Let `I = int_(1)^(3)sqrt(x^(4)+x^(2))`, dx then

Options

A. `I gt 6sqrt(10)`
B. `I lt 2sqrt(2)`
C. `2sqrt(2)lt I lt 6sqrt(10)`
D. `I lt 1`

Answer

To solve the integral \( I = \int_{1}^{3} \sqrt{x^4 + x^2} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We start with the expression under the square root:
\[ \sqrt{x^4 + x^2} \]
We can factor out \( x^2 \):
\[ \sqrt{x^4 + x^2} = \sqrt{x^2(x^2 + 1)} = x \sqrt{x^2 + 1} \]
Thus, we can rewrite the integral:
\[ I = \int_{1}^{3} x \sqrt{x^2 + 1} \, dx \]

### Step 2: Use substitution
Let \( t = x^2 + 1 \). Then, differentiating gives:
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} \]
Now we need to change the limits of integration. When \( x = 1 \):
\[ t = 1^2 + 1 = 2 \]
When \( x = 3 \):
\[ t = 3^2 + 1 = 10 \]
Now we substitute \( dx \) and change the limits:
\[ I = \int_{2}^{10} x \sqrt{t} \cdot \frac{dt}{2x} = \frac{1}{2} \int_{2}^{10} \sqrt{t} \, dt \]

### Step 3: Integrate
The integral of \( \sqrt{t} \) is:
\[ \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} \]
Thus, we have:
\[ I = \frac{1}{2} \cdot \frac{2}{3} \left[ t^{3/2} \right]_{2}^{10} = \frac{1}{3} \left[ 10^{3/2} - 2^{3/2} \right] \]

### Step 4: Calculate the limits
Now we calculate \( 10^{3/2} \) and \( 2^{3/2} \):
\[ 10^{3/2} = 10 \sqrt{10} \quad \text{and} \quad 2^{3/2} = 2 \sqrt{2} \]
Thus, we can write:
\[ I = \frac{1}{3} \left[ 10 \sqrt{10} - 2 \sqrt{2} \right] \]

### Step 5: Final expression
Now we have the final expression for the integral:
\[ I = \frac{10 \sqrt{10} - 2 \sqrt{2}}{3} \]