The value of `int_(1)^(4) e^(sqrt(x))dx`, is

Options

A. `e^(2)`
B. `2e^(2)`
C. `4e^(2)`
D. `3e^(2)`

Answer

To solve the integral \( \int_{1}^{4} e^{\sqrt{x}} \, dx \), we will use substitution and integration by parts. Let's go through the solution step by step.

### Step 1: Substitution
Let \( t = \sqrt{x} \). Then, we differentiate both sides to find \( dx \):
\[ x = t^2 \implies dx = 2t \, dt \]

### Step 2: Change the limits of integration
When \( x = 1 \):
\[ t = \sqrt{1} = 1 \]
When \( x = 4 \):
\[ t = \sqrt{4} = 2 \]

### Step 3: Rewrite the integral
Now we can rewrite the integral in terms of \( t \):
\[ \int_{1}^{4} e^{\sqrt{x}} \, dx = \int_{1}^{2} e^{t} (2t) \, dt = 2 \int_{1}^{2} t e^{t} \, dt \]

### Step 4: Integration by parts
We will use integration by parts, where we let:
- \( u = t \) (thus \( du = dt \))
- \( dv = e^{t} dt \) (thus \( v = e^{t} \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int t e^{t} \, dt = t e^{t} - \int e^{t} \, dt \]
\[ = t e^{t} - e^{t} + C \]

### Step 5: Evaluate the integral
Now we can evaluate \( \int_{1}^{2} t e^{t} \, dt \):
\[ \int_{1}^{2} t e^{t} \, dt = \left[ t e^{t} - e^{t} \right]_{1}^{2} \]

Calculating the upper limit:
\[ = \left[ 2 e^{2} - e^{2} \right] = e^{2} \]

Calculating the lower limit:
\[ = \left[ 1 e^{1} - e^{1} \right] = e - e = 0 \]

### Step 6: Combine results
Now substituting back into our expression:
\[ \int_{1}^{2} t e^{t} \, dt = e^{2} - 0 = e^{2} \]
Thus,
\[ 2 \int_{1}^{2} t e^{t} \, dt = 2 e^{2} \]

### Final Answer
The value of the integral \( \int_{1}^{4} e^{\sqrt{x}} \, dx \) is:
\[ \boxed{2 e^{2}} \]