 Hello, Namaste, Myself, MS Bhasargaon, Assistant Professor, Department of Humanities and Sciences, Walsh Chandrishnath Technology, Solapur. Now learning outcome, at the end of this session students will be able to express the given function in 4-year series in the interval 0 to 2L. Now the 4-year series of f of x in the interval 0 to 2L is given by Euler's formulae that is f of x is equal to a naught plus summation of n equal to 1 to infinity a n cos n pi x by L plus b n sin of n pi x by L, where a naught is equal to 1 by 2L integration from 0 to 2L f of x dx and a n is equal to 1 by L integration from 0 to 2L f of x cos of n pi x by L dx and b n is equal to 1 by L integration from 0 to 2L f of x sin n pi x by L dx. Now pause the video for a while and find the 4-year constant a n for f of x equal to 1 by 2 into bracket pi minus x in the interval 0 to 2. I hope you have completed. Now if you compare with 0 to 2L here 2L is equal to 2 therefore L is equal to 1. Now we can calculate a n is equal to 1 by L integration from 0 to 2L f of x cos of n pi x by L dx which is equal to here L is 1 therefore integration from 0 to 2L f of x is 1 by 2 into bracket pi minus x cos of n pi x dx. Now using generalized law of integration by part taking u as a pi minus x and u as a cos n pi x we get taking 1 by 2 outside which is equal to 1 by 2 pi minus x as it is an integration of cos of n pi x sin n pi x by n pi minus the derivative of pi minus x is minus 1 into integration of sin n pi x by n pi is minus cos n pi x by n square pi square with a limit 0 to 2. Now putting the upper limit x equal to 2 here sin up to n pi is 0 therefore first term is 0 that is minus into minus into minus that you get minus cos up to n pi by n square pi square minus of lower limit again here sin 0 is 0 therefore first term is 0 and putting here x equal to 0 your cos 0 minus cos 0 upon n square pi square which is equal to 1 half into bracket cos up to n pi is 1 that is minus 1 upon n square pi square and this minus minus plus cos 0 is 1 that is 1 by n square pi square which is equal to 0 therefore the value of n is 0. Now we will see one more example find the Fourier series of f of x equal to x when 0 is less than x is less than 1 and equal to 1 minus x when 1 is less than x is less than 2. Now here you can compare the limit with 0 to 2 L here 2 L is equal to 2 therefore L is equal to 1. Now the Fourier series of f of x with period 2 L is equal to 2 is given by now putting here L equal to 1 in this general formula we get f of x equal to a naught plus summation of n equal to 1 to infinity a n cos n pi x plus b n sin n pi x. Now where a naught is equal to 1 by 2 L integration from 0 to 2 L f of x dx which is equal to 1 by 2 in the limit 0 to 1 f of x is x that is integration from 0 to 1 x dx plus integration from 1 to 2 f of x is 1 minus x that is 1 minus x dx. Now integrating with respect to x which is equal to 1 by 2 integration of x is x square by 2 with the limit 0 to 1 plus integration of 1 minus x is x minus x square by 2 with the limit 1 to 2 which is equal to 1 half putting x equal to 1 that you get 1 half and for minus lower limit you get 0 plus now putting x equal to 2 here 2 minus 4 by 2 means 2 2 minus 2 minus half lower limit that is putting x equal to 1 1 minus 1 by 2 which is equal to 1 by 2 into bracket you get 1 by 2 and simplification of this is 1 by 2 that is 1 by 2 minus 1 by 2 which is equal to 0. Therefore, here a naught is 0 now we will calculate a n a n is equal to 1 by L integration from 0 to 2 L f of x cos n pi x by L dx which is equal to 1 by 1 integration from 0 to 1 x cos n pi x dx plus integration from 1 to 2 1 minus x cos of n pi x dx. Now again here using integration by part which is equal to keeping x as it is and integration of cos n pi x is sin n pi x by n pi minus the derivative of x is 1 and integration of sin of n pi x by n pi is minus cos n pi x by n square pi square with the limit 0 to 1 plus now we will integrate the second integral keeping 1 minus x as it is and integration of cos n pi x is sin n pi x by n pi minus the derivative of 1 minus x is minus 1 and integration of sin n pi x by n pi is minus cos n pi x by n square pi square with a limit 1 to 2. Now we will put the limit for the first term that is upper limit x equal to 1 if you put x equal to 1 you get sin of n pi 0 therefore first term is 0 that is minus minus plus cos of n pi by n square pi square minus of lower limit putting x equal to 0 again first term is 0 and this minus minus plus cos 0 is 1 that is 1 by n square pi square plus now the second term putting x equal to 2 again here we get sin of 2 n pi that is 0 and minus into minus into minus we get minus then cos of 2 n pi by n square pi now minus of lower limit putting x equal to 1 the first term will be 0 because 1 minus 1 and sin of n pi is also 0 then minus into minus into minus that we get the minus sign putting x equal to 1 means cos of n pi by n square pi square now which is equal to by simplifying here cos of n pi is minus 1 raise to n that is minus 1 raise to n by n square pi square minus 1 by n square pi square and cos of 2 n pi is 1 therefore we get minus 1 by n square pi square and this minus minus plus and cos of n pi is minus 1 raise to n by n square pi square which is equal to by simplifying we get 2 by n square pi square into bracket first we write minus 1 raise to n minus 1 now b n is equal to 1 by l integration from 0 to 2 l f of x sin n pi x by l dx which is equal to 1 by 1 integration from 0 to 1 x sin n pi x dx plus integration from 1 to 2 1 minus x into sin n pi x dx again using integration by part keeping x as it is and integration of sin of n pi x is minus cos n pi x by n pi minus derivative of x is 1 and integration of minus cos n pi x by n pi is minus sin n pi x by n square pi square with the limit 0 to 1 now in the second integral keeping 1 minus x as it is and integration of sin n pi x is minus cos n pi x by n pi minus the derivative of 1 minus x is minus 1 and integration of minus cos n pi x by n pi is minus sin n pi x by n square pi square with the limit 1 to 2 now putting x upper limit x equal to 1 you get 1 into minus cos n pi by n pi minus minus plus sin of n pi is 0 minus of lower limit we have to put x equal to 0 because here multiplied by x is there therefore first term is 0 and sin 0 is also 0 therefore lower limit will be 0 plus now second term putting x equal to 2 that is 1 minus 2 that is minus 1 minus cos of 2 n pi by n pi and minus sin of 2 n pi is 0 and now minus of lower limit now putting x equal to 1 first term is 0 because 1 minus 1 is 0 and sin of n pi is 0 now simplifying this which is equal to get minus cos of n pi is minus 1 raise to n divide by n pi and this minus minus plus cos of 2 n pi is 1 that is you get plus 1 by n pi taking 1 by n pi common you get which is equal to 1 by n pi into bracket 1 minus minus 1 raise to n now substituting the values of a naught a n and b n you get f of x is equal to 2 by pi square summation of n equal to 1 to infinity 1 by n square into bracket minus 1 raise to n minus 1 cos n pi x plus 1 by pi into summation of n equal to 1 to infinity 1 by n into bracket 1 minus minus 1 raise to n sin n pi x now references higher engineering mathematics by Dr. B. S. Grievall. Thank you.