Question

Given f(x) = e^(-x) where x is greater than or equal to zero...and a function g(x) such that for every integer k greater than or equal to zero , its graph is a straight line joining [k,f(k)] and [k+1,f(k+1)] ... find the area between the two curves f(x) and g(x) ?

Answer 1

please help...ideas would definitely help...:)

Answer 2

(y - f(k))/(x -k) = [f(k+1)-f(k)]/

Answer 3

find out y which is g(x)

Answer 4

put in the values of f(k+1) and f(k)

Answer 5

we get \[g(x)=(x-k)(e^{-k-1}-e^{-k})+e^{-k}\]
But g(x) has a different definition for every interval [k,k+1)...how will u use it to integrate then?

Answer 6

now evaluate f(x) - g(x)

Answer 7

g(x) will have a different definition in [0,1), different in [1.2) and so on...but u need to integrate g(x)-f(x) from (0,+infty)...how will u do so with this changing g(x)?

Answer 8

just take a general definiton from k to k+1, then put k as 0, then evaluate the sum frm 0 to 1, then put it as 1 and evaluate frm 1 to 2..this way ull get a series or something

Answer 9

do u get a series?

Answer 10

maybe

Answer 11

do and see...

Answer 12

u do it..ur question

Answer 13

i dont get a series...do u get?

Answer 14

1st term: \[\frac{3}{2e}-\frac{1}{2}\]
2nd term: \[\frac{1}{e}\times(\frac{3}{2e}-\frac{1}{2})\]
yups..thnx thats a geometric series... :)