 Hello, Namaste, Myself, M. S. Bhastragam, Assistant Professor, Department of Humanities and Sciences, Alchanist of Technology, Sholapur. Now, lowering outcome, at the end of the session, students will be able to express the given function in four-year series in the interval minus l to l. Now, the four-year series of f of x in the interval minus l to l is given by f of x is equal to a0 plus summation of n equal to 1 to infinity an cos n pi x by l plus bn sin n pi x by l, where a0 is equal to 1 by 2l, integration from minus l to l f of x dx, and an is equal to 1 by l, integration from minus l to l f of x cos n pi x by l dx, and bn is equal to 1 by l, integration from minus l to l f of x sin n pi x by l dx. Now, these are the Euler's formulae to find the four-year constants a0, a9 and bn. Now, pause the video and find bn for the function f of x is equal to 0 when minus 5 is less than x is less than 0, and equal to 3 when 0 is less than x is less than 5. I hope you have completed here, f of x equal to 0 when minus 5 is less than x is less than 0, and equal to 3 when 0 is less than x is less than 5. Now, comparing minus 5 to 5 with minus l to l, here you get l is equal to 5. Now, bn is equal to 1 by l, integration from minus l to l f of x sin of n pi x by l dx, which is equal to here 1 by l means 1 by 5 l is 5, integration from minus 5 to 0, and here f of x is 0 that is 0 into sin n pi x by 5 dx plus integration from 0 to 5 in 0 to 5 f of x is 3. Therefore, here 3 into sin of n pi x by 5 dx, which is equal to 3 by 5 taking 3 common from the second term, and the value of the first integral is 0, because here your integrant is 0. Now, integration of sin of n pi x by 5 is minus 5 by n pi cos of n pi x by 5 with a limit 0 to 5, which is equal to 3 by 5. Now, here putting x is equal to 5 we get minus 5 by n pi cos of n pi, and minus up lower limit that is minus minus plus, and putting x equal to 0 here cos 0 is 1 that is plus 5 by n pi, which is equal to 3 by n pi into bracket 1 minus minus 1 raise to n. Therefore, bn is equal to 3 by n pi into bracket 1 minus minus 1 raise to n. Now, we will see the example, find the Fourier series of f of x equal to x when minus 1 is less than x is less than 0, and equal to x plus 2 when 0 is less than x is less than 1. Now, here comparing minus 1 to 1 with minus l to l we find that here l is 1. The Fourier series of f of x in the interval minus l to l is given by 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 bn sin n pi x by l. Here l is equal to 1 by putting l equal to 1 that is here we get f of x is 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, we will calculate the Fourier constant first of all a naught a naught is equal to here 1 by 2 l integration from minus l to l f of x dx which is equal to here l is 1 that is 1 by 2 and integration from minus 1 to 0 here f of x is x that is x dx plus integration from 0 to 1 f of x is x plus 2 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 minus 1 to 0 plus now integrating x plus 2 we get x square by 2 plus 2 x with the limit 0 to 1. Now, in the first part putting upper limit that is x equal to 0 we get 0 minus half lower limit that is x equal to minus 1 means minus 1 whole square by 2 plus now putting upper limit x equal to 1 that is 1 by 2 and plus 2 minus half lower limit putting x equal to 0 we get the value of the lower limit is 0. Now, which is equal to 1 by 2 into bracket this minus 1 by 2 plus 1 by 2 and plus 2 that is minus 1 by 2 and plus 1 by 2 will get cancelled that is 2 by 2 which is equal to 1. Now, we will calculate a n a n is equal to 1 by l integration from minus l to l f of x cos n pi x by l dx which is equal to here l is 1 that is 1 by 1 integration from minus 1 to 0 f of x is x that is x into cos of n pi x dx plus integration from 0 to 1 x plus 2 cos of n pi x dx. Now, first of all we will integrate the first integral taking u as a x and v as a cos n pi x by using generalized rule of integration by part. First keeping x as it is and integration of cos of n pi x is sin n pi x by n 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 minus 1 to 0 plus now integrating the second integral taking u as a x plus 2 and v as a cos n pi x. First we will keeping x plus 2 as it is and integration of cos n pi x is sin n pi x by n pi minus the derivative of x plus 2 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. Now, in the first part first we will put the upper limit x equal to 0 that is putting x equal to 0 in the first term as we know sin 0 is 0 and this into 0 that we get 0 and this minus minus plus we have cos 0 is 1 that you get 1 upon n square pi square. Now, minus of lower limit putting x equal to minus 1 that is sin of minus n pi 0 therefore, the first term is 0 and this minus minus plus we get cos of minus n pi that is what we get cos of n pi divided by n square pi square. Now, plus now considering the second term putting x equal to 1 again sin of n pi 0 therefore, the first term is 0. Now, minus into minus plus putting x equal to 1 we get cos of n pi by n square pi square. Now, minus of lower limit putting x equal to 0 because of sin 0 first term is 0 and this minus minus plus value of cos 0 is 1 that we get plus 1 by n square pi square which is equal to 1 by n square pi square and this minus 1 by n square pi square we will get cancel then minus cos n pi by n square pi square and plus cos n pi by n square pi square we will get cancel therefore, the value of n is 0. Now, we will calculate b n b n is equal to 1 by l integration from minus l to l f of x sin of n pi x by l dx which is equal to 1 by 1 integration from minus 1 to 0 x into sin n pi x dx plus integration from 0 to 1 x plus to sin of n pi x dx. Now, again here integrating by part by using the generalized rule of integration by part taking u as a x and v as a sin n pi x first we keep x as it is an integration of sin n pi x is minus cos n pi x by n pi minus the 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 minus 1 to 0. Now, integrating the second integral taking x plus 2 as a u and sin n pi x as a v first we keeping x plus 2 as it is an integration of sin of n pi x is minus cos n pi x by n pi minus the derivative of x plus 2 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, putting the upper limit first here in the first part putting x equal to 0 because of multiplication of x here first term is 0 then minus minus plus sin 0 is again 0. Therefore, by putting upper limit we get the value as a 0 then minus of lower limit that is minus. Now, putting x equal to minus 1 we get minus cos of n pi by n pi and minus minus plus putting x equal to minus 1 again what we get here is 0. Now, for the second part putting x equal to 1 here then we get 1 plus 2 that is 3 minus sin minus 3 by n pi cos of n pi this minus minus plus sin of n pi is 0. Now, minus of lower limit now putting x equal to 0 we get here minus 2 by n pi and cos 0 as a 1 and sin of 0 as 0. Now, which is equal to minus of minus 1 is to n by n pi because cos of n pi is minus 1 is to n then minus 3 by n pi into cos of n pi is minus 1 is to n and minus minus plus 2 by n pi now taking 2 by n pi common we get which is equal to 2 by n pi into bracket 1 minus 2 into minus 1 is to n. Now, substituting the values of a naught a n and b n we get f of x is equal to 1 plus 2 by pi summation of n equal to 1 to infinity 1 minus 2 into minus 1 is to n by n sin n pi x. Now, references higher engineering mathematics by Dr. B. S. Gravel. Thank you.