Find the number of values of `x` satisfying `int_(0)^(x)t^(2)sin (x-t)dt=x^(2)` in `[0, 100]`.

Answer

To solve the equation

\[ \int_{0}^{x} t^{2} \sin(x - t) \, dt = x^{2} \]

we will use integration by parts.

### Step 1: Set up the integral
Let \( I(x) = \int_{0}^{x} t^{2} \sin(x - t) \, dt \).

### Step 2: Use integration by parts
We will apply integration by parts where we let:
- \( u = t^{2} \) (first function)
- \( dv = \sin(x - t) \, dt \) (second function)

Then, we need to find \( du \) and \( v \):
- \( du = 2t \, dt \)
- To find \( v \), we integrate \( dv \):
\[ v = -\cos(x - t) \]

### Step 3: Apply integration by parts
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:

\[ I(x) = \left[ -t^{2} \cos(x - t) \right]_{0}^{x} + \int_{0}^{x} 2t \cos(x - t) \, dt \]

Calculating the boundary term:

\[ -t^{2} \cos(x - t) \bigg|_{0}^{x} = -x^{2} \cos(0) + 0 = -x^{2} \]

So,

\[ I(x) = -x^{2} + 2 \int_{0}^{x} t \cos(x - t) \, dt \]

### Step 4: Compute the remaining integral
Now we need to compute \( \int_{0}^{x} t \cos(x - t) \, dt \) using integration by parts again.

Let:
- \( u = t \)
- \( dv = \cos(x - t) \, dt \)

Then:
- \( du = dt \)
- \( v = \sin(x - t) \)

Applying integration by parts again:

\[ \int t \cos(x - t) \, dt = \left[ t \sin(x - t) \right]_{0}^{x} - \int \sin(x - t) \, dt \]

Calculating the boundary term:

\[ t \sin(x - t) \bigg|_{0}^{x} = x \sin(0) - 0 = 0 \]

So we have:

\[ \int_{0}^{x} t \cos(x - t) \, dt = -\int_{0}^{x} \sin(x - t) \, dt \]

### Step 5: Solve the integral
Now, \( \int_{0}^{x} \sin(x - t) \, dt = -\cos(x - t) \bigg|_{0}^{x} = -\cos(0) + \cos(x) = 1 - \cos(x) \).

Thus,

\[ \int_{0}^{x} t \cos(x - t) \, dt = -(1 - \cos(x)) = \cos(x) - 1 \]

### Step 6: Substitute back into \( I(x) \)
Now substituting back into \( I(x) \):

\[ I(x) = -x^{2} + 2(\cos(x) - 1) = -x^{2} + 2\cos(x) - 2 \]

### Step 7: Set the equation
Now we set \( I(x) \) equal to \( x^{2} \):

\[ -x^{2} + 2\cos(x) - 2 = x^{2} \]

This simplifies to:

\[ 2\cos(x) - 2 = 2x^{2} \]

or

\[ \cos(x) = x^{2} + 1 \]

### Step 8: Analyze the function
We need to find the number of solutions to \( \cos(x) = x^{2} + 1 \) in the interval \( [0, 100] \).

### Step 9: Determine the range
The function \( \cos(x) \) oscillates between -1 and 1, while \( x^{2} + 1 \) is always greater than or equal to 1 for \( x \geq 0 \). Therefore, \( \cos(x) \) cannot equal \( x^{2} + 1 \) for any \( x \geq 0 \).

### Conclusion
Thus, there are no values of \( x \) satisfying the equation in the interval \([0, 100]\).

### Final Answer
The number of values of \( x \) satisfying the equation is **0**.

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