Let f be a non-negative function defined on the interval [0,1] . If `int_0^x sqrt(1-(f'(t))^(2))dt=int_0^x f(t)dt , 0 lexle1` and `f(0)=0` then the value of `lim_(x to0)int_0^xf(t)/x^2 dt` is

Options

A. `1/2`
B. `-1/2`
C. 2
D. 1

Answer

To solve the problem step by step, we start with the given equation:

\[ \int_0^x \sqrt{1 - (f'(t))^2} \, dt = \int_0^x f(t) \, dt \]

for \( x \in [0, 1] \) and \( f(0) = 0 \). We need to find:

\[ \lim_{x \to 0} \frac{1}{x^2} \int_0^x f(t) \, dt \]

### Step 1: Differentiate both sides with respect to \( x \)

Using the Fundamental Theorem of Calculus, we differentiate both sides:

\[ \sqrt{1 - (f'(x))^2} = f(x) \]

### Step 2: Square both sides

Squaring both sides gives us:

\[ 1 - (f'(x))^2 = f(x)^2 \]

Rearranging this, we find:

\[ (f'(x))^2 = 1 - f(x)^2 \]

### Step 3: Take the square root

Taking the square root of both sides, we have:

\[ f'(x) = \pm \sqrt{1 - f(x)^2} \]

Since \( f(x) \) is non-negative, we take the positive root:

\[ f'(x) = \sqrt{1 - f(x)^2} \]

### Step 4: Use separation of variables to solve for \( f(x) \)

We can separate variables:

\[ \frac{df}{\sqrt{1 - f^2}} = dx \]

Integrating both sides, we get:

\[ \sin^{-1}(f) = x + C \]

### Step 5: Apply the initial condition

Using the initial condition \( f(0) = 0 \):

\[ \sin^{-1}(0) = 0 + C \implies C = 0 \]

Thus, we have:

\[ \sin^{-1}(f) = x \implies f(x) = \sin(x) \]

### Step 6: Substitute \( f(t) \) into the limit expression

Now we substitute \( f(t) = \sin(t) \) into our limit expression:

\[ \lim_{x \to 0} \frac{1}{x^2} \int_0^x \sin(t) \, dt \]

### Step 7: Evaluate the integral

The integral of \( \sin(t) \) is:

\[ \int_0^x \sin(t) \, dt = -\cos(t) \bigg|_0^x = -\cos(x) + \cos(0) = 1 - \cos(x) \]

### Step 8: Substitute back into the limit

Substituting this back into our limit gives:

\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} \]

### Step 9: Apply L'Hôpital's Rule

Since both the numerator and denominator approach 0 as \( x \to 0 \), we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{\sin(x)}{2x} = \lim_{x \to 0} \frac{1}{2} = \frac{1}{2} \]

### Final Answer

Thus, the value of the limit is:

\[ \frac{1}{2} \]