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

• Feb 20th 2009, 12:24 PM
fattydq
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?
• Feb 20th 2009, 12:58 PM
Krizalid
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 $\displaystyle \int_{-4}^{4}{\big(\left| 2x \right|-x^{2}+8\big)\,dx}=2\int_{0}^{4}{\big(\left| 2x \right|-x^{2}+8\big)\,dx}=\frac{160}{3}.$
• Feb 20th 2009, 01:01 PM
fattydq
OK thanks but could you explain more clearly how you knew the first function was greater than the second on that interval?
• Feb 20th 2009, 01:02 PM
Krizalid
Graph it or just pick any number of the interval (not whose make the functions zero) and you'll see.