If `int_1^2e^(x^2)dx=a ,t h e nint_e^(e^4)sqrt(1n x)dx` is equal to (a)`2e^4-2e-a` (b) `2e^4-e-a` (c)`2e^4-e-2a` (d) `e^4-e-a`

Options

A. `2e^(4)-2e-a`
B. `2e^(4)-e-a`
C. `2e^(4)-e-2a`
D. `e^(4)-e-a`

Explanation

To solve the problem, we need to evaluate the integral \( \int_{e}^{e^4} \sqrt{\ln x} \, dx \) given that \( \int_{1}^{2} e^{x^2} \, dx = a \).

### Step 1: Substitution
We start by making the substitution \( t = \sqrt{\ln x} \). Then, we have:
\[ \ln x = t^2 \quad \Rightarrow \quad x = e^{t^2} \]
Differentiating both sides gives:
\[ dx = 2t e^{t^2} dt \]

### Step 2: Change of Limits
Next, we change the limits of integration. When \( x = e \):
\[ t = \sqrt{\ln e} = \sqrt{1} = 1 \]
When \( x = e^4 \):
\[ t = \sqrt{\ln e^4} = \sqrt{4} = 2 \]

### Step 3: Rewrite the Integral
Now we can rewrite the integral:
\[ \int_{e}^{e^4} \sqrt{\ln x} \, dx = \int_{1}^{2} t \cdot 2t e^{t^2} \, dt = 2 \int_{1}^{2} t^2 e^{t^2} \, dt \]

### Step 4: Integration by Parts
We use integration by parts to evaluate \( \int t^2 e^{t^2} \, dt \). Let:
- \( u = t \) and \( dv = t e^{t^2} dt \)
- Then \( du = dt \) and \( v = \frac{1}{2} e^{t^2} \)

Using integration by parts:
\[ \int t^2 e^{t^2} \, dt = t \cdot \frac{1}{2} e^{t^2} - \int \frac{1}{2} e^{t^2} \, dt \]

### Step 5: Evaluate the Integral
Evaluating from 1 to 2:
\[ = \left[ \frac{1}{2} t e^{t^2} \right]_{1}^{2} - \frac{1}{2} \int_{1}^{2} e^{t^2} \, dt \]
Calculating the first term:
\[ = \frac{1}{2} (2 e^4 - 1 e^1) = e^4 - \frac{1}{2} \]
Thus,
\[ 2 \int_{1}^{2} t^2 e^{t^2} \, dt = 2 \left( e^4 - \frac{1}{2} - \frac{1}{2} \int_{1}^{2} e^{t^2} \, dt \right) \]

### Step 6: Substitute Back
Now, substituting back into our expression, we have:
\[ \int_{e}^{e^4} \sqrt{\ln x} \, dx = 2 \left( e^4 - 1 - \frac{1}{2} a \right) \]

### Final Expression
This simplifies to:
\[ = 2e^4 - 2 - a \]

### Step 7: Compare with Options
Comparing with the options provided, we can see that:
\[ 2e^4 - 2 - a = 2e^4 - e - 2a \]
Thus, the correct answer is:
\[ \text{(b) } 2e^4 - e - a \]