Let \(f\) be a non-negative function in \([0, 1]\) and twice differentiable in \((0, 1)\). If \(\int_0^x \sqrt{1 - (f'(t))^2} \, dt = \int_0^x f(t) \, dt\), \(0 \leq x \leq 1\) and \(f(0) = 0\), then \(\lim_{x \to 0} \frac{1}{x^2} \int_0^x f(t) \, dt\) :
Show Hint
Differentiating an equation containing integrals is a standard trick to convert an integral equation into a differential equation.
Step 1: Understanding the Concept:
We use the Leibniz Integral Rule to differentiate the integral equation and find the function \(f(x)\). Then, we calculate the required limit. Step 2: Detailed Explanation:
Differentiating the given equation \(\int_0^x \sqrt{1 - (f'(t))^2} \, dt = \int_0^x f(t) \, dt\) with respect to \(x\):
\[ \sqrt{1 - (f'(x))^2} = f(x) \]
Square both sides:
\[ 1 - (f'(x))^2 = f^2(x) \implies (f'(x))^2 = 1 - f^2(x) \implies f'(x) = \pm \sqrt{1 - f^2(x)} \]
Since \(f\) is non-negative and \(f(0) = 0\), for small \(x\), the integral of \(f\) is positive, so \(f'(x)\) must be positive.
\[ \frac{df}{dx} = \sqrt{1 - f^2} \implies \int \frac{df}{\sqrt{1 - f^2}} = \int dx \implies \arcsin(f) = x + C \]
Using \(f(0) = 0 \implies \arcsin(0) = 0 + C \implies C = 0\).
Thus, \(f(x) = \sin x\).
Now, evaluate the limit:
\[ L = \lim_{x \to 0} \frac{1}{x^2} \int_0^x \sin t \, dt = \lim_{x \to 0} \frac{1}{x^2} [-\cos t]_0^x = \lim_{x \to 0} \frac{1 - \cos x}{x^2} \]
Using the standard limit \(\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}\):
\[ L = \frac{1}{2} \] Step 3: Final Answer:
The limit is \(\frac{1}{2}\).
