The value of `lim_(xrarr2)int_(2)^(x)(3t^(2))/(x-2)dt` is

Options

A. 10
B. 12
C. 8
D. 16

Answer

To solve the limit problem \( \lim_{x \to 2} \int_{2}^{x} \frac{3t^2}{x-2} dt \), we will follow these steps:

### Step 1: Set up the integral
We start with the expression:
\[ \lim_{x \to 2} \int_{2}^{x} \frac{3t^2}{x-2} dt \]

### Step 2: Integrate the function
Since the integrand is \( \frac{3t^2}{x-2} \), we can factor out \( \frac{1}{x-2} \) from the integral:
\[ \lim_{x \to 2} \frac{1}{x-2} \int_{2}^{x} 3t^2 dt \]

### Step 3: Compute the definite integral
Now we compute the integral \( \int_{2}^{x} 3t^2 dt \):
\[ \int 3t^2 dt = t^3 + C \]
Thus,
\[ \int_{2}^{x} 3t^2 dt = [t^3]_{2}^{x} = x^3 - 2^3 = x^3 - 8 \]

### Step 4: Substitute back into the limit
Now substituting back, we have:
\[ \lim_{x \to 2} \frac{x^3 - 8}{x-2} \]

### Step 5: Factor the numerator
Notice that \( x^3 - 8 \) can be factored using the difference of cubes:
\[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{x - 2} \]

### Step 6: Cancel the common terms
We can cancel \( (x - 2) \) from the numerator and denominator (as long as \( x \neq 2 \)):
\[ \lim_{x \to 2} (x^2 + 2x + 4) \]

### Step 7: Evaluate the limit
Now we can directly substitute \( x = 2 \):
\[ 2^2 + 2(2) + 4 = 4 + 4 + 4 = 12 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{12} \]

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