Let's evaluate each integral by interpreting the areas involved.

Problem 76

$\int_0^3 (3 - x) \, dx$ The integrand, $3 - x$, represents a line with a y-intercept of 3 and a slope of -1, so it crosses the x-axis at $x = 3$. From $x = 0$ to $x = 3$, this line forms a right triangle with the x-axis.

• Base of the triangle = 3 (from $x = 0$ to $x = 3$)
• Height of the triangle = 3 (the y-intercept)

The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$: $\int_0^3 (3 - x) \, dx = \frac{1}{2} \times 3 \times 3 = \frac{9}{2} = 4.5$

Problem 77

$\int_2^3 (3 - x) \, dx$ Here, we are looking at the same line $3 - x$ but only on the interval $x = 2$ to $x = 3$.

• Base of this smaller triangle = 1 (from $x = 2$ to $x = 3$)
• Height of the triangle = 1 (since $3 - 2 = 1$ at $x = 2$ and $3 - 3 = 0$ at $x = 3$)

The area of this triangle is: $\int_2^3 (3 - x) \, dx = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} = 0.5$

Problem 78

$\int_{-3}^3 (3 - |x|) \, dx$ The function $3 - |x|$ is symmetric about the y-axis. This function forms a triangle with vertices at $(-3, 0)$, $(0, 3)$, and $(3, 0)$.

• Base of the triangle = 6 (from $x = -3$ to $x = 3$)
• Height of the triangle = 3 (the maximum value at $x = 0$)

The area of this triangle is: $\int_{-3}^3 (3 - |x|) \, dx = \frac{1}{2} \times 6 \times 3 = 9$

Final Answers

1. 76: $\int_0^3 (3 - x) \, dx = 4.5$
2. 77: $\int_2^3 (3 - x) \, dx = 0.5$
3. 78: $\int_{-3}^3 (3 - |x|) \, dx = 9$

Would you like further details or explanations on any of these steps?

Related Questions

1. How would you compute the integral if the bounds or function changed, such as $\int_1^4 (3 - x) \, dx$?
2. What if the function had a different slope, such as $\int_0^3 (4 - 2x) \, dx$?
3. How would we interpret areas for integrals involving $|x|$ over different bounds, like $\int_{-2}^2 (5 - |x|) \, dx$?
4. Can you apply similar geometric interpretations to integrals of non-linear functions, such as $\int_0^3 (3 - x^2) \, dx$?
5. What are some other common functions where area formulas simplify the computation of integrals?

Tip

When evaluating integrals over linear functions, visualize the graph. This often helps you identify geometric shapes (triangles, rectangles) for quick area calculations.