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The image presents two problems asking for the derivative \( G'(x) \) using the Fundamental Theorem of Calculus. Here's the breakdown for both:

### Problem 17:
\[
G(x) = \int_1^x 2t \, dt
\]
We need to find \( G'(x) \). By the **Fundamental Theorem of Calculus**, if \( G(x) = \int_a^x f(t) \, dt \), then:
\[
G'(x) = f(x)
\]
In this case, \( f(t) = 2t \). Therefore:
\[
G'(x) = 2x
\]

### Problem 18:
\[
G(x) = \int_x^1 2t \, dt
\]
Again, we use the **Fundamental Theorem of Calculus**. For integrals with the upper limit being a constant and the lower limit being \( x \), we use the following rule:
\[
G'(x) = -f(x)
\]
Here, \( f(t) = 2t \), so:
\[
G'(x) = -2x
\]

Thus, the derivatives for both problems are:
- For problem 17: \( G'(x) = 2x \)
- For problem 18: \( G'(x) = -2x \)

Would you like any further details on the steps, or do you have any questions?

### Follow-up questions:
1. How does the Fundamental Theorem of Calculus apply to integrals with variable limits?
2. What happens when the bounds of the integral are reversed, as in problem 18?
3. Can you explain why the derivative in problem 18 includes a negative sign?
4. How would the solution change if the integrand were a more complex function, like \( t^2 \)?
5. Can you apply this concept to solve similar problems with more general functions?

**Tip:** When the bounds of an integral are swapped, the integral's value becomes negative.

In Problems 17-26, find G'(x). Problem 17: G(x) = ∫(1 to x) 2t dt; Problem 18: G(x) = ∫(x to 1) 2t dt.

Find the derivative of integrals in Problems 17 and 18 using the Fundamental Theorem of Calculus. Learn step-by-step how to apply this theorem to solve G'(x) for the integrals ∫(1 to x) 2t dt and ∫(x to 1) 2t dt. Suitable for advanced calculus students in Grades 11-12.

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