Let's solve each integral by interpreting the graph in terms of geometric areas. The areas **above the x-axis are positive**, and areas **below the x-axis are negative**. Let's break the calculations down step by step. --- ### **Step 1: Understand the graph** We see the piecewise linear graph of \( f(x) \) from \( x = 0 \) to \( x = 9 \). Vertices (approximate from the graph): - (0, 0) - (1, 1) - (3, 1) - (5, -1) - (7, -2) - (9, -1) --- ### **Step 2: Calculate areas for each interval** #### 1. \(\int_0^2 f(x)\,dx\) From \(x=0\) to \(x=1\): Triangle from (0,0) to (1,1) - Area = \( \frac{1}{2} \times 1 \times 1 = 0.5 \) From \(x=1\) to \(x=2\): Rectangle (height = 1, width = 1) - Area = \( 1 \times 1 = 1 \) Total area from \(x=0\) to \(x=2\): \(0.5 + 1 = \boxed{1.5}\) --- #### 2. \(\int_0^5 f(x)\,dx\) Sum areas piece by piece: - From \(x=0\) to \(x=1\): Triangle, area = 0.5 - From \(x=1\) to \(x=3\): Rectangle, area = 2 (height 1, width 2) - From \(x=3\) to \(x=5\): Trapezoid (top base = 1, bottom base = -1, height = 2) Trapezoid area = \( \frac{1}{2} \times (1 + (-1)) \times 2 = 0 \) (since bases add to zero, area=0) **Sum up:** \(0.5 + 2 = \boxed{2.5}\) --- #### 3. \(\int_5^7 f(x)\,dx\) From \(x=5\) to \(x=7\): Straight line from (-1) to (-2). This forms a trapezoid: - Height: 2 (from 5 to 7) - Bases: -1 and -2 Area = \( \frac{1}{2} \times (-1 + -2) \times 2 = \frac{1}{2} \times (-3) \times 2 = -3 \) So, \(\boxed{-3}\) --- #### 4. \(\int_0^9 f(x)\,dx\) Add up all areas from each interval: - \(0 \to 1\): Triangle, area = 0.5 - \(1 \to 3\): Rectangle, area = 2 - \(3 \to 5\): Trapezoid: \( \frac{1}{2} \times (1 + (-1)) \times 2 = 0 \) - \(5 \to 7\): Trapezoid: \( -3 \) (already calculated) - \(7 \to 9\): Trapezoid: bases (-2 and -1), height 2 Area = \( \frac{1}{2} \times (-2 + -1) \times 2 = \frac{1}{2} \times (-3) \times 2 = -3 \) Sum: \(0.5 + 2 + 0 + (-3) + (-3) = 2.5 - 6 = \boxed{-3.5}\) --- ## **Final Answers:** 1. \(\int_0^2 f(x)\,dx = \boxed{1.5}\) 2. \(\int_0^5 f(x)\,dx = \boxed{2.5}\) 3. \(\int_5^7 f(x)\,dx = \boxed{-3}\) 4. \(\int_0^9 f(x)\,dx = \boxed{-3.5}\) Let me know if you need detailed breakdown with sketches!