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Chapter 7 Additional Integration Topics Section 1 Area Between Curves Learning Objectives for Section 7.1 Area Between Two Curves The student will be able to: Day 1: • Determine the unsigned area between a curve and the x axis. • Determine the unsigned area between two curves. Day 2: • Solve applications involving area between curves Barnett/Ziegler/Byleen Business Calculus 12e 2 Area “Between Curves” In the previous chapter, we found that the definite integral 𝑏 𝑓 𝑥 𝑑𝑥 𝑎 represents the accumulated sum of the signed areas between the graph of f(x) and the x-axis over the interval [a, b]. In this chapter: • We will treat all areas as positive, even though they may fall below the x-axis. • We will find the total unsigned area between two curves. Barnett/Ziegler/Byleen Business Calculus 12e 3 No Negative Areas! The unsigned area between the graph of a negative function and the x axis is equal to the definite integral of the negative of the function. A b a [ f ( x) ] dx B 𝐴=− c b B a A 𝑏 𝑂𝑅 y = f (x) b c 𝑓 𝑥 𝑑𝑥 𝑎 f ( x) dx Barnett/Ziegler/Byleen Business Calculus 12e 4 Area Between f(x) and x-axis The area between f (x) and the x axis can be found as follows: If f (x) is above the x axis: Area = b a f ( x) dx If f (x) is below the x axis: Area = b a [ f ( x) ] dx 𝑏 𝑂𝑅 𝐴𝑟𝑒𝑎 = − 𝑓 𝑥 𝑑𝑥 𝑎 Barnett/Ziegler/Byleen Business Calculus 12e 5 Example 1 Find the area bounded by y = 4 – x2 x-axis Find x-intercepts: 1x4 -2, 2 2 𝐴𝑟𝑒𝑎 = 4 𝑓 𝑥 𝑑𝑥 − 1 2 𝑓 𝑥 𝑑𝑥 2 4 4 − 𝑥 2 𝑑𝑥 − = 1 𝑥3 2 = 4𝑥 − − 3 1 = 4 − 𝑥 2 𝑑𝑥 2 𝑥3 4𝑥 − 3 8 1 8− − 4− 3 3 − 4 2 64 8 16 − − 8− 3 3 Barnett/Ziegler/Byleen Business Calculus 12e 37 = 3 6 Example 2 Find the area bounded by y = x2 – 9 Find x-intercepts: x-axis -3, 3 Over the interval: [2, 4] 3 𝐴𝑟𝑒𝑎 = − 4 𝑓 𝑥 𝑑𝑥 + 2 3 3 4 𝑥 2 − 9 𝑑𝑥 + =− 3 2 𝑥 2 − 9 𝑑𝑥 3 4 −𝑥 2 + 9 𝑑𝑥 + = 2 𝑥3 3 = − + 9𝑥 + 3 2 = 𝑓 𝑥 𝑑𝑥 𝑥 2 − 9 𝑑𝑥 3 3 𝑥 4 − 9𝑥 3 3 8 −9 + 27 − − + 18 + 3 Barnett/Ziegler/Byleen Business Calculus 12e 64 − 36 − 9 − 27 3 =6 7 Area Between Two Curves Theorem 1. If f (x) g(x) over the interval [a, b], then the area bounded by y = f (x) and y = g(x) for a x b is given by A f ( x) g ( x) dx b f (x) A a b g (x) a Note: The graphs do not need to be above the xaxis; however, the graph of f must be above (or equal) to the graph of g throughout the interval [a, b]. Barnett/Ziegler/Byleen Business Calculus 12e 8 Example 3 Find the area bounded by f(x) = x2 – 1 and g(x) = 3. Interval isn’t given so you need to find where the two graphs intersect. (use “CALC Intersect” function on calculator or do it algebraically) (-2, 3) & (2, 3) 2 𝐴𝑟𝑒𝑎 = 2 3 − 𝑥 2 − 1 𝑑𝑥 𝑔 𝑥 − 𝑓 𝑥 𝑑𝑥 = −2 −2 2 4 − 𝑥 2 𝑑𝑥 = −2 3 𝑥 2 8 8 = 4𝑥 − = 8− − −8 + 3 −2 3 3 Barnett/Ziegler/Byleen Business Calculus 12e 32 = 3 9 Example 4 Find the area bounded by 𝑓 𝑥 = 5 − 𝑥 2 and 𝑔 𝑥 = 2 − 2𝑥 Finding intersection points algebraically: 5 − 𝑥 2 = 2 − 2𝑥 𝑥 2 − 2𝑥 − 3 = 0 (𝑥 − 3)(𝑥 + 1) = 0 𝑥 = −1, 3 3 𝑓 𝑥 − 𝑔(𝑥) 𝑑𝑥 −1 Barnett/Ziegler/Byleen Business Calculus 12e 10 Example 4 (continued) 3 5 − 𝑥 2 − (2 − 2𝑥) 𝑑𝑥 −1 3 3 + 2𝑥 − 𝑥 2 𝑑𝑥 −1 3 3 𝑥 3𝑥 + 𝑥 2 − 3 −1 1 9 + 9 − 9 − −3 + 1 + 3 Barnett/Ziegler/Byleen Business Calculus 12e 32 = 3 11 Example 5 Find the area bounded by 𝑓 𝑥 = 2𝑥 2 and 𝑔 𝑥 = 4 − 2𝑥 [-2, 2] 𝑓(𝑥) 𝑔(𝑥) 1 2 𝑔 𝑥 − 𝑓(𝑥) 𝑑𝑥 + −2 𝑓 𝑥 − 𝑔(𝑥) 𝑑𝑥 1 1 2 4 − 2𝑥 − 2𝑥 2 𝑑𝑥 + 2𝑥 2 − (4 − 2𝑥) 𝑑𝑥 −2 3 1 2𝑥 4𝑥 − 𝑥 2 − + 3 −2 1 2𝑥 3 − 4𝑥 + 𝑥 2 3 Barnett/Ziegler/Byleen Business Calculus 12e 2 1 38 = 3 12 Example 6 Find the area bounded by 𝑦 = 25 − 𝑥 2 and y = 0; -5 x 5 5 25 − 𝑥 2 𝑑𝑥 𝐴𝑟𝑒𝑎 = −5 1 2 𝐴𝑟𝑒𝑎 = 𝜋𝑟 2 1 2 = 𝜋5 2 25 = 𝜋 2 You don’t know how to evaluate these types of integrals yet, so use geometry. Barnett/Ziegler/Byleen Business Calculus 12e 13 Homework Barnett/Ziegler/Byleen Business Calculus 12e 14