ah one more integral

[legacyofpiracy]

sorry bout this, just wanted to check my work with someone here because the homework is telling me I did something wrong. I followed Stapel's method (two posts down) on a similar problem and I must have mixed something up.

 8
S  |(2x-6)|  h<0
 h

alright so i graphed this function on my calculator and set in the limits to the best of my knowledge

         :                10
    | \  :             /|
    |   \:          /   |   
    |    :\       /     |
    |    :  \  /        |
<-----:---------------->
    k   :    3          8             
         :

so i determined the formula of the right triangle to be 1/2((5)(10)) and the left to be 1/2((k-3)(|2k-6|)

so then i just added these two formula's together, but i am told this is wrong. Can anyone see where i went wrong?

 

[stapel]

...Stapel's method (two posts down)

Since the posts' positions fluctuate according to "freshness", it might be useful to post replies within the originating thread, or else post a link to that thread. Thank you.

 8
S  |(2x-6)|  h<0
 h

I'm sorry, but I don't understand. Shouldn't the integral end with "dx", not "h < 0"? What are the instructions for the exercise?

Eliz.

 

[pka]

The anti-derivative of \(\displaystyle 2x - 6\) is \(\displaystyle F(x) = x^2 - 6x\).
Thus, if h<0 then \(\displaystyle \int\limits_h^8 {\left| {2x - 6} \right|}dx = \int\limits_h^3 {\left( { - 2x + 6} \right)dx} + \int\limits_3^8 {\left( {2x - 6} \right)dx} = F(h) + F(8) - 2F(3)\)

 

[legacyofpiracy]

whoops sorry I forgot to add the dx to the end of the function. The h<0 is separate :roll:

also my apologies for the separate post, I will remember next time

 

[Gene]

From what you drew and wrote Staple would have used 3-x, not x-3. If k (or h?) = -1 you would get a negative area for your first term. You want it to be negative only if k>3.