calculate the area of the region bounded by curves f(x)=x^2 +5x and g(x)=2-x^2

Sunday, November 3, 2013

calculate the area of the region bounded by curves f(x)=x^2 +5x and g(x)=2-x^2

First we need to find the intersection points of the
curves.

==> f(x)=
g(x)

==> $\displaystyle{{x}}^{{2}}+{5}{x}={2}-$
x^2

$\displaystyle=\Rightarrow{2}{{x}}^{{2}}+$
5x -2 = 0

$\displaystyle=\Rightarrow$
x1= (-5 + sqrt(25+16))/4 =
(-5+sqrt41)/4

class="AM"> $\displaystyle=\Rightarrow{x}{2}=$
(-5-sqrt(41))/4

type="image/svg+xml"
src="/jax/includes/tinymce/jscripts/tiny_mce/plugins/asciisvg/js/d.svg"
sscr="-7.5,7.5,-5,5,1,1,1,1,1,300,200,func,2-x^2,null,0,0,-7,7,black,1,none,func,x^2 +
5x,null,0,0,-7,7,red,1,none"/>

class="AM"> Now we know that the area bounded by the curve
is:

$\displaystyle{A}=\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}-\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}=$
int g(x)-f(x) dx

$\displaystyle{A}=$
int 2-x^2 - (x^2 +5x) dx = int -2x^2 - 5x +2
dx

$\displaystyle{A}=-\frac{{{2}{{x}}^{{3}}}}{{3}}-$
(5x^2) /2 + 2x

class="AM"> Now we will find the area bounded by (-5+sqrt41)/4 and
(-5-sqrt41)/4.

$\displaystyle{A}{1}=-\frac{{2}}{{3}}$
((-5+sqrt41)/4)^3 - 5/2 ((-5+sqrt41)/4)^2
+2(-5+sqrt41)/4

$\displaystyle{A}{2}=$
-2/3 ((-5-sqrt41)/4)^3 -5/2 ((-5-sqrt41)/4)^2 +2
(-5-sqrt41)/4

class="AM"> The area bounded is A = A1 -
A2

$\displaystyle{A}=-\frac{{2}}{{3}}{{\left(\frac{{-{5}+\sqrt{{41}}}}{{4}}\right)}}^{{3}}+\frac{{2}}{{3}}$
((-5+sqrt41)/3)^3 +(25sqrt41)/2 +
sqrt41

class="AM">

Help Notes

Sunday, November 3, 2013