1. ## Sketching the Graph of an Interval (Help Please)

Given first sketch the graph y = f(x) on the given interval.
Then find the integral of
f using your knowledge of area formulas for

rectangles, triangles and circles.

2. Originally Posted by evansf
Given first sketch the graph y = f(x) on the given interval.

Then find the integral of
f using your knowledge of area formulas for
rectangles, triangles and circles.

$\int_{-2}^4 |3 - |3x||dx = \int_{-2}^4 |3 - 3|x||dx = \int_{-2}^4 3|1 - |x||dx$.. now, this can be separated to

$\int_{-2}^4 |3 - |3x||dx = ... = \int_{-2}^0 3|1 - |x||dx + \int_{0}^4 3|1 - |x||dx$..

we know, in the first addend, x is negative, hence we take |x| = -x; on the other hand, in the second addend, x is positive hence |x| = x

so the integral would be

$\int_{-2}^0 3|1 - |x||dx + \int_{0}^4 3|1 - |x||dx = \int_{-2}^0 3|1 + x|dx + \int_{0}^4 3|1 - x|dx$

now, we have to check where |1-x| is positive or negative and |1+x| is positive or negative..

for |1-x|, we take 1-x if 1-x >= 0 or x <= 1; we take x-1 otherwise

for |1+x|, we take 1+x if 1+x >= 0 or x >= -1; we take -x-1 otherwise

so now,

$... = \int_{-2}^{-1} 3(-1 - x)dx + \int_{-1}^{0} 3(1 + x)dx + \int_{0}^1 3(1 - x)dx + \int_{1}^4 3(x - 1)dx$

so can you do it now??

3. ## How do you integrate it?

...

4. they are just linear functions, then they can easily be integrated.. Ü

5. Originally Posted by evansf
...
You might try posting a message in the message box, rather than just in the title.

Take a look at your graph. The integral is the total area between the function and the x axis. (Areas under the x axis count as negative.) As your function is composed of sections of straight lines, your areas are triangles. Easy to find the area of these.

-Dan