Homework Sourseintegral minf ex2 12 solve for mSolutionintegral minf ex2
integral [m,inf] e^((-x)/2) = 1/2
solve for m
solve for m
Solution
integral [m,inf] e^((-x)/2) = 1/2
integration gives us e^-x/2 /(-1/2)
putting limits m and infinity we get
2e^-m/2 = 1/2
so -m/2 = -ln4
or m= 2.77
let me know if you have any confusion
![integral [m,inf] e^((-x)/2) = 1/2 solve for mSolutionintegral [m,inf] e^((-x)/2) = 1/2 integration gives us e^-x/2 /(-1/2) putting limits m and infinity we get](/WebImages/20/integral-minf-ex2-12-solve-for-msolutionintegral-minf-ex2-1044826-1761543321-0.webp "integral [m,inf] e^((-x)/2) = 1/2 solve for mSolutionintegral [m,inf] e^((-x)/2) = 1/2 integration gives us e^-x/2 /(-1/2) putting limits m and infinity we get")
