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Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 78

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 78. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | 23.2 Characteristics of the Area Function 751 SOLUTION Let s begin with -2A f x because none of the domain is to the left of the anchor point. -2A f x is increasing on -2 2 and decreasing on 2 6 . _2Af x is concave down on -2 6 because f is decreasing. -2 graph of _2Af sign of f 6 Finding a Formula for -2Af x . First consider -2Af x for x on the interval -2 2 . -2Af x is the area of the trapezoid bounded by f t 2 - t the t-axis t -2 and t x. See Figure 23.10. Its area is 1 1 - hl h2 b 4 2 - x x - -2 1 2 6 - x x 2 12 2 12 - x2 4x x2 ------F 2x 6. 2 2 -2Af x --y 2x 6 on -2 2 . For x on 2 6 to flnd -2Af x we must subtract from the area of the triangle above the t-axis the area of the appropriate triangle below. See Figure 23.11. 752 CHAPTER 23 The Area Function and Its Characteristics area of triangle above the t-axis - area of triangle below the t-axis 1 1 _2 4 4 _ x - 2 x - 2 8 - 2 x2 - 4x 4 x2 8 - 2x - 2 x 1 2 2x 6 x2 So -2Af x - X2 2x 6 on the entire domain given. Once we have found -2Af x finding 0Af x and 1 Af x is not difficult because the functions differ only by additive constants. Finding a Formula for 0Af x . 0Af x jox f t dt. To relate 0Af x to -2Af x we use the splitting interval property j f t dt j f t dt j f t dt. i f t dt i f t dt i f t dt 2 -2 J0 -2Af x y f t dt oAf x where j 2 f t dt is the area of the trapezoid shaded in Figure 23.12. j-2 f t dt 2 4 2 2 6 -2Af x 6 oAf x So oAf x -2 Af x - 6. 2 2 - 2Af x - 2x 6 therefore 0Af x - 2x. 23.2 Characteristics of the Area Function 753 Figure 23.12 Finding a Formula for 1 Af x . Similarly ff2 f t dt f 2 f t dt ff f t dt. 2Af x y f t dt 1Af x where f22 f t dt is the area shaded in Figure 23.13. -2Af 0 f -2 -1 Figure 23.13 t 1 2Af 1 2 4 1 3 15 15 2Af x 1Af x So 15 1Af x 2Af x . 2Af x X- 2x 6 therefore 1Af x X- 2x 1.5. r2 Alternatively we could have found 1 Af x by starting with 1 Af x 2x K. Knowing that 1 Af 1 0 we can solve for .