SOLUTION: Please help me solve this equation: Consider the function {{{ f(x) = e^(-x)/(1-e^-5) }}} {{{ 0<=x<=5 }}} a) Verify that {{{ int( f(x), dx, 0, 5 ) =1 }}} b) Find {{{ int ( x f(

Algebra ->  Test -> SOLUTION: Please help me solve this equation: Consider the function {{{ f(x) = e^(-x)/(1-e^-5) }}} {{{ 0<=x<=5 }}} a) Verify that {{{ int( f(x), dx, 0, 5 ) =1 }}} b) Find {{{ int ( x f(      Log On


   

Question 355597: Please help me solve this equation:
Consider the function +f%28x%29+=+e%5E%28-x%29%2F%281-e%5E-5%29+ +0%3C=x%3C=5+
a) Verify that +int%28+f%28x%29%2C+dx%2C+0%2C+5+%29+=1+
b) Find +int+%28+x+f%28x%29%2C+dx%2C+0%2C+5+%29+
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
+f%28x%29+=+e%5E%28-x%29%2F%281-e%5E%28-5%29%29+ +0%3C=x%3C=5+
For both parts it helps to recognize that the denominator of f(x) is a constant. So we can rewrite the function as:
+f%28x%29+=+%281%2F%281-e%5E%28-5%29%29%29%2Ae%5E%28-x%29

a) Verify that +int%28+f%28x%29%2C+dx%2C+0%2C+5+%29+=1+
+int%28+%281%2F%281-e%5E%28-5%29%29%29%2Ae%5E%28-x%29%2C+dx%2C+0%2C+5+%29
We can factor out the constant from the integrand:
+%281%2F%281-e%5E%28-5%29%29%29%2Aint%28+e%5E%28-x%29%2C+dx%2C+0%2C+5+%29
Now the integral is easy to find:
+%281%2F%281-e%5E%28-5%29%29%29%2A%28-e%5E%28-x%29%29+ evaluated from 0 to 5:
+%281%2F%281-e%5E%28-5%29%29%29%2A%28-e%5E%28-5%29+-+%28-e%5E%28-0%29%29%29+
since e%5E-0+=+1
+%281%2F%281-e%5E%28-5%29%29%29%2A%28-e%5E%28-5%29+-+%28-1%29%29+
+%281%2F%281-e%5E%28-5%29%29%29%2A%28-e%5E%28-5%29+%2B+1%29+
which simplifies to 1 (since %28-e%5E%28-5%29+%2B+1%29+=+%281-e%5E%28-5%29%29%29.

b) Find +int+%28+x+f%28x%29%2C+dx%2C+0%2C+5+%29+
+int+%28+x+%2A%281%2F%281-e%5E-5%29%29%2Ae%5E%28-x%29%2C+dx%2C+0%2C+5+%29+
Again we can factor out the constant:
+%281%2F%281-e%5E-5%29%29%2Aint+%28+x+%2Ae%5E%28-x%29%2C+dx%2C+0%2C+5+%29+
This integral is a job for Integration by Parts. If u = x and dv = e%5E%28-x%29%2Adx. This makes v = -e%5E%28-x%29 and du = dx. Substituting u and dv we get:
+%281%2F%281-e%5E-5%29%29%2Aint+%28+u%2C+dv%2C+0%2C+5+%29+
(Technically the boundary values, 0 to 5, change, too. But since I am going to substitute back in for u and v I am not going to spend the time to make this adjustment. You, however, should not skip this step.)
By the rule for Integration by Parts, int%28u%2C+dv%29+=+u%2Av+-+int%28v%2C+du%29, we get:

The remaining integral is easy:
+%281%2F%281-e%5E%28-5%29%29%29%2A%28x%2A%28-e%5E%28-x%29%29+-+e%5E%28-x%29%29
Evaluating this from 0 to 5:


+%281%2F%281-e%5E%28-5%29%29%29%2A%28%28-5%2A%28e%5E%28-5%29%29+-+e%5E%28-5%29%29+%2B+1%29
+%281%2F%281-e%5E%28-5%29%29%29%2A%28-6%2A%28e%5E%28-5%29%29+%2B+1%29
+%28-6%2A%28e%5E%28-5%29%29+%2B+1%29%2F%281-e%5E%28-5%29%29
If we multiply the numerator and denominator by e%5E5 we get:
+%28-6+%2B+e%5E5%29%2F%28e%5E5+-+1%29