Thread: calculate the area enclosed by the curve & line

1. calculate the area enclosed by the curve & line

calculate the area enclosed by the curve y=12-4x+3x² & line y=5 and the coordinates x=-2 and x=3
steps on how to complete would be great, figure i have to make the line and curve equal to each other which is already done because they both equal y, then i got stuck finding the values for x for them both when you factor out the quadratic... most likely going in the wrong direction but made a start.
thanks

2. Re: calculate the area enclosed by the curve & line

$\displaystyle A=\int\limits _{-2}^{3}(\!3\,{x}^{2}-4\,x+12){dx}-5^2=60$

3. Re: calculate the area enclosed by the curve & line

i have to make the line and curve equal to each other
No, you don't. You would only do that to determine the points where they cross if you were asked to find the entire area they bound in order to find the limits of integration. Here they do not bound a region- you also have x= -2 and x= 3 as boundaries so those are the limits of integration.

4. Re: calculate the area enclosed by the curve & line

it should of been posted in there, simple error .. its all the same rubbish in the end anyway.
thanks for the help though

5. Re: calculate the area enclosed by the curve & line

Hello, bobmarly12345!

Calculate the area enclosed by the curve $\displaystyle y\:=\:3x^2-4x+12$ and the lines $\displaystyle y=5,\;x=-2,\;x=3$

Figure i have to make the line and curve equal to each other. . Why?
. . They gave us the left and right limits.

I don't suppose you made a sketch . . .

Code:
              |
*       |           *
|
*      |          *
*     |         *
::*   |       *::
::::::* * *::::::
::::::|::::::::::
::::::|::::::::::
- - - - + - - - - - -
:     |5        :
:     |         :
:     |         :
----+-----+---------+----
-2     |         3
|
The parabola is above the line.

$\displaystyle A \;=\;\int^3_{\text{-}2}\big[(3x^2-4x+12) - 5\big]\,dx \;=\;\int^3_{\text{-}2}\left(3x^2 - 4x + 7\right)\,dx$