Let's solve each question one at a time.

Question 4

We are given: $G(x) = \int_3^x (t + 1)^3 \, dt$ and asked to compute $G'(7)$.

Using the Fundamental Theorem of Calculus: $G'(x) = \frac{d}{dx} \int_3^x (t + 1)^3 \, dt = (x + 1)^3$ since differentiating the integral with respect to the upper limit $x$ replaces $t$ with $x$ in the integrand.

Thus, $G'(7) = (7 + 1)^3 = 8^3 = 512.$ Answer: $G'(7) = 512$.

Question 5

We are asked to evaluate: $\int_0^5 f(x) \, dx$ using the provided graph of $f(x)$.

To do this, we will calculate the areas under the curve from $x = 0$ to $x = 5$, noting that areas above the x-axis are positive and areas below are negative.

1. From $x = 0$ to $x = 1$: The area is a triangle with base 1 and height 1, so: $\text{Area} = \frac{1}{2} \times 1 \times 1 = 0.5$

2. From $x = 1$ to $x = 3$: The area is a triangle with base 2 and height 2, so: $\text{Area} = \frac{1}{2} \times 2 \times 2 = 2$

3. From $x = 3$ to $x = 4$: The area is a triangle below the x-axis with base 1 and height 1, so: $\text{Area} = -\frac{1}{2} \times 1 \times 1 = -0.5$

4. From $x = 4$ to $x = 5$: The area is a triangle below the x-axis with base 1 and height 2, so: $\text{Area} = -\frac{1}{2} \times 1 \times 2 = -1$

Adding these areas together: $\int_0^5 f(x) \, dx = 0.5 + 2 - 0.5 - 1 = 1$ Answer: $\int_0^5 f(x) \, dx = 1$.

Do you need further clarification on any step?

Related Questions

1. How would the solution to Question 4 change if the lower limit of the integral was not constant?
2. What is the significance of the Fundamental Theorem of Calculus in computing derivatives of integral functions?
3. How would the answer to Question 5 change if the graph had different values on the y-axis?
4. What method could we use to approximate integrals if we only had a table of values instead of a graph?
5. How does the concept of signed area apply to interpreting definite integrals with portions below the x-axis?

Tip

For integrals involving graphs, always remember that areas below the x-axis contribute negatively to the integral.