`int_(0)^(1)(x^(2)+3x+2)/(sqrt(x))dx=`

Options

A. `(23)/(5)`
B. `(32)/(5)`
C. `-(32)/(5)`
D. `(5)/(32)`

Answer

To solve the definite integral

\[ I = \int_{0}^{1} \frac{x^2 + 3x + 2}{\sqrt{x}} \, dx, \]

we can break it down into simpler parts using the linearity of integration.

### Step 1: Split the Integral

We can separate the integral into three parts:

\[ I = \int_{0}^{1} \frac{x^2}{\sqrt{x}} \, dx + 3 \int_{0}^{1} \frac{x}{\sqrt{x}} \, dx + 2 \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx. \]

### Step 2: Simplify Each Integral

Now, we simplify each term:

1. For the first integral:
\[ \frac{x^2}{\sqrt{x}} = x^{2 - \frac{1}{2}} = x^{\frac{3}{2}}. \]
So,
\[ \int_{0}^{1} \frac{x^2}{\sqrt{x}} \, dx = \int_{0}^{1} x^{\frac{3}{2}} \, dx. \]

2. For the second integral:
\[ \frac{x}{\sqrt{x}} = x^{1 - \frac{1}{2}} = x^{\frac{1}{2}}. \]
So,
\[ 3 \int_{0}^{1} \frac{x}{\sqrt{x}} \, dx = 3 \int_{0}^{1} x^{\frac{1}{2}} \, dx. \]

3. For the third integral:
\[ \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}. \]
So,
\[ 2 \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx = 2 \int_{0}^{1} x^{-\frac{1}{2}} \, dx. \]

### Step 3: Evaluate Each Integral

Now we evaluate each integral using the formula

\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C. \]

1. For \(\int_{0}^{1} x^{\frac{3}{2}} \, dx\):
\[ = \left[ \frac{x^{\frac{5}{2}}}{\frac{5}{2}} \right]_{0}^{1} = \left[ \frac{2}{5} x^{\frac{5}{2}} \right]_{0}^{1} = \frac{2}{5} (1) - 0 = \frac{2}{5}. \]

2. For \(3 \int_{0}^{1} x^{\frac{1}{2}} \, dx\):
\[ = 3 \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{1} = 3 \left[ \frac{2}{3} x^{\frac{3}{2}} \right]_{0}^{1} = 3 \cdot \frac{2}{3} (1) - 0 = 2. \]

3. For \(2 \int_{0}^{1} x^{-\frac{1}{2}} \, dx\):
\[ = 2 \left[ \frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right]_{0}^{1} = 2 \cdot 2 \left[ x^{\frac{1}{2}} \right]_{0}^{1} = 4 (1) - 0 = 4. \]

### Step 4: Combine the Results

Now, we combine all the results:

\[ I = \frac{2}{5} + 2 + 4 = \frac{2}{5} + \frac{10}{5} + \frac{20}{5} = \frac{32}{5}. \]

### Final Answer

Thus, the value of the integral is

\[ \boxed{\frac{32}{5}}. \]