If `f` is continuously differentiable function then `int_(0)^(1.5) [x^2] f'(x) dx` is

Options

A. `f(1.5) - f(sqrt2) - f(1)`
B. `f(1.5) + f(sqrt2) + f(1)`
C. `2f(1.5) + f (sqrt2) + f(1)`
D. `2f(1.5) - f(sqrt2) - f(1)`

Answer

To solve the integral \( \int_{0}^{1.5} x^2 f'(x) \, dx \), we can use integration by parts. The integration by parts formula is given by:

\[ \int u \, dv = uv - \int v \, du \]

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = x^2 \) (which means \( du = 2x \, dx \))
- \( dv = f'(x) \, dx \) (which means \( v = f(x) \))

### Step 2: Apply the integration by parts formula

Using the integration by parts formula, we have:

\[ \int_{0}^{1.5} x^2 f'(x) \, dx = \left[ x^2 f(x) \right]_{0}^{1.5} - \int_{0}^{1.5} f(x) \cdot 2x \, dx \]

### Step 3: Evaluate the boundary term

Now we need to evaluate the boundary term \( \left[ x^2 f(x) \right]_{0}^{1.5} \):

\[ \left[ x^2 f(x) \right]_{0}^{1.5} = (1.5)^2 f(1.5) - (0)^2 f(0) = 2.25 f(1.5) - 0 = 2.25 f(1.5) \]

### Step 4: Write the integral

Now we can write the expression:

\[ \int_{0}^{1.5} x^2 f'(x) \, dx = 2.25 f(1.5) - 2 \int_{0}^{1.5} x f(x) \, dx \]

### Step 5: Final expression

Thus, the final expression for the integral \( \int_{0}^{1.5} x^2 f'(x) \, dx \) is:

\[ \int_{0}^{1.5} x^2 f'(x) \, dx = 2.25 f(1.5) - 2 \int_{0}^{1.5} x f(x) \, dx \]