Question 355597
{{{ f(x) = e^(-x)/(1-e^(-5)) }}} {{{ 0<=x<=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(x) = (1/(1-e^(-5)))*e^(-x)}}}<br>
a) Verify that {{{ int( f(x), dx, 0, 5 ) =1 }}}
{{{ int( (1/(1-e^(-5)))*e^(-x), dx, 0, 5 )}}}
We can factor out the constant from the integrand:
{{{ (1/(1-e^(-5)))*int( e^(-x), dx, 0, 5 )}}}
Now the integral is easy to find:
{{{ (1/(1-e^(-5)))*(-e^(-x)) }}} evaluated from 0 to 5:
{{{ (1/(1-e^(-5)))*(-e^(-5) - (-e^(-0))) }}}
since {{{e^-0 = 1}}}
{{{ (1/(1-e^(-5)))*(-e^(-5) - (-1)) }}}
{{{ (1/(1-e^(-5)))*(-e^(-5) + 1) }}}
which simplifies to 1 (since {{{(-e^(-5) + 1) = (1-e^(-5)))}}}.<br>
b) Find {{{ int ( x f(x), dx, 0, 5 ) }}}
{{{ int ( x *(1/(1-e^-5))*e^(-x), dx, 0, 5 ) }}}
Again we can factor out the constant:
{{{ (1/(1-e^-5))*int ( x *e^(-x), dx, 0, 5 ) }}}
This integral is a job for Integration by Parts. If u = x and dv = {{{e^(-x)*dx}}}. This makes v = {{{-e^(-x)}}} and du = dx. Substituting u and dv we get:
{{{ (1/(1-e^-5))*int ( u, dv, 0, 5 ) }}}
(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(u, dv) = u*v - int(v, du)}}}, we get:
{{{ (1/(1-e^(-5)))*(x*(-e^(-x)) - int ( (-e^(-x)), dx, 0, 5 )) }}}
The remaining integral is easy:
{{{ (1/(1-e^(-5)))*(x*(-e^(-x)) - e^(-x))}}}
Evaluating this from 0 to 5:
{{{ (1/(1-e^(-5)))*((5*(-e^(-5)) - e^(-5)) - (0*(-e^(-0)) - e^(-0)))}}}
{{{ (1/(1-e^(-5)))*((-5*(e^(-5)) - e^(-5)) - (-1))}}}
{{{ (1/(1-e^(-5)))*((-5*(e^(-5)) - e^(-5)) + 1)}}}
{{{ (1/(1-e^(-5)))*(-6*(e^(-5)) + 1)}}}
{{{ (-6*(e^(-5)) + 1)/(1-e^(-5))}}}
If we multiply the numerator and denominator by {{{e^5}}} we get:
{{{ (-6 + e^5)/(e^5 - 1)}}}