$If $m\in\left(0,\frac{5}{6}\right)$. Then Calculate value of $m$ in \\\\$\int\limits_{0}^{2}\left|\frac{1}{3}x^3+mx^2-2x-2m-\frac{1}{3}\right|dx=4$
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66Consider
f(x) = \dfrac{x^3}{3} + m x^2 -2x -2m -\frac{1}{3}
It is easy to see that f(0) and f(2) are both negative. At the same time f''(x) = 2(x+m) > 0 for all x in [0,2].
Therefore, f(x) is convex. And so f(x) ≤ 0 for all x in [0,2].
So
|f(x)| = -f(x)
so the integration turns out to be
\int_0^2 \left(2x+2m + \frac{1}{3}-\frac{x^3}{3}-mx^2\right)\ \mathrm dx
which easily evaluates to \dfrac{4m+10}{3}
Hence, solving for m we get
m =12
