6.1. USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH
341
(e) A moving particle is traveling along the curve given by y = f (x) = 0.1x 2 + 1,
and does so at a constant rate of 7 cm/sec, where both x and y are measured
in cm (that is, the curve y = f (x) is the path along which the object actually
travels; the curve is not a “position function”). Find the position of the particle
when t = 4 sec, assuming that when t = 0, the particle’s location is (0, f (0)).
⊳
Summary
In this section, we encountered the following important ideas:
• To find the area between two curves, we think about slicing the region into thin
rectangles. If, for instance, the area of a typical rectangle on the interval x = a to x = b
is given by A rect = (g(x) − f (x))△x, then the exact area of the region is given by the
definite integral
A =
b
a
(g(x) − f (x)) dx.
• The shape of the region usually dictates whether we should use vertical rectangles of
thickness △x or horizontal rectangles of thickness △y. We desire to have the height of
the rectangle governed by the difference between two curves: if those curves are best
thought of as functions of y, we use horizontal rectangles, whereas if those curves are
best viewed as functions of x, we use vertical rectangles.
• The arc length, L, along the curve y = f (x) from x = a to x = b is given by
L =
b
a
1 + f ′ (x) 2 dx.
Exercises
1. Find the exact area of each described region.
(a) The finite region between the curves x = y(y − 2) and x = −(y − 1)(y − 3).
(b) The region between the sine and cosine functions on the interval [
π
4 ,
3π
4 ].
(c) The finite region between x = y 2 − y − 2 and y = 2x − 1.
(d) The finite region between y = mx and y = x 2 − 1, where m is a positive
constant.
2. Let f (x) = 1 − x 2 and g(x) = ax 2 − a, where a is an unknown positive real number.
For what value(s) of a is the area between the curves f and g equal to 2?
3. Let f (x) = 2 − x 2 . Recall that the average value of any continuous function f on an
interval [a, b] is given by
1
b−a
b
a
f (x) dx.
341
(e) A moving particle is traveling along the curve given by y = f (x) = 0.1x 2 + 1,
and does so at a constant rate of 7 cm/sec, where both x and y are measured
in cm (that is, the curve y = f (x) is the path along which the object actually
travels; the curve is not a “position function”). Find the position of the particle
when t = 4 sec, assuming that when t = 0, the particle’s location is (0, f (0)).
⊳
Summary
In this section, we encountered the following important ideas:
• To find the area between two curves, we think about slicing the region into thin
rectangles. If, for instance, the area of a typical rectangle on the interval x = a to x = b
is given by A rect = (g(x) − f (x))△x, then the exact area of the region is given by the
definite integral
A =
b
a
(g(x) − f (x)) dx.
• The shape of the region usually dictates whether we should use vertical rectangles of
thickness △x or horizontal rectangles of thickness △y. We desire to have the height of
the rectangle governed by the difference between two curves: if those curves are best
thought of as functions of y, we use horizontal rectangles, whereas if those curves are
best viewed as functions of x, we use vertical rectangles.
• The arc length, L, along the curve y = f (x) from x = a to x = b is given by
L =
b
a
1 + f ′ (x) 2 dx.
Exercises
1. Find the exact area of each described region.
(a) The finite region between the curves x = y(y − 2) and x = −(y − 1)(y − 3).
(b) The region between the sine and cosine functions on the interval [
π
4 ,
3π
4 ].
(c) The finite region between x = y 2 − y − 2 and y = 2x − 1.
(d) The finite region between y = mx and y = x 2 − 1, where m is a positive
constant.
2. Let f (x) = 1 − x 2 and g(x) = ax 2 − a, where a is an unknown positive real number.
For what value(s) of a is the area between the curves f and g equal to 2?
3. Let f (x) = 2 − x 2 . Recall that the average value of any continuous function f on an
interval [a, b] is given by
1
b−a
b
a
f (x) dx.