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(A) Draw the Line and Use Geometry to Find the Area Under This Line, Above

(A) Draw the Line and Use Geometry to Find the Area Under This Line, Above

DISCOVERY PROJECT AREA FUNCTIONS ■ 1

4.3 DISCOVERY PROJECT: AREA FUNCTIONS

This project can be completed 1. (a) Draw the y ෇ 2t ϩ 1 and use to find the area under this line, above the anytime after you have studied t-axis, and between the vertical lines t ෇ 1 and t ෇ 3 . Section 4.3 in the textbook. (b) If x Ͼ 1 , let A͑x͒ be the area of the region that lies under the line y ෇ 2t ϩ 1 between t ෇ 1 and t ෇ x . Sketch this region and use geometry to find an expression for A͑x͒ . (c) Differentiate the area A͑x͒ . What do you notice? ഛ ഛ ␲ ͑ ͒ ෇ xx ͑ ͒ 2. (a) If 0 x , let A x 0 sin t dt. A x represents the area of a region. Sketch that region. (b) Use the Evaluation Theorem to find an expression for A͑x͒ . (c) Find AЈ͑x͒ . What do you notice? (d) If x is any number between 0 and ␲ and h is a small positive number, then A͑x ϩ h͒ Ϫ A͑x͒ represents the area of a region. Describe and sketch the region. (e) Draw a that approximates the region in part (d). By comparing the areas of these two regions, show that A͑x ϩ h͒ Ϫ A͑x͒ Ϸ sin x h (f) Use part (e) to give an intuitive explanation for the result of part (c).

; 3. (a) Draw the graph of the function f ͑x͒ ෇ cos͑x 2 ͒ in the viewing rectangle ͓0, 2͔ by ͓Ϫ1.25, 1.25͔. (b) If we define a new function t by

x t͑x͒ ෇ y cos͑t 2 ͒ dt 0 then t͑x͒ is the area under the graph of f from 0 to x [until f ͑x͒ becomes negative, at which t͑x͒ becomes a difference of areas]. Use part (a) to determine the value of x at which t͑x͒ starts to decrease. [Unlike the in Problem 2, it is impossible to evaluate the integral defining t to obtain an explicit expression for t͑x͒ .] (c) Use the integration command on your calculator or computer to estimate t͑0.2͒ ,t͑0.4͒ , t͑0.6͒,...,t͑1.8͒ ,t͑2͒ . Then use these values to sketch a graph of t . (d) Use your graph of t from part (c) to sketch the graph of tЈ using the interpretation of tЈ͑x͒ as the slope of a tangent line. How does the graph of tЈ compare with the graph off ? 4. Suppose f is a continuous function on the interval ͓a, b͔ and we define a new function t by the equation

x t͑x͒ ෇ y f ͑t͒ dt a Based on your results in Problems 1–3, conjecture an expression for tЈ͑x͒ . 2013, Cengage Learning. All rights reserved. © Copyright 2 ■ DISCOVERY PROJECT AREA FUNCTIONS

SOLUTIONS

1. (a) (b)

As in part (a), 1 1 Area of = (b1 + b2)h = (3 + 7)2 2 2 A(x)=1 [3 + (2x +1)](x 1) =10square units 2 − 1 = 2 (2x +4)(x 1) Or: − =(x +2)(x 1) Area of rectangle + area of − = x2 + x 2 units 1 − = brhr + 2 btht 1 =(2)(3)+2 (2)(4) = 10 square units

(c) A0(x)=2x +1.Thisisthey-coordinate of the point (x, 2x +1)on the given line.

2. (a) (b) A(x)= x sin tdt=[ cos t]x 0 − 0 = U cos x ( 1) = 1 cos x − − − −

(c) A0(x)=sinx.Thisisthey-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 =sint An approximating rectangle is shown in the figure. − from t = x to t = x + h. It has height sin x, width h, and area h sin x,so 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.Ash approaches 0, the quotient

approaches the instantaneous rate of change — namely, A0(x). So the result of part (c), A0(x)=sinx,is geometrically plausible. 2013, Cengage Learning. All rights reserved. © Copyright DISCOVERY PROJECT AREA FUNCTIONS ■ 3

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.   x 2 (c) g(x)= 0 cos t dt. Using an integration command, (d)Wesketchthegraphofg0 using the method of

we find thatU g(0) = 0, g(0.2) 0.200, Example 1 in Section 2.2. The graphs of g0(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, g0(x)=f(x). ≈ ≈ ≈ g(1.6) 0.826, g(1.8) 0.635,andg(2.0) 0.461. ≈ ≈ ≈

x 4. In Problems 1 and 2, we showed that if g (x)= a f(t) dt,theng0(x)=f(x), for the functions f(t)=2t +1and U f(t)=sint. In Problem 3 we guessed that the same is true for f(t)=cos(t2), based on visual evidence. So we

conjecture that g0(x)=f(x) for any continuous function f. This turns out to be true and is proved in Section 4.4

(the Fundamental Theorem of ). 2013, Cengage Learning. All rights reserved. © Copyright