# Find the area

• Jan 26th 2009, 07:24 PM
serix
Find the area
y=8x2x3+x
y=x2+11x

Find The Total area enclosed by the graphs of these 2 functions.

Can anyone show me how to solve this without graphing it, we need to make it into an integral and I can't seem to figure out the answer, any help would be much appreciated.
• Jan 27th 2009, 12:28 AM
Showcase_22
I might be wrong, but I think you can do it like this:

$8x^2-x^3+x=x^2+11x$

$7x^2-x^3-10x=0$

$x^2-7x+10=0$

$(x-2)(x-5)=0$

$x=2 \ or \ 5$.

So we want $\int_{2}^{5} x^2-7x+10dx$.
• Jan 27th 2009, 06:38 AM
Krizalid
Quote:

Originally Posted by Showcase_22

$7x^2-x^3-10x=0$

$x^2-7x+10=0$

Don't ever do that again.

Your work was for nice way, so we have to split the given area into two integrals:

$\int_{0}^{2}{\alpha (x)\,dx}+\int_{2}^{5}{\beta (x)\,dx}.$
Where $\alpha,\beta$ are some functions to being integrated.

Now, put $f(x)=8x^2-x^3+x$ and $g(x)=x^2+11x.$ Note that $\alpha(x)=g(x)-f(x)\ge0$ on $[0,2]$ and $\beta(x)=f(x)-g(x)\ge0$ on $[2,5].$ Finally, the required area is $\mathcal A=\int_{0}^{2}{\big(g(x)-f(x)\big)\,dx}+\int_{2}^{5}{\big(f(x)-g(x)\big)\,dx}.$