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