# Math Help - Area between Curves

1. ## Area between Curves

Hey I have a calc exam Thursday and have been practicing for it. I came across this area problem that has me stumped, usually area is no problem for me but I can't seem to figure this one out.

I figured that you would break the area into two from 0 to 1 and 1 to 3 since the function and y=127 intersect at x=1 but when I went to integrate everything the computer told me that my answer was wrong. Please help

2. The question, as posted, doesn't make sense to me. The graph of the cubic, the x-axis, and x= 1 form a single region. I don't see what "y= 127" has to do with that.

3. Hello, rawkstar!

Since you didn't show your work, we have no idea where your error is.

$\text{Consider the region bounded by the graph of:}$
$y\:=\:x^3 + 18x^2 + 108x,\;(0 \le x \le 1)\,\text{ on the left,}$
$x\text{-axis below, }\,y = 127\text{ above, and }x = 3\text{ on the right.}$

$\text{(1) Find the area of the region.}$

The graph looks like this:

Code:

|
127+       * - - - - - - - *
|      .|:::::::::::::::|
|     .*|:::::::::::::::|
|   .*B:|::::::A::::::::|
| .*::::|:::::::::::::::|
- - * - - - + - - - + - - - + -
|       1       2       3
|

Area $\,A$ is a 2-by-127 rectangle: . $A = 254$

Area $\,B$ requires an integral: . $\displaystyle B \;=\;\int^1_0(x^3 + 18x^2 + 108x)\,dx$

We have: . $B \;=\;\frac{1}{4}x^4 + 6x^3 + 54x^2\,\bigg]^1_0 \;=\;\frac{1}{4} + 6 + 54$
. . . . . . . . $B \;=\;60\frac{1}{4}$

Therefore, the total area is: . $254 + 60\frac{1}{4} \;=\;314\frac{1}{4}\text{ units}^2.$