Finding the area of region between two graphs

**Sundevils** · December 3rd 2007, 12:18 PM

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.

**Krizalid** · December 3rd 2007, 12:58 PM

Nasty, but possible.

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:

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

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

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

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

**angel.white** · December 3rd 2007, 02:06 PM

Originally Posted by Krizalid

Nasty, but possible.

Let's find where these curves intersect:

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

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

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

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

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

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

**colby2152** · December 3rd 2007, 02:45 PM

Originally Posted by angel.white

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

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

Find the points where $f(x) = g(x)$
$x^2-19x+100 =-x^2+19x+30$
$2x^2-38x+70 = 0$
$x^2-19x+35 = 0$

**angel.white** · December 3rd 2007, 05:27 PM

Originally Posted by colby2152

Find the points where $f(x) = g(x)$
$x^2-19x+100 =-x^2+19x+30$
$2x^2-38x+70 = 0$
$x^2-19x+35 = 0$

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

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