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The problem defines \( g(x) = \int_0^x f(t) \, dt \), where \( f(t) \) is the function represented by the graph. This integral represents the area under the curve of \( f(t) \) from \( t = 0 \) to \( t = x \).

### (a) Evaluate \( g(0) \) and \( g(6) \).

- **\( g(0) \)**: Since \( g(x) = \int_0^x f(t) \, dt \), \( g(0) \) represents the integral from 0 to 0, which is always 0:
  \[
  g(0) = \int_0^0 f(t) \, dt = 0.
  \]

- **\( g(6) \)**: To compute \( g(6) \), we need to find the area under the curve of \( f(t) \) from 0 to 6. From the graph, we can visually estimate this area by dividing it into segments. It looks like:
  - From 0 to 3, the area is mostly above the x-axis and positive.
  - From 3 to 6, the area is below the x-axis and negative.
  
  Let's approximate these areas based on the graph:
  - Area from 0 to 3 (positive): Approximate as a trapezoid with base 3 and average height around 1.5, giving an approximate area of \( 3 \times 1.5 = 4.5 \).
  - Area from 3 to 6 (negative): Approximate as a triangle with base 3 and height around -2, giving an approximate area of \( \frac{1}{2} \times 3 \times (-2) = -3 \).

  Thus, we estimate:
  \[
  g(6) \approx 4.5 + (-3) = 1.5.
  \]

### (b) Estimate \( g(x) \) for \( x = 1, 2, 3, 4, 5 \).

To estimate \( g(x) \) at these points, we'll consider the areas under the curve up to these values of \( x \).

- \( g(1) \): The area from 0 to 1 looks like a small trapezoid, so:
  \[
  g(1) \approx 0.5.
  \]

- \( g(2) \): The area from 0 to 2 is larger, so:
  \[
  g(2) \approx 2.
  \]

- \( g(3) \): The area from 0 to 3 is approximately the trapezoid as estimated earlier:
  \[
  g(3) \approx 4.5.
  \]

- \( g(4) \): The area starts to include negative values after 3. Subtracting the triangle from 3 to 4:
  \[
  g(4) \approx 4.5 + (-1) = 3.5.
  \]

- \( g(5) \): The area from 3 to 5 is mostly negative:
  \[
  g(5) \approx 4.5 + (-2) = 2.5.
  \]

### (c) On what interval is \( g \) increasing?

The function \( g(x) \) is increasing when the graph of \( f(x) \) is above the x-axis (i.e., when \( f(x) > 0 \)). From the graph, \( f(x) \) is positive from \( x = 0 \) to \( x = 3 \), so \( g(x) \) is increasing on the interval \( (0, 3) \).

### (d) Where does \( g \) have a maximum?

The function \( g(x) \) reaches a maximum where the area under the curve stops increasing and starts decreasing. This happens at \( x = 3 \), where \( f(x) \) changes from positive to negative.

### (e) Sketch a rough graph of \( g(x) \).

The graph of \( g(x) \) starts at 0 at \( x = 0 \), increases up to \( x = 3 \), and then decreases slightly as the negative area is added between \( x = 3 \) and \( x = 6 \).

### (f) Use the graph from part (e) to sketch the graph of \( g'(x) \). Compare with the graph of \( f \).

Since \( g'(x) = f(x) \), the graph of \( g'(x) \) should be identical to the graph of \( f(x) \).

---

Would you like more details on any part? Here are 5 related questions you might find useful:

1. What are other techniques for estimating the area under a curve?
2. How can we improve the accuracy of our area estimations?
3. What happens to \( g(x) \) when \( f(x) \) has a discontinuity?
4. How does the Fundamental Theorem of Calculus relate to this problem?
5. How can we compute exact values for \( g(x) \) if \( f(x) \) is given algebraically?

**Tip**: Always visualize the function you're integrating over to better understand the behavior of its integral!

Let g(x) = ∫0^x f(t) dt, where f is the function whose graph is shown. (a) Evaluate g(0) and g(6). (b) Estimate g(x) for x = 1, 2, 3, 4, 5. (c) On what interval is g increasing? (d) Where does g have a maximum? (e) Sketch a rough graph of g. (f) Use the graph from part (e) to sketch the graph of g'(x). Compare with the graph of f.

Explore how to calculate g(x) from the graph of f(x) by finding the area under the curve. Estimate g(x) values, identify intervals where g(x) increases, and determine its maximum. Learn how g'(x) relates to f(x) by applying the Fundamental Theorem of Calculus.

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