Ex 7.4, 23 - Chapter 7 Class 12 Integrals
Last updated at July 11, 2018 by Teachoo
Transcript
Ex 7.4, 23 5 + 3 2 + 4 + 10 5 + 3 2 + 4 + 10 =5 + 3 5 2 + 4 + 10 = 5 2 2 + 6 5 2 + 4 + 10 = 5 2 2 + 4 4 + 6 5 2 + 4 + 10 = 5 2 2 + 4 2 + 4 + 10 + 5 2 14 5 2 + 4 + 10 = 5 2 2 + 4 2 + 4 + 10 +7 2 + 4 + 10 Solving I1= 5 2 2 + 4 2 + 4 + 10 . Let 2 + 4 + 10= Diff both sides w.r.t.x 2 +4+0= = 2 + 4 Thus, our equation becomes I1= 5 2 2 + 4 2 + 4 + 10 . Putting the value of 2 +4 +10 = and = 2 + 4 I1= 5 2 2 + 4 . I1= 5 2 2 + 4 . 2 + 4 I1= 5 2 1 . I1= 5 2 1 1 2 . I1= 5 2 1 2 . I1= 5 2 1 2 + 1 1 2 + 1 + 1 I1= 5 2 1 2 1 2 + 1 I1=5 1 2 + 1 I1=5 + 1 I1=5 2 +4 +10 + 1 Solving I2= 7 2 + 4 + 10 . I2=7 1 2 + 2 2 + 10 . I2=7 1 2 + 2 2 + 2 2 2 2 + 10 . I2=7 1 + 2 2 2 2 + 10 . I2=7 1 + 2 2 4 + 10 . I2=7 1 + 2 2 + 6 . I2=7 1 + 2 2 + 6 2 . I2=7 +2+ +2 2 + 6 2 + 2 I2=7 +2+ 2 +4 +4+6 + 2 I2=7 +2+ 2 +4 +10 + 2 Putting the values of I1 and I2 in (1) 5 + 3 2 + 4 + 10 . = 5 2 2 + 4 2 + 4 + 10 . 7 2 + 4 + 10 . =5 2 +4 +10 + 1 7 +2+ 2 +4 +10 + 2 = + + + + + + +
Ex 7.4
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Ex 7.4, 23 You are here
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About the Author
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.