Show that the line integral is independent of path and evaluate the integral: C

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integ (1-ye^-x)dx + e^-x dy in C from (0,1) to (1,2) 1. Consider a direct path. -------------------------------------- y = x + 1 => dy = dx integ (1 - (x+1)e^-x ) dx + e^-x dx from x = 0 to 1 integ 1 - xe^-x dx from 0 to 1 let -x = t => -dx = dt integ -(1 + te^t) dt from 0 to -1 - (t + e^t(t-1)) from 0 to -1 [-(-1 + e^(-1)(-1-1))] - [-(0 + e^0(0-1))] [1 + 2e^-1]+[-1] = 2/e 2. Consider it from a. (0,1) ---> (0,2) , b. (0,2) ----> (1,2) a. x = 0 line ----------------- dx = 0 [since no change in x] integ e^0 dy from y = 1 to 2 = integ dy from y = 1 to 2 = y from 1 to 2 = 2-1 = 1 b. y = 2 line ------------------- dy = 0 , as no change in y I = integ (1 - 2e^-x) dx from x = 0 to 1 = x + 2e^-x from 0 to 1 = 1 + 2/e - 2 = 2/e - 1 Total I = a + b = 1 + 2/e - 1= 2/e so we see no matter what path we take we get the same integral i.e '2/e' hence its path independent

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