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: Evaluate the integral

We start by evaluating the integral

\[ \int_0^1 t \sin(x + \pi t) \, dt. \]

Using integration by parts, let:

- \( u = t \) → \( du = dt \)
- \( dv = \sin(x + \pi t) dt \) → \( v = -\frac{1}{\pi} \cos(x + \pi t) \)

Using the integration by parts formula:

\[ \int u \, dv = uv - \int v \, du, \]

we have:

\[ \int_0^1 t \sin(x + \pi t) \, dt = \left[ -\frac{t}{\pi} \cos(x + \pi t) \right]_0^1 + \frac{1}{\pi} \int_0^1 \cos(x + \pi t) \, dt. \]

### Step 2: Evaluate the boundary terms

Calculating the boundary terms:

\[ \left[ -\frac{t}{\pi} \cos(x + \pi t) \right]_0^1 = -\frac{1}{\pi} \cos(x + \pi) + 0 = \frac{1}{\pi} \cos(x). \]

### Step 3: Evaluate the remaining integral

Now we need to evaluate

\[ \int_0^1 \cos(x + \pi t) \, dt. \]

This integral can be computed as follows:

\[ \int_0^1 \cos(x + \pi t) \, dt = \left[ \frac{1}{\pi} \sin(x + \pi t) \right]_0^1 = \frac{1}{\pi} (\sin(x + \pi) - \sin(x)) = \frac{1}{\pi} (-\sin(x) - \sin(x)) = -\frac{2}{\pi} \sin(x). \]

### Step 4: Combine results

Combining the results from the boundary terms and the integral, we have:

\[ \int_0^1 t \sin(x + \pi t) \, dt = \frac{1}{\pi} \cos(x) - \frac{2}{\pi^2} \sin(x). \]

### Step 5: Substitute back into \(f(x)\)

Now substituting back into \(f(x)\):

\[ f(x) = \pi^2 \left( \frac{1}{\pi} \cos(x) - \frac{2}{\pi^2} \sin(x) \right) = \pi \cos(x) - 2 \sin(x). \]

### Step 6: Find the maximum value

To find the maximum value of

\[ f(x) = \pi \cos(x) - 2 \sin(x), \]

we can express this in the form \(a \cos(x) + b \sin(x)\) where \(a = \pi\) and \(b = -2\). The maximum value of \(a \cos(x) + b \sin(x)\) is given by

\[ \sqrt{a^2 + b^2}. \]

Calculating this gives:

\[ \sqrt{\pi^2 + (-2)^2} = \sqrt{\pi^2 + 4}. \]

### Final Answer

Thus, the maximum value of the function \(f(x)\) is

\[ \sqrt{\pi^2 + 4}. \]