Hard Exam Question From Vietnam - Mind Your Decisions - Free Printable

I need to solve this integral equation:

$$\int_4^8 \frac{[f'(x)]^2}{[f(x)]^4} dx = 1$$

Let me work through this step by step.

Step 1: Recognize the pattern

Looking at the integrand $\frac{[f'(x)]^2}{[f(x)]^4}$, I can rewrite this as:
$$\left(\frac{f'(x)}{[f(x)]^2}\right)^2$$

Step 2: Use substitution

Let me try a substitution. Let $u = f(x)$. Then $du = f'(x)dx$, which means $dx = \frac{du}{f'(x)}$.

But wait, this gets complicated because we have $[f'(x)]^2$ in the numerator.

Let me think differently. Notice that:
$$\frac{d}{dx}\left(-\frac{1}{f(x)}\right) = -\left(-\frac{1}{[f(x)]^2}\right)f'(x) = \frac{f'(x)}{[f(x)]^2}$$

So if I let $g(x) = -\frac{1}{f(x)}$, then $g'(x) = \frac{f'(x)}{[f(x)]^2}$.

Therefore:
$$\frac{[f'(x)]^2}{[f(x)]^4} = \left(\frac{f'(x)}{[f(x)]^2}\right)^2 = [g'(x)]^2$$

Step 3: Rewrite the integral

The integral becomes:
$$\int_4^8 [g'(x)]^2 dx = 1$$

where $g(x) = -\frac{1}{f(x)}$.

Step 4: Consider what function would satisfy this

For $\int_4^8 [g'(x)]^2 dx = 1$, I need to find a function $g(x)$ such that when I square its derivative and integrate from 4 to 8, I get 1.

One simple approach: if $g'(x)$ is constant, say $g'(x) = c$, then:
$$\int_4^8 c^2 dx = c^2 \cdot (8-4) = 4c^2 = 1$$
$$c^2 = \frac{1}{4}$$
$$c = \pm\frac{1}{2}$$

So $g'(x) = \frac{1}{2}$ or $g'(x) = -\frac{1}{2}$.

This means $g(x) = \frac{x}{2} + C$ or $g(x) = -\frac{x}{2} + C$.

Step 5: Find f(x)

Since $g(x) = -\frac{1}{f(x)}$, we have:
$$-\frac{1}{f(x)} = \frac{x}{2} + C$$
$$f(x) = -\frac{1}{\frac{x}{2} + C} = -\frac{2}{x + 2C}$$

Or with the negative case:
$$-\frac{1}{f(x)} = -\frac{x}{2} + C$$
$$f(x) = \frac{1}{\frac{x}{2} - C} = \frac{2}{x - 2C}$$

Let me use $A = 2C$ for simplicity:
$$f(x) = \frac{2}{x - A}$$

Step 6: Verify the solution

Let me verify that $f(x) = \frac{2}{x - A}$ works for any constant $A$:

$f'(x) = -\frac{2}{(x-A)^2}$

$$\frac{[f'(x)]^2}{[f(x)]^4} = \frac{\left(-\frac{2}{(x-A)^2}\right)^2}{\left(\frac{2}{x-A}\right)^4} = \frac{\frac{4}{(x-A)^4}}{\frac{16}{(x-A)^4}} = \frac{4}{16} = \frac{1}{4}$$

So:
$$\int_4^8 \frac{1}{4} dx = \frac{1}{4} \cdot (8-4) = \frac{1}{4} \cdot 4 = 1$$ ✓

Final Answer:

$$f(x) = \frac{2}{x - A}$$

where $A$ is any constant (as long as $x \neq A$ for $x \in [4,8]$).

Alternatively, this can be written as:

$$f(x) = \frac{k}{x - A}$$

where $k = 2$ and $A$ is any constant.