Quote:

Originally Posted by

**Showcase_22**

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

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

Don't ever do that again.

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

$\displaystyle \int_{0}^{2}{\alpha (x)\,dx}+\int_{2}^{5}{\beta (x)\,dx}.$

Where $\displaystyle \alpha,\beta$ are some functions to being integrated.

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