# Thread: Area of region of plane between the graph

1. ## 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

2. 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.

3. 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

4. combine like terms ... that's all.

then integrate.

5. oh ok thanks again!

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

would area=28?

6. 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?

7. 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