Consider the function defined on `[0,1] rarr R , f(x) =(sinx- xcosx)/(x^(2)), if x ne 0 and f(0) =0`
`int _(0)^(1)f(x)` is equal to

Options

A. A. `1- sin (1)`
B. B. `sin (1)-1`
C. C. ` sin (1)`
D. D. `- sin(1)`

Answer

To solve the integral \( \int_0^1 f(x) \, dx \) where \( f(x) = \frac{\sin x - x \cos x}{x^2} \) for \( x \neq 0 \) and \( f(0) = 0 \), we can break down the problem step by step.

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int_0^1 f(x) \, dx = \int_0^1 \frac{\sin x - x \cos x}{x^2} \, dx \]

### Step 2: Split the Integral
We can split the integral into two parts:
\[ \int_0^1 f(x) \, dx = \int_0^1 \frac{\sin x}{x^2} \, dx - \int_0^1 \frac{\cos x}{x} \, dx \]

### Step 3: Use Integration by Parts on the First Integral
For the first integral \( \int_0^1 \frac{\sin x}{x^2} \, dx \), we will use integration by parts. Let:
- \( u = \sin x \) (thus \( du = \cos x \, dx \))
- \( dv = \frac{1}{x^2} \, dx \) (thus \( v = -\frac{1}{x} \))

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int_0^1 \frac{\sin x}{x^2} \, dx = \left[-\frac{\sin x}{x}\right]_0^1 + \int_0^1 \frac{\cos x}{x} \, dx \]

### Step 4: Evaluate the Boundary Terms
Evaluating the boundary terms:
- At \( x = 1 \): \( -\frac{\sin(1)}{1} = -\sin(1) \)
- At \( x = 0 \): We need to find the limit:
\[ \lim_{x \to 0} -\frac{\sin x}{x} = -1 \quad \text{(using } \lim_{x \to 0} \frac{\sin x}{x} = 1\text{)} \]

Thus, we have:
\[ \left[-\frac{\sin x}{x}\right]_0^1 = -\sin(1) + 1 \]

### Step 5: Combine the Results
Now substituting back into our integral:
\[ \int_0^1 f(x) \, dx = (-\sin(1) + 1) + \int_0^1 \frac{\cos x}{x} \, dx - \int_0^1 \frac{\cos x}{x} \, dx \]
The two \( \int_0^1 \frac{\cos x}{x} \, dx \) terms cancel each other out.

### Step 6: Final Result
Thus, we find:
\[ \int_0^1 f(x) \, dx = 1 - \sin(1) \]

### Final Answer
The value of the integral \( \int_0^1 f(x) \, dx \) is:
\[ \boxed{1 - \sin(1)} \]