Graphical minimization

$\begin{aligned} I(k) & = \int_0^4 |x(4-x)-k|\ dx & \small \blue{\text{Let }u=2-x \implies du = - dx} \\ & = \int_{-2}^2 |(2-u)(2+u) - k | \ du & \small \blue{\text{Since the integrand}} \\ & = 2 \int_0^2 |4-k-u^2 | \ du & \small \blue{\text{is an even function}} \end{aligned}$

For $k \le 0$ ,

$\begin{aligned} I(k\le 0) & = 2 \int_0^2 (4-k-u^2) \ du = 2 \left[4u-ku-\frac {u^3}3 \right]_0^2 = \frac {32}3 - 4k \end{aligned}$

$\implies \min(I(k \le0)) = \dfrac {32}3$ , when $k=0$ .

For $0 < k <4$ ,

$\begin{aligned} I(0<k<4) & = 2 \int_0^2 (4-k-u^2) \ du \\ & = 2 \int_0^{\sqrt{4-k}} (4-k-u^2) \ du + 2 \int_{\sqrt{4-k}}^2 (u^2 - 4+ k) \ du \\ & = \frac 83(4-k)^\frac 32 + \frac {16}3 - 4(4-k) \end{aligned}$

To find $\min(I(0<k<4))$ , $\dfrac {dI(k)}{dk} = - 4\sqrt{4-k} + 4$ . Putting $\dfrac {dI(k)}{dk} = 0$ , $\implies \sqrt{4-k} = 1 \implies k = 3$ . Therefore $\min (I(0<k<4)) = I(3) = 4$ .

For $k \ge 4$ ,

$\begin{aligned} I(k\ge 4) & = 2 \int_0^2 (u^2-4+k) \ du = 2 \left[\frac {u^3}3 -4u + ku \right]_0^2 = 4k - \frac {32}3 \end{aligned}$

$\implies \min(I(k \ge 4)) = \dfrac {16}3$ .

Therefore $\min(I(k)) = 4$ , when $k = \boxed 3$ .

We are going to use the following formula without proof. Searching "Leibniz integral rule" will let you find a proof of this result online.

Let $I(\alpha) = \int_{a(\alpha)}^{b(\alpha)}{f(x, \alpha) dx}$ .

Then

$\frac{dI}{d \alpha} = f(b, \alpha)\frac{db}{d \alpha} - f(a, \alpha) \frac{da}{d \alpha} + \int_{a(\alpha)}^{b(\alpha)}{ \frac{\partial f(x, \alpha)}{\partial \alpha} dx }$ .

Furthermore, it is clear that the required value of $k$ is between 0 and 4 (this can be seen graphically).

For $0 \leq k \leq 4$ , the line $y=k$ intersects the parabola $y=x(4-x)$ at $x = 2 \pm \sqrt{4-k}$ . Between these two roots, the parabola is above the line - and thus $x(4-x) - k \geq 0$ , and elsewhere in $[0, 4]$ we have $x(4-x) - k \leq 0$ . Thus,

$F(k) = \int_{2-\sqrt{4-k}}^{2+\sqrt{4-k}}{(x(4-x) - k) \, dx} + \int_{0}^{2-\sqrt{4-k}}{(k-x(4-x)) \, dx} + \int_{2+\sqrt{4-k}}^{4}{(k-x(4-x)) \, dx}.$

Using the above result,

$\frac{dF}{dk} = -\int_{2-\sqrt{4-k}}^{2+\sqrt{4-k}}{dx} + \int_{0}^{2-\sqrt{4-k}}{dx} + \int_{2+\sqrt{4-k}}^{4}{dx}$ .

Setting this equal to $0$ ,

$4 = 4 \sqrt{4-k}$ so $k = 3$