Hello,
I would be most grateful if someone could check my work.
$\displaystyle
f(x)= (x-3)(e^\frac{-x}{4}-1)
$
Find the area enclosed by the graph of this function and the x axis between x=0 and x=6.
As the graph crosses the x-axis at x=0 and x=3 all values of x are positive between x=0 and x=3 all values of x>3 are negative.
It is necessary to to find the positive area bounded by x=0 and x=3 and subtract it from the negative area bounded by x=3 and x=6.
So
$\displaystyle
\int^{3}_{0}(x-3)(e^\frac{-x}{4}-1)dx -\int^{6}_{3}(x-3)(e^\frac{-x}{4}-1)dx
$
After hopefully performing the integration correctly I get an answer of 4.1319