Area between Curves

**rawkstar** · March 15th 2011, 01:30 PM

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

**HallsofIvy** · March 15th 2011, 02:15 PM

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.

**Soroban** · March 15th 2011, 03:24 PM

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: . $254 + 60\frac{1}{4} \;=\;314\frac{1}{4}\text{ units}^2.$

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