DISCOVERY PROJECT 4.3 AREA FUNCTIONS ■ 1 DISCOVERY PROJECT: AREA FUNCTIONS This project can be completed anytime after you have studied Section 4.3 in the textbook. 1. (a) Draw the line y 苷 2t ⫹ 1 and use geometry to find the area under this line, above the t -axis, and between the vertical lines t 苷 1 and t 苷 3. (b) If x ⬎ 1, let A共x兲 be the area of the region that lies under the line y 苷 2t ⫹ 1 between t 苷 1 and t 苷 x . Sketch this region and use geometry to find an expression for A共x兲. (c) Differentiate the area function A共x兲. What do you notice? 2. (a) If 0 艋 x 艋 ␲, let A共x兲苷 (b) (c) (d) (e) x0x sin t dt . A共x兲 represents the area of a region. Sketch that region. Use the Evaluation Theorem to find an expression for A共x兲. Find A⬘共x兲. What do you notice? If x is any number between 0 and ␲ and h is a small positive number, then A共x ⫹ h兲 ⫺ A共x兲 represents the area of a region. Describe and sketch the region. Draw a rectangle that approximates the region in part (d). By comparing the areas of these two regions, show that A共x ⫹ h兲 ⫺ A共x兲 ⬇ sin x h (f) Use part (e) to give an intuitive explanation for the result of part (c). 2 ; 3. (a) Draw the graph of the function f 共x兲苷 cos共x 兲 in the viewing rectangle 关0, 2兴 by 关⫺1.25, 1.25兴. (b) If we define a new function t by x t共x兲苷 y cos共t 2 兲 dt 0 then t共x兲 is the area under the graph of f from 0 to x [until f 共x兲 becomes negative, at which point t共x兲 becomes a difference of areas]. Use part (a) to determine the value of x at which t共x兲 starts to decrease. [Unlike the integral in Problem 2, it is impossible to evaluate the integral defining t to obtain an explicit expression for t共x兲.] (c) Use the integration command on your calculator or computer to estimate t共0.2兲, t共0.4兲, t共0.6兲, . . . , t共1.8兲, t共2兲. Then use these values to sketch a graph of t. (d) Use your graph of t from part (c) to sketch the graph of t⬘ using the interpretation of t⬘共x兲 as the slope of a tangent line. How does the graph of t⬘ compare with the graph of f ? 4. Suppose f is a continuous function on the interval 关a, b兴 and we define a new function t by the equation x t共x兲苷 y f 共t兲 dt a Copyright © 2013, Cengage Learning. All rights reserved. Based on your results in Problems 1–3, conjecture an expression for t⬘共x兲. 2 ■ DISCOVERY PROJECT AREA FUNCTIONS SOLUTIONS 1. (a) (b) As in part (a), Area of trapezoid = 12 (b1 + b2 )h = 12 (3 + 7)2 A(x) = 12 [3 + (2x + 1)](x − 1) = 10 square units = 12 (2x + 4)(x − 1) Or: = (x + 2)(x − 1) Area of rectangle + area of triangle = x2 + x − 2 square units = br hr + 12 bt ht = (2)(3) + 12 (2)(4) = 10 square units (c) A0 (x) = 2x + 1. This is the y-coordinate of the point (x, 2x + 1) on the given line. 2. (a) (b) A(x) = Ux 0 sin t dt = [− cos t]x0 = − cos x − (−1) = 1 − cos x (c) A0 (x) = sin x. This is the y-coordinate of the point (x, sin x) on the given curve. (d) (e) A(x + h) − A(x) is the area under the curve y = sin t from t = x to t = x + h. An approximating rectangle is shown in the figure. It has height sin x, width h, and area h sin x, so Copyright © 2013, Cengage Learning. All rights reserved. A(x + h) − A(x) ≈ h sin x A(x + h) − A(x) ≈ sin x. h ⇒ (f ) Part (e) says that the average rate of change of A is approximately sin x. As h approaches 0, the quotient approaches the instantaneous rate of change — namely, A0 (x). So the result of part (c), A0 (x) = sin x, is geometrically plausible. DISCOVERY PROJECT AREA FUNCTIONS 3. (a) f(x) = cos x2 (b) g(x) starts to decrease at that value of x where cos t2 changes from positive to negative; that is, at about x = 1.25. Ux (d) We sketch the graph of g 0 using the method of (c) g(x) = 0 cos t2 dt. Using an integration command, we find that g(0) = 0, g(0.2) ≈ 0.200, Example 1 in Section 2.2. The graphs of g 0 (x) g(0.4) ≈ 0.399, g(0.6) ≈ 0.592, g(0.8) ≈ 0.768, and f (x) look alike, so we guess that g(1.0) ≈ 0.905, g(1.2) ≈ 0.974, g(1.4) ≈ 0.950, g 0 (x) = f (x). g(1.6) ≈ 0.826, g(1.8) ≈ 0.635, and g(2.0) ≈ 0.461. 4. In Problems 1 and 2, we showed that if g (x) = Ux a f (t) dt, then g 0 (x) = f (x), for the functions f(t) = 2t + 1 and f (t) = sin t. In Problem 3 we guessed that the same is true for f (t) = cos(t2 ), based on visual evidence. So we conjecture that g 0 (x) = f (x) for any continuous function f . This turns out to be true and is proved in Section 4 .4 Copyright © 2013, Cengage Learning. All rights reserved. (the Fundamental Theorem of Calculus). ■ 3