# Area of region of plane between the graph

• Apr 13th 2009, 04:54 PM
Jim Marnell
Area of region of plane between the graph
Find the area of the region of the plane between the graphs of
$y=9-x^2$ and
$y=1+2x$

would $f(x)=9-x^2$
and $g(x)=1+2x$

and then set them equal? $9-x^2=1+2x$

confused on how to do this problem
• Apr 13th 2009, 05:01 PM
skeeter
Quote:

Originally Posted by Jim Marnell
Find the area of the region of the plane between the graphs of
$y=9-x^2$ and
$y=1+2x$

would $f(x)=9-x^2$
and $g(x)=1+2x$

and then set them equal? $9-x^2=1+2x$

confused on how to do this problem

$9-x^2 = 1+2x$

$0 = x^2+2x-8$

$0 = (x+4)(x-2)$

intersection points are $x = -4$ and $x = 2$

$A = \int_{-4}^2 (9-x^2) - (1+2x) \, dx$

simplify the integrand, integrate, and use the FTC to find the area.
• Apr 13th 2009, 05:21 PM
Jim Marnell
I'm not sure how to simplify the integrand, probably because im not sure what it means to do that but i can integrate and perform the Fundamental Theorum. Can any1 show me how to simplify the integrand
• Apr 13th 2009, 05:24 PM
skeeter
combine like terms ... that's all.

then integrate.
• Apr 13th 2009, 05:27 PM
Jim Marnell
oh ok thanks again!

so simplified it would just be $-x^2+8+2x$

would area=28?
• Apr 13th 2009, 07:52 PM
mr fantastic
Quote:

Originally Posted by Jim Marnell
oh ok thanks again!

so simplified it would just be $-x^2+8 {\color{red}+} 2x$ Mr F says: No. Please take greater care with your algebra. The red plus is wrong.

would area=28? Mr F says: Given the above mistake, I assume 28 will also be wrong.

The first thing you should have done in this whole question is draw a diagram. Did you do that?
• Apr 14th 2009, 08:21 AM
Jim Marnell
that plus error was my fault, i had it right on paper but typed it in wrong. But yes i had to draw the diagram to go along with the problem but i did it after i completed the problem but definitely would have helped if i drew it before i did the problem