# Thread: Area of region bounded by curves...where am I going wrong?

1. ## Area of region bounded by curves...where am I going wrong?

The point of intersection for x would be 4 so my bounds would be 0 and 4 since the absolute value makes one bound 0. Then I subtract the two, sincd 2x is the upper It'd be 2x-(x^2-8) which becomes 2x-x^2+8 and after integration it would be 2x^2/2-x^3/x+8x. I plug in four and get 80/3 every time. Where am I going wrong?

2. A certain function must be integrated for $\displaystyle -4\le x\le4.$ Put $\displaystyle f(x)=|2x|,\,g(x)=x^2-8$ and note that $\displaystyle f(x)\ge g(x)$ on $\displaystyle [-4,4]$ and the integral is $\displaystyle \int_{-4}^{4}{\big(\left| 2x \right|-x^{2}+8\big)\,dx}=2\int_{0}^{4}{\big(\left| 2x \right|-x^{2}+8\big)\,dx}=\frac{160}{3}.$

3. OK thanks but could you explain more clearly how you knew the first function was greater than the second on that interval?

4. Graph it or just pick any number of the interval (not whose make the functions zero) and you'll see.