if ` int_(0)^f(x) 4x^(3)dx=g(x)(x-2)` if `f(2)=6` and `f'(2)=(1)/(48)` then find `lim_(x to 2) g(x)`

Options

A. 18
B. 17
C. 20
D. 19

Answer

To solve the problem, we need to analyze the given integral and the conditions provided.

The problem states that:

\[ \int_{0}^{f(x)} 4x^3 \, dx = g(x)(x-2) \]

We need to find \( \lim_{x \to 2} g(x) \) given that \( f(2) = 6 \) and \( f'(2) = \frac{1}{48} \).

### Step 1: Evaluate the integral

First, we evaluate the integral:

\[ \int_{0}^{f(x)} 4x^3 \, dx \]

Using the power rule for integration, we have:

\[ \int 4x^3 \, dx = x^4 + C \]

Thus,

\[ \int_{0}^{f(x)} 4x^3 \, dx = [x^4]_{0}^{f(x)} = (f(x))^4 - 0^4 = (f(x))^4 \]

### Step 2: Set up the equation

Now we can rewrite the equation:

\[ (f(x))^4 = g(x)(x-2) \]

### Step 3: Substitute \( x = 2 \)

Next, we substitute \( x = 2 \) into the equation:

\[ (f(2))^4 = g(2)(2-2) \]

Since \( 2 - 2 = 0 \), the right-hand side becomes \( 0 \). Therefore, we have:

\[ (f(2))^4 = 0 \]

### Step 4: Calculate \( f(2) \)

From the problem, we know that \( f(2) = 6 \). Therefore:

\[ (6)^4 = 0 \]

This is not possible, which indicates that we need to analyze the limit as \( x \) approaches \( 2 \).

### Step 5: Differentiate both sides

To find \( g(x) \), we differentiate both sides with respect to \( x \):

Using the chain rule on the left side:

\[ \frac{d}{dx}[(f(x))^4] = 4(f(x))^3 f'(x) \]

On the right side, we apply the product rule:

\[ \frac{d}{dx}[g(x)(x-2)] = g'(x)(x-2) + g(x) \]

Setting these equal gives us:

\[ 4(f(x))^3 f'(x) = g'(x)(x-2) + g(x) \]

### Step 6: Substitute \( x = 2 \) again

Now we substitute \( x = 2 \):

\[ 4(f(2))^3 f'(2) = g'(2)(2-2) + g(2) \]

The left-hand side becomes:

\[ 4(6)^3 \cdot \frac{1}{48} = 4 \cdot 216 \cdot \frac{1}{48} = \frac{864}{48} = 18 \]

The right-hand side simplifies to:

\[ g(2) \]

Thus, we have:

\[ g(2) = 18 \]

### Final Answer

Therefore, the limit we are looking for is:

\[ \lim_{x \to 2} g(x) = 18 \]