If the area of the region (x,y):0≤y≤min2x,6x−x2 is A, then 12A is equal to \_\_\_\_\_\_\_\_.
Answer: 304
Solutions · 1
Official solution
2x and 6x−x2 meet where x2−4x=0⇒x=0,4; for 0<x<4, 6x−x2>2x (since 6x−x2−2x=x(4−x)>0), so min=2x there. For 4<x<6, 6x−x2<2x, so min=6x−x2 (valid until 6x−x2=0 at x=6).
\[A=\int_0^4 2x\,dx+\int_4^6(6x-x^2)\,dx = 16+\left[3x^2-\frac{x^3}{3}\right]_4^6=16+\left(36-\frac{80}{3}\right)=16+\frac{28}{3}=\frac{76}{3}.\]
\[12A=12\times\frac{76}{3}=304.\]