Let a and b respectively be the points of local maximum and local minimum of the function f(x)=2x3−3x2−12x. If A is the total area of the region bounded by y=f(x), the x-axis and the lines x=a and x=b, then 4A is equal to \_\_\_\_\_\_\_\_.
Answer: 114
Solutions · 1
Official solution
f′(x)=6x2−6x−12=6(x+1)(x−2), so f′(x)=0 at x=−1,2; a=−1 (local max), b=2 (local min).
\[A = \int_{-1}^{0}f(x)\,dx + \int_0^2\bigl(-f(x)\bigr)\,dx\]
where the sign change accounts for f crossing zero at x=0.
\[= \left[\frac{x^4}{2}-x^3-6x^2\right]_{-1}^0 + \left[6x^2+x^3-\frac{x^4}{2}\right]_0^2 = \frac{114}{4}\]
4A=114