Let `f(x)=int _( x )^(2) (dy)/(sqrt(1+ y ^(3))).` The value of the integral `int _(0)^(2) xf (x ) dx` is equal to:

Options

A. 1
B. `1/3`
C. `4/3`
D. `2/3`

Answer

To solve the problem, we need to evaluate the integral

\[ I = \int_0^2 x f(x) \, dx \]

where

\[ f(x) = \int_x^2 \frac{dy}{\sqrt{1 + y^3}}. \]

### Step 1: Use Integration by Parts

We can use integration by parts, where we let:

- \( u = f(x) \)
- \( dv = x \, dx \)

Then, we have:

- \( du = f'(x) \, dx \)
- \( v = \frac{x^2}{2} \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:

\[ I = \left[ f(x) \cdot \frac{x^2}{2} \right]_0^2 - \int_0^2 \frac{x^2}{2} f'(x) \, dx. \]

### Step 2: Evaluate the Boundary Terms

Now, we evaluate the boundary terms:

\[ \left[ f(x) \cdot \frac{x^2}{2} \right]_0^2 = f(2) \cdot \frac{2^2}{2} - f(0) \cdot \frac{0^2}{2}. \]

Calculating \( f(2) \):

\[ f(2) = \int_2^2 \frac{dy}{\sqrt{1 + y^3}} = 0, \]

and \( f(0) \) is not needed since it is multiplied by zero.

Thus, the boundary term evaluates to:

\[ 0 - 0 = 0. \]

### Step 3: Find \( f'(x) \)

Next, we need to find \( f'(x) \). By the Fundamental Theorem of Calculus, we have:

\[ f'(x) = -\frac{1}{\sqrt{1 + x^3}}. \]

### Step 4: Substitute \( f'(x) \) into the Integral

Now, substituting \( f'(x) \) into the integral, we have:

\[ I = 0 - \int_0^2 \frac{x^2}{2} \left(-\frac{1}{\sqrt{1 + x^3}}\right) \, dx = \frac{1}{2} \int_0^2 \frac{x^2}{\sqrt{1 + x^3}} \, dx. \]

### Step 5: Evaluate the Integral

Now we need to evaluate

\[ \int_0^2 \frac{x^2}{\sqrt{1 + x^3}} \, dx. \]

Using a substitution \( u = 1 + x^3 \), we have \( du = 3x^2 \, dx \) or \( dx = \frac{du}{3x^2} \). When \( x = 0 \), \( u = 1 \), and when \( x = 2 \), \( u = 9 \).

Thus, the integral becomes:

\[ \int_1^9 \frac{x^2}{\sqrt{u}} \cdot \frac{du}{3x^2} = \frac{1}{3} \int_1^9 u^{-1/2} \, du. \]

Evaluating this integral gives:

\[ \frac{1}{3} \cdot \left[ 2u^{1/2} \right]_1^9 = \frac{2}{3} \left( 3 - 1 \right) = \frac{2}{3} \cdot 2 = \frac{4}{3}. \]

### Step 6: Final Calculation of \( I \)

Now substituting back into the expression for \( I \):

\[ I = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}. \]

Thus, the value of the integral \( \int_0^2 x f(x) \, dx \) is

\[ \boxed{\frac{2}{3}}. \]