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The given problem defines a function \( f(x) \) as:

\[
f(x) = \int_0^x \left( t^3 + 3t^2 + 2 \right) \, dt
\]

We are tasked with finding \( f''(x) \), the second derivative of \( f(x) \).

### Step 1: Differentiate \( f(x) \)

Using the Fundamental Theorem of Calculus, we know that the derivative of a function defined as an integral from a constant to \( x \) is simply the integrand evaluated at \( x \). Thus:

\[
f'(x) = t^3 + 3t^2 + 2 \Big|_{t=x} = x^3 + 3x^2 + 2
\]

### Step 2: Find \( f''(x) \)

Now, we differentiate \( f'(x) \) to obtain \( f''(x) \):

\[
f''(x) = \frac{d}{dx} \left( x^3 + 3x^2 + 2 \right)
\]

Using the power rule:

\[
f''(x) = 3x^2 + 6x
\]

Thus, the second derivative is:

\[
f''(x) = 3x^2 + 6x
\]

### Final Answer:
\[
f''(x) = 3x^2 + 6x
\]

Do you need further details or have any questions?

Here are 5 related questions:
1. How does the Fundamental Theorem of Calculus apply to differentiation of integrals?
2. What would the third derivative \( f'''(x) \) be for this function?
3. What is the significance of finding higher-order derivatives like \( f''(x) \)?
4. How would this process change if the integrand were a more complex function?
5. Can you explain how to integrate \( t^3 + 3t^2 + 2 \) directly to verify \( f(x) \)?

**Tip:** When differentiating an integral with a variable upper limit, the integrand can often be directly used for the first derivative.

If f(x) = ∫ from 0 to x of (t^3 + 3t^2 + 2) dt, then f''(x) = ?

Explore how to find the second derivative f''(x) when f(x) is defined as an integral from 0 to x of (t^3 + 3t^2 + 2) dt. This calculus problem covers concepts of differentiation and the Fundamental Theorem of Calculus, making it suitable for students in Grades 11-12.

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