The value of the integral `I=inte^(x)(sinx+cosx)dx` is equal to `e^(x).f(x)+C, C` being the constant of integration. Then the maximum value of `y=f(x^(2)), AA x in R` is equal to

Options

A. 0
B. `-1
C. 1
D. `(1)/(2)`

Answer

To solve the integral \( I = \int e^x (\sin x + \cos x) \, dx \) and find the maximum value of \( y = f(x^2) \), where \( f(x) \) is derived from the integral, we will follow these steps:

### Step 1: Identify the integral format
The integral \( I = \int e^x (\sin x + \cos x) \, dx \) can be recognized as fitting the format of integration by parts or a known integral form.

### Step 2: Apply integration by parts
We can use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = e^x \)
- \( dv = \sin x + \cos x \, dx \)

Then, we differentiate and integrate:
- \( du = e^x \, dx \)
- \( v = -\cos x + \sin x \)

### Step 3: Calculate the integral
Now substituting into the integration by parts formula:
\[ I = e^x (-\cos x + \sin x) - \int (-\cos x + \sin x) e^x \, dx \]
This results in:
\[ I = e^x (-\cos x + \sin x) + \int e^x (\sin x + \cos x) \, dx \]
This means we have:
\[ I = e^x (-\cos x + \sin x) + I \]
To isolate \( I \), we can rearrange:
\[ I - I = e^x (-\cos x + \sin x) \]
Thus, we can conclude that:
\[ I = e^x f(x) + C \]
where \( f(x) = -\cos x + \sin x \).

### Step 4: Find \( f(x^2) \)
Next, we need to evaluate \( f(x^2) \):
\[ f(x^2) = -\cos(x^2) + \sin(x^2) \]

### Step 5: Determine the maximum value of \( y = f(x^2) \)
To find the maximum value of \( y = -\cos(x^2) + \sin(x^2) \), we analyze the range of \( \sin(x^2) \) and \( -\cos(x^2) \):
- The maximum value of \( \sin(x^2) \) is 1.
- The maximum value of \( -\cos(x^2) \) is also 1.

Therefore, the maximum value of \( y \) occurs when both \( \sin(x^2) \) and \( -\cos(x^2) \) are at their maximum:
\[ \text{Maximum value of } y = 1 + 1 = 2. \]

### Final Answer
The maximum value of \( y = f(x^2) \) is \( 2 \).