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!