(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 line y ෇ 2t ϩ 1 and use geometry 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 function 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 rectangle 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 point t͑x͒ becomes a difference of areas]. Use part (a) to determine the value of x at which t͑x͒ starts to decrease. [Unlike the integral 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

SOLUTIONS

1. (a) (b)

As in part (a), 1 1 Area of trapezoid = (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 triangle − = x2 + x 2 square units 1 − = brhr + 2 btht 1 =(2)(3)+2 (2)(4) = 10 square units

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

(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