wouldn't letters d, e, and f be the same answer??? since your taking the area over the same interval???
Hello, asilvester635!
Wouldn't letters d, e, and f be the same answer?
. . Certainly not!
Did you realize that some answers are negative?
Didn't you make any sketches?
$\displaystyle (e)\;\int^6_{\text{-}4} |f(x)|\,dx$
All the regions below the x-axis are reflected upward!
We want the area of the shaded region.Code:2| ..* * *.. * *:::::|:::::* *:::* *.. *:::::::|:::::::* *:::::::* ::::*.. *::::::::|::::::::* *:::::::::::* ::::::::*. ::::::::::|:::::::::: *:::::::::::::::* - - + - - - - - * - - - - * - - - - * - - - - + - - - - * - - -4 -2 | 2 4 6 |
$\displaystyle (f)\;\int^6_{\text{-}4}[f(x)+2]\,dx$
The entire graph is moved 2 units upward!
We want the area under the graph.Code:| 4+ * | *:::* | *:::::::* | *:::::::::::* | *:::::::::::::::* .* 2+ *:::::::::::::::::::* .*:::| | |:::::::::::::::::::| .*:::::::|* | *|:::::::::::::::::::| *:::::::::::|:*. | .*:|:::::::::::::::::::| |:::::::::::|:::*.. | ..*:::|:::::::::::::::::::| - - + - - - - - + - - - * * * - - - + - - - - + - - - - + - - -4 -2 | 2 4 6 |
Got it?
holy crap... yes i got it.... do we minus the area if it's below the x-axis??? like for the small triangle's area under the x-axis is 1 and he semicircle's area is 2pi and that big triangle above the x-axis area is 4....
Hello, asilvester635!
Do we subtract the area if it's below the x-axis?
Yes . . . and that is the "shortcut" way of looking at it.
Actually, since you are working with areas,
. . you should have been aware of this long ago.
Consider the area bounded by: $\displaystyle y \:=\:2x-4$ and the coordinate axes.
The integral is: .$\displaystyle A \;=\;\int^2_0(2x-4)\,dx$Code:| | / ----+-----*---- |::::/2 |:::/ |::/ |:/ |/ * /| |
We have: .$\displaystyle x^2 - 4x\,\bigg|^2_0 \;=\;(4-8) - (0-0) \;=\;{\color{red}-}4$ . (negative four)
If the region is below the x-axis, the value of the integral is negative.
. . (You can call it "negative area" if you like.)