Given `overset(2)underset(1)inte^(x^(2))dx=a`, the value of `overset(e^(4))underset(e )int sqrt(log_(e )x)dx`, is

Options

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

Answer

To solve the problem step by step, we will first evaluate the given integral and then find the required value.

### Step 1: Evaluate the given integral
We are given:
\[ \int_{1}^{2} e^{x^2} \, dx = a \]

### Step 2: Change of variable for the first integral
Let \( t = x^2 \). Then, we have:
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2\sqrt{t}} \]
Now, we need to change the limits of integration:
- When \( x = 1 \), \( t = 1^2 = 1 \)
- When \( x = 2 \), \( t = 2^2 = 4 \)

Thus, the integral becomes:
\[ \int_{1}^{2} e^{x^2} \, dx = \int_{1}^{4} e^t \cdot \frac{dt}{2\sqrt{t}} = \frac{1}{2} \int_{1}^{4} \frac{e^t}{\sqrt{t}} \, dt = a \]

### Step 3: Evaluate the second integral
We need to find:
\[ \int_{e}^{e^4} \sqrt{\ln x} \, dx \]

### Step 4: Change of variable for the second integral
Let \( u = \ln x \). Then, we have:
\[ x = e^u \quad \Rightarrow \quad dx = e^u \, du \]
Now, we change the limits of integration:
- When \( x = e \), \( u = \ln e = 1 \)
- When \( x = e^4 \), \( u = \ln(e^4) = 4 \)

Thus, the integral becomes:
\[ \int_{e}^{e^4} \sqrt{\ln x} \, dx = \int_{1}^{4} \sqrt{u} \cdot e^u \, du \]

### Step 5: Evaluate the new integral using integration by parts
Let:
- \( v = \sqrt{u} \) and \( dv = \frac{1}{2\sqrt{u}} \, du \)
- \( dw = e^u \, du \) and \( w = e^u \)

Using integration by parts:
\[ \int v \, dw = vw - \int w \, dv \]
We have:
\[ \int_{1}^{4} \sqrt{u} \cdot e^u \, du = \left[ \sqrt{u} \cdot e^u \right]_{1}^{4} - \int_{1}^{4} e^u \cdot \frac{1}{2\sqrt{u}} \, du \]

### Step 6: Calculate the boundary terms
Calculating the boundary terms:
\[ \left[ \sqrt{u} \cdot e^u \right]_{1}^{4} = \left( \sqrt{4} \cdot e^4 \right) - \left( \sqrt{1} \cdot e^1 \right) = 2e^4 - e \]

### Step 7: Substitute back into the integral
Now, we have:
\[ \int_{1}^{4} \sqrt{u} \cdot e^u \, du = 2e^4 - e - \frac{1}{2} \int_{1}^{4} e^u \cdot \frac{1}{\sqrt{u}} \, du \]
The last integral is equal to \( a \) (from Step 2):
\[ \int_{1}^{4} e^u \cdot \frac{1}{\sqrt{u}} \, du = 2a \]

### Step 8: Final expression
Thus, we have:
\[ \int_{1}^{4} \sqrt{u} \cdot e^u \, du = 2e^4 - e - a \]

### Final Answer
The value of the integral \( \int_{e}^{e^4} \sqrt{\ln x} \, dx \) is:
\[ 2e^4 - e - a \]