MHB Stardock's question at Yahoo Answers regarding minimizing a definite integral

[MarkFL]

Here is the question:

Please help me with this integral. Let f(a)?

. 2
=∫ | x (x-a) | dx for 0 ≤ a ≤2
. 0

1) find the function f(a)
2) Find the minimum of f(a)

note: try to explain process the better you can, please

I have posted a link there to this topic so the OP can find my work.

 

[MarkFL]

Hello stardock,

We are given:

$$f(a)=\int_0^2|x(x-a)|\,dx$$ where $$0\le a\le2$$

1.) The expression within the absolute value bars in the integrand is negative on the interval $(0,a)$ and positive on the interval $(a,2)$. It is a parabola opening upwards with the roots $x=0,\,a$. Hence, we may write:

$$f(a)=\int_0^a ax-x^2\,dx+\int_a^2 x^2-ax\,dx$$

Applying the anti-derivative form of the FTOC, there results:

$$f(a)=\left[\frac{a}{2}x^2-\frac{1}{3}x^3 \right]_0^a+\left[\frac{1}{3}x^3-\frac{a}{2}x^2 \right]_a^2$$

$$f(a)=\frac{a^3}{2}-\frac{a^3}{3}+\frac{2^3}{3}-2a-\frac{a^3}{3}+\frac{a^3}{2}$$

$$f(a)=\frac{a^3}{3}-2a+\frac{8}{3}$$

$$f(a)=\frac{1}{3}\left(a^3-6a+8 \right)$$

2.) To find the absolute minimum of $f(a)$ for $a$ on the interval $[0,2]$, we want to find any relative extrema within the interval, then evaluate $f(a)$ at these critical values and also at the end-points of the interval.

Differentiating with respect to $a$, and equating to zero we find:

$$f'(a)=\frac{1}{3}\left(3a^2-6 \right)=a^2-2=0$$

Thus, the critical value within the given interval is:

$$a=\sqrt{2}$$

Using the second derivative test, we find:

$$f''(a)=2a$$

$$f''\left(\sqrt{2} \right)=2\sqrt{2}>0$$

Thus, we have determined, since the function $f(a)$ is concave up at this critical value, that the extremum associated with this critical value is a minimum.

So, we want to compare the function's value at the end-points and at this critical value:

$$f(0)=\frac{1}{3}\left(0^3-6(0)+8 \right)=\frac{8}{3}$$

$$f\left(\sqrt{2} \right)=\frac{1}{3}\left(\left(\sqrt{2} \right)^3-6\left(\sqrt{2} \right)+8 \right)=\frac{8-4\sqrt{2}}{3}$$

$$f(2)=\frac{1}{3}\left(2^3-6(2)+8 \right)=\frac{4}{3}$$

Now, since:

$$\frac{8-4\sqrt{2}}{3}<\frac{4}{3}<\frac{8}{3}$$

we have determined:

$$f_{\min}=f\left(\sqrt{2} \right)=\frac{8-4\sqrt{2}}{3}=\frac{4}{3}\left(2-\sqrt{2} \right)$$

We could also have determined that the relative extremum we found in the given interval is also the absolute minimum by observing that:

$$f'(0)=0^2-2=-2$$ function is decreasing here.

$$f'(2)=2^2-2=2$$ function is increasing here.

Hence, we know at the left end-point of the interval, the function is decreasing, and continues to decrease until $a=\sqrt{2}$ at which point the function turns, then begins to increase, and continues to do so throughout the remainder of the interval, and so the relative minimum must be the absolute minimum on the given interval.