +2 3 ( 1) 2 =3 uc1 1 + 9 2 1 2 1 6 +3 ( 1) =3 uc1 1 + 3 2 1 2 1 2 +3 1 =3 uc1 1 + 3 2 1 1 2 1 +3 1 =3 uc1 + 3 2 1 1 2 1 +3 1 =3 uc1 1+ 3 2 1 1 1 2 1 +3 1 1 =3 uc1 1 + 3 2 uc1 1 1 2 1 + uc1 3 1 1 =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 Putting =1 =4 = = 4 1 = 3 & = Hence we can write it as 1 4 = 4 1 uc1 1 1 + 1+ + 1+2 + + 1+ 1 = 1 =1 1+ =1+ 1+2 =1+2 .. 1+ 1 =1+ 1 Thus, 1 4 =3 uc1 1 1+ 1+ + 1+2 + + 1+ 1 =3 uc1 1 1+1+1+ +1+ +2 + + 1 =3 uc1 1 + 1+2+ +( 1) =3 uc1 1 + 1 2 =3 uc1 + 1 2 =3 uc1 1+ 1 2 =3 uc1 1+ 3 1 2 =3 uc1 1+ 3 2 1 =3 uc1 1+ 3 2 1 1 =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 2 1 4 = 21 15 2 = 42 15 2 =

Ex 7.8

Chapter 7 Class 12 Integrals
Serial order wise 