use the fundamental theorem of calculus part 2  to evaluate the integral 0 to x of 3 plus sin of t and differentiate

To solve the problem, we'll use **The Fundamental Theorem of Calculus, Part 2**. It states that:

\[
\frac{d}{dx} \left( \int_{a}^{x} f(t) \, dt \right) = f(x)
\]

Here, we are given the integral:

\[
\int_0^x (3 + \sin(t)) \, dt
\]

### Step 1: Apply the Fundamental Theorem of Calculus, Part 2

By Part 2 of the theorem, if you differentiate this integral with respect to \( x \), you can replace the dummy variable \( t \) with \( x \), leading to:

\[
\frac{d}{dx} \left( \int_0^x (3 + \sin(t)) \, dt \right) = 3 + \sin(x)
\]

Thus, the derivative of the integral from \( 0 \) to \( x \) of \( 3 + \sin(t) \) is:

\[
3 + \sin(x)
\]

### Final Answer:
\[
\boxed{3 + \sin(x)}
\]

Let me know if you'd like a further breakdown or any clarifications!

Here are 5 related questions to expand your understanding:
1. How does Part 1 of the Fundamental Theorem of Calculus differ from Part 2?
2. Can you compute the definite integral \( \int_0^x (3 + \sin(t)) \, dt \) directly?
3. How would the answer change if the limits of integration were from \( a \) to \( x \) instead of 0 to \( x \)?
4. What happens if we differentiate \( \int_a^b (3 + \sin(t)) \, dt \) with respect to \( x \) when both \( a \) and \( b \) are constants?
5. Can you apply the same process if the function inside the integral were more complex, such as \( \cos(t^2) \)?

**Tip:** The Fundamental Theorem of Calculus is a powerful tool for connecting differentiation and integration, making it easier to evaluate integrals and their derivatives efficiently.

Explore how to use the Fundamental Theorem of Calculus Part 2 to evaluate the integral of (3 + sin(t)) from 0 to x and differentiate the result. This detailed solution covers integration, differentiation, and trigonometric functions. Suitable for advanced high school students.

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