# Some reassurance

• Jul 3rd 2006, 12:21 PM
macca101
Some reassurance
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
• Jul 3rd 2006, 12:46 PM
Jameson
• Jul 3rd 2006, 12:50 PM
Jameson
Also, if you're allowed to do these problems on the calculator I would suggest using the absolute value button. It's for you to note that total area means positive and negative area, not just the integral, but if you need a quick calculation, I know on the TI models you can enter $\displaystyle \int |f(x)|dx$
• Jul 3rd 2006, 01:18 PM
macca101
Thanks for for that Jameson :)
• Jul 3rd 2006, 02:46 PM
ThePerfectHacker
Quote:

Originally Posted by Jameson
Also, if you're allowed to do these problems on the calculator I would suggest using the absolute value button. It's for you to note that total area means positive and negative area, not just the integral, but if you need a quick calculation, I know on the TI models you can enter $\displaystyle |\int f(x)dx|$

No, Jameson it is.
$\displaystyle \int_a^b|f(x)|dx$