(a) Draw the line -axis, and between the vertical lines

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