Maximum vlaue of the function `f (x) = pi^(2) int _(0)^(1)t sin (x+ pi t ) dt` over all real number x:

Options

A. `sqrt(pi^(2)+1)`
B. `sqrt(pi^(2)+2)`
C. `sqrt(pi^(2)+3)`
D. `sqrt(pi^(2)+4)`

Answer

To find the maximum value of the function \( f(x) = \pi^2 \int_0^1 t \sin(x + \pi t) \, dt \), we will follow these steps:

### Step 1: Set up the integral
We start with the function:
\[ f(x) = \pi^2 \int_0^1 t \sin(x + \pi t) \, dt \]

### Step 2: Use integration by parts
We will apply integration by parts, where we let:
- \( u = t \) (hence \( du = dt \))
- \( dv = \sin(x + \pi t) \, dt \) (hence \( v = -\frac{1}{\pi} \cos(x + \pi t) \))

Using integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ f(x) = \pi^2 \left[ -\frac{t}{\pi} \cos(x + \pi t) \bigg|_0^1 + \frac{1}{\pi} \int_0^1 \cos(x + \pi t) \, dt \right] \]

### Step 3: Evaluate the boundary term
Evaluating the boundary term:
\[ -\frac{t}{\pi} \cos(x + \pi t) \bigg|_0^1 = -\frac{1}{\pi} \cos(x + \pi) + 0 = \frac{1}{\pi} \cos(x) \]

### Step 4: Evaluate the integral of cosine
Now we need to evaluate the integral:
\[ \int_0^1 \cos(x + \pi t) \, dt \]
This can be computed as:
\[ \int_0^1 \cos(x + \pi t) \, dt = \frac{1}{\pi} \sin(x + \pi t) \bigg|_0^1 = \frac{1}{\pi} (\sin(x + \pi) - \sin(x)) = \frac{1}{\pi} (-\sin(x) - \sin(x)) = -\frac{2}{\pi} \sin(x) \]

### Step 5: Combine results
Putting it all together:
\[ f(x) = \pi^2 \left( \frac{1}{\pi} \cos(x) - \frac{2}{\pi^2} \sin(x) \right) = \pi \cos(x) - 2 \sin(x) \]

### Step 6: Find the maximum value
To find the maximum value of \( f(x) = \pi \cos(x) - 2 \sin(x) \), we can use the formula for the maximum of a linear combination of sine and cosine:
\[ R = \sqrt{a^2 + b^2} \]
where \( a = \pi \) and \( b = -2 \):
\[ R = \sqrt{\pi^2 + (-2)^2} = \sqrt{\pi^2 + 4} \]

Thus, the maximum value of \( f(x) \) is:
\[ \sqrt{\pi^2 + 4} \]

### Final Answer
The maximum value of the function \( f(x) \) is:
\[ \sqrt{\pi^2 + 4} \]

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