# Finding the area of region between two graphs

• Dec 3rd 2007, 12:18 PM
Sundevils
Finding the area of region between two graphs
Having trouble doint this problem

Consider the functions f(x)=x^2-19x+100 and g(x)=-x^2+19x+30. Find the area of the region between the two graphs.

Any help would be appreciated.
• Dec 3rd 2007, 12:58 PM
Krizalid
Nasty, but possible.

Quote:

Originally Posted by Sundevils
Consider the functions f(x)=x^2-19x+100 and g(x)=-x^2+19x+30. Find the area of the region between the two graphs.

Let's find where these curves intersect:

$\displaystyle f(x)=g(x)\implies x^2-19x+35=0.$

So we need to integrate a certain function on the interval $\displaystyle \left[ {\frac{{19 - \sqrt {221} }} {2},\frac{{19 + \sqrt {221} }} {2}} \right].$ The thing is, that in such interval $\displaystyle f(x)<g(x),$ then the integral to compute is

$\displaystyle \int_b^a {\left( {38x - 2x^2 - 70} \right)\,dx},$

where $\displaystyle \left[ {a,b} \right] = \left[ {\frac{{19 + \sqrt {221} }} {2},\frac{{19 - \sqrt {221} }} {2}} \right].$
• Dec 3rd 2007, 02:06 PM
angel.white
Quote:

Originally Posted by Krizalid
Nasty, but possible.

Let's find where these curves intersect:

$\displaystyle f(x)=g(x)\implies x^2-19x+35=0.$

So we need to integrate a certain function on the interval $\displaystyle \left[ {\frac{{19 - \sqrt {221} }} {2},\frac{{19 + \sqrt {221} }} {2}} \right].$ The thing is, that in such interval $\displaystyle f(x)<g(x),$ then the integral to compute is

$\displaystyle \int_b^a {\left( {38x - 2x^2 - 70} \right)\,dx},$

where $\displaystyle \left[ {a,b} \right] = \left[ {\frac{{19 + \sqrt {221} }} {2},\frac{{19 - \sqrt {221} }} {2}} \right].$

I did the problem the same, but I had a different interval. Can you explain this part:

$\displaystyle f(x)=g(x)\implies x^2-19x+35=0.$
• Dec 3rd 2007, 02:45 PM
colby2152
Quote:

Originally Posted by angel.white
I did the problem the same, but I had a different interval. Can you explain this part:

$\displaystyle f(x)=g(x)\implies x^2-19x+35=0.$

Find the points where $\displaystyle f(x) = g(x)$
$\displaystyle x^2-19x+100 =-x^2+19x+30$
$\displaystyle 2x^2-38x+70 = 0$
$\displaystyle x^2-19x+35 = 0$
• Dec 3rd 2007, 05:27 PM
angel.white
Quote:

Originally Posted by colby2152
Find the points where $\displaystyle f(x) = g(x)$
$\displaystyle x^2-19x+100 =-x^2+19x+30$
$\displaystyle 2x^2-38x+70 = 0$
$\displaystyle x^2-19x+35 = 0$

Oh, I did come out with the same answer, I just didn't simplify it as much, so it appeared different (Tmi)