If the integral `I=int(2x^(2))/(4+x^(2))dx=2x-f(x)+c`, where `f(2)=pi`, then the minimum value of `y=f(x)AA x in [-2, 2]` is (where, c is the constant of integration)

Options

A. 0
B. `-pi`
C. `2pi`
D. `-4pi`

Answer

To solve the integral \( I = \int \frac{2x^2}{4+x^2} \, dx \) and find the function \( f(x) \) such that \( I = 2x - f(x) + c \) where \( f(2) = \pi \), we will follow these steps:

### Step 1: Simplify the Integral
We start with the integral:
\[ I = \int \frac{2x^2}{4+x^2} \, dx \]

### Step 2: Use Substitution
We can simplify the integral by breaking it down. Notice that:
\[ \frac{2x^2}{4+x^2} = 2 - \frac{8}{4+x^2} \]
Thus, we can rewrite the integral as:
\[ I = \int \left( 2 - \frac{8}{4+x^2} \right) \, dx \]

### Step 3: Integrate Each Term
Now we can integrate each term separately:
1. The integral of \( 2 \) is:
\[ \int 2 \, dx = 2x \]
2. The integral of \( \frac{8}{4+x^2} \) can be solved using the formula for the integral of \( \frac{1}{a^2 + x^2} \):
\[ \int \frac{8}{4+x^2} \, dx = 8 \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) = 4 \tan^{-1}\left(\frac{x}{2}\right) \]

### Step 4: Combine the Results
Putting it all together, we have:
\[ I = 2x - 4 \tan^{-1}\left(\frac{x}{2}\right) + C \]

### Step 5: Identify \( f(x) \)
From the equation \( I = 2x - f(x) + c \), we can identify:
\[ f(x) = 4 \tan^{-1}\left(\frac{x}{2}\right) \]

### Step 6: Find \( f(2) \)
We know that \( f(2) = \pi \):
\[ f(2) = 4 \tan^{-1}\left(\frac{2}{2}\right) = 4 \tan^{-1}(1) = 4 \cdot \frac{\pi}{4} = \pi \]
This confirms our function \( f(x) \).

### Step 7: Find the Minimum Value of \( f(x) \) on the Interval \([-2, 2]\)
To find the minimum value of \( f(x) \) on the interval \([-2, 2]\), we need to evaluate \( f(x) \) at the endpoints and check for critical points.

1. Calculate \( f(-2) \):
\[ f(-2) = 4 \tan^{-1}\left(-1\right) = 4 \cdot \left(-\frac{\pi}{4}\right) = -\pi \]
2. Calculate \( f(0) \):
\[ f(0) = 4 \tan^{-1}(0) = 0 \]
3. Calculate \( f(2) \):
\[ f(2) = \pi \]

### Step 8: Compare Values
Now we compare the values:
- \( f(-2) = -\pi \)
- \( f(0) = 0 \)
- \( f(2) = \pi \)

The minimum value of \( f(x) \) on the interval \([-2, 2]\) is:
\[ \text{Minimum value} = -\pi \]

### Final Answer
The minimum value of \( y = f(x) \) for \( x \in [-2, 2] \) is \( -\pi \).