AREA OF A SURFACE OF REVOLUTION

AREA OF A SURFACE OF REVOLUTION
cut
A surface of revolution is formed when a curve is rotated about a line. Such a surface is
the lateral boundary of a solid of revolution of the type discussed in Sections 7.2 and 7.3.
We want to define the area of a surface of revolution in such a way that it corresponds
to our intuition. If the surface area is A, we can imagine that painting the surface would
require the same amount of paint as does a flat region with area A.
Let’s start with some simple surfaces. The lateral surface area of a circular cylinder with
radius r and height h is taken to be A 2 rh because we can imagine cutting the cylinder and unrolling it (as in Figure 1) to obtain a rectangle with dimensions 2 r and h.
Likewise, we can take a circular cone with base radius r and slant height l , cut it along
the dashed line in Figure 2, and flatten it to form a sector of a circle with radius l and central angle 2 rl. We know that, in general, the area of a sector of a circle with radius
l and angle is 12 l 2 (see Exercise 67 in Section 6.2) and so in this case it is
h
r
h
A 12 l 2 12 l 2
2πr
2r
l
rl
Therefore, we define the lateral surface area of a cone to be A rl.
FIGURE 1
2πr
cut
l
¨
r
l
FIGURE 2
What about more complicated surfaces of revolution? If we follow the strategy we used
with arc length, we can approximate the original curve by a polygon. When this polygon
is rotated about an axis, it creates a simpler surface whose surface area approximates the
actual surface area. By taking a limit, we can determine the exact surface area.
The approximating surface, then, consists of a number of bands, each formed by rotating a line segment about an axis. To find the surface area, each of these bands can be
considered a portion of a circular cone, as shown in Figure 3. The area of the band (or frustum of a cone) with slant height l and upper and lower radii r1 and r2 is found by subtracting the areas of two cones:
l¡
r¡
1
l
A r2l1 l r1l1 r2 r1l1 r2 l
From similar triangles we have
l1
l1 l
r1
r2
r™
which gives
r2 l1 r1l1 r1l
FIGURE 3
or
r2 r1l1 r1l
Putting this in Equation 1, we get
A r1l r2 l
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or
2
A 2 rl
where r 12 r1 r2 is the average radius of the band.
1
2 ■ AREA OF A SURFACE OF REVOLUTION
y
y=ƒ
0
a
b
x
(a) Surface of revolution
y
P¸
0
Pi-1
Pi
a
Now we apply this formula to our strategy. Consider the surface shown in Figure 4,
which is obtained by rotating the curve y f x, a x b, about the x-axis, where f is
positive and has a continuous derivative. In order to define its surface area, we divide the
interval a, b into n subintervals with endpoints x0, x1, . . . , xn and equal width x, as we
did in determining arc length. If yi f x i , then the point Pi x i, yi lies on the curve. The
part of the surface between x i1 and x i is approximated by taking the line segment Pi1Pi
and rotating it about the x-axis. The result is a band with slant height l Pi1Pi and aver1
age radius r 2 yi1 yi so, by Formula 2, its surface area is
yi
b
x
Pi s1 f xi* 2 x
i1
FIGURE 4
As in the proof of Theorem 7.4.2, we have
P
(b) Approximating band
yi1 yi
Pi1Pi
2
2
Pn
where xi* is some number in x i1, x i . When x is small, we have yi f x i f xi* and
also yi1 f x i1 f xi*, since f is continuous. Therefore
2
yi1 yi
Pi1Pi 2 f xi* s1 f xi* 2 x
2
and so an approximation to what we think of as the area of the complete surface of revolution is
n
2 f x* s1 f x*
3
i
i
2
x
i1
This approximation appears to become better as n l and, recognizing (3) as a Riemann
sum for the function tx 2 f x s1 f x 2, we have
n
lim
2 f x* s1 f x*
n l i1
i
i
2
b
x y 2 f x s1 f x 2 dx
a
Therefore, in the case where f is positive and has a continuous derivative, we define the
surface area of the surface obtained by rotating the curve y f x, a x b, about
the x-axis as
4
b
S y 2 f x s1 f x 2 dx
a
With the Leibniz notation for derivatives, this formula becomes
5
b
S y 2 y
a
1
dy
dx
2
dx
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If the curve is described as x ty, c y d, then the formula for surface area
becomes
6
S
y
d
c
2 y
1
dx
dy
2
dy
and both Formulas 5 and 6 can be summarized symbolically, using the notation for arc
AREA OF A SURFACE OF REVOLUTION ■ 3
length given in Section 7.4, as
S y 2 y ds
7
For rotation about the y-axis, the surface area formula becomes
S y 2 x ds
8
where, as before, we can use either
ds 1
dy
dx
2
dx
ds or
1
2
dx
dy
dy
These formulas can be remembered by thinking of 2 y or 2 x as the circumference of a
circle traced out by the point x, y on the curve as it is rotated about the x-axis or y-axis,
respectively (see Figure 5).
y
y
(x, y)
y
x
circumference=2πx
circumference=2πy
0
FIGURE 5
(x, y)
x
0
(a) Rotation about x-axis: S=j 2πy ds
x
(b) Rotation about y-axis: S=j 2πx ds
EXAMPLE 1 The curve y s4 x 2, 1 x 1, is an arc of the circle x 2 y 2 4.
Find the area of the surface obtained by rotating this arc about the x-axis. (The surface is
a portion of a sphere of radius 2. See Figure 6.)
SOLUTION We have
y
dy
x
12 4 x 2 122x dx
s4 x 2
and so, by Formula 5, the surface area is
1
x
S
y
1
1
2 y
1
1
2 y s4 x 2
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1
1
2 y s4 x 2
FIGURE 6
Figure 6 shows the portion of the sphere
whose surface area is computed in Example 1.
■ ■
1
1
dy
dx
2
1
dx
x2
dx
4 x2
2
dx
s4 x 2
4 y 1 dx 4 2 8
1
4 ■ AREA OF A SURFACE OF REVOLUTION
■ ■ Figure 7 shows the surface of revolution
whose area is computed in Example 2.
y
EXAMPLE 2 The arc of the parabola y x 2 from 1, 1 to 2, 4 is rotated about the
y-axis. Find the area of the resulting surface.
SOLUTION 1 Using
(2, 4)
y x2
dy
2x
dx
and
y=≈
we have, from Formula 8,
0
1
2
y 2 x ds
S
x
FIGURE 7
y
2
1
2 x
dy
dx
1
2
dx
2
2 y x s1 4x 2 dx
1
Substituting u 1 4x 2, we have du 8x dx. Remembering to change the limits of
integration, we have
S
As a check on our answer to Example 2,
notice from Figure 7 that the surface area
should be close to that of a circular cylinder with
the same height and radius halfway between
the upper and lower radius of the surface:
2 1.53 28.27. We computed that
the surface area was
■ ■
(17 s17 5 s5 ) 30.85
6
which seems reasonable. Alternatively, the surface area should be slightly larger than the area
of a frustum of a cone with the same top and bottom edges. From Equation 2, this is
2 1.5(s10 ) 29.80.
4
y
17
5
4
su du [ 23 u 32 ]175
(17s17 5s5 )
6
SOLUTION 2 Using
x sy
dx
1
dy
2sy
and
we have
4
S y 2 x ds y 2 x
1
4
2 y sy
1
1
4
(17s17 5s5 )
6
17
5
1
dx
dy
2
dy
1
4
dy y s4y 1 dy
1
4y
y
(where u 1 4y )
su du
(as in Solution 1)
EXAMPLE 3 Find the area of the surface generated by rotating the curve y e x,
0 x 1, about the x-axis.
Another method: Use Formula 6 with
x ln y.
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■ ■
SOLUTION Using Formula 5 with
y ex
and
dy
ex
dx
AREA OF A SURFACE OF REVOLUTION ■ 5
we have
1
S y 2 y
0
2
dy
dx
1
e
2 y s1 u 2 du
4
■ ■
0
(where u e x )
1
2 y
1
dx 2 y e x s1 e 2x dx
sec 3 d
(where u tan and tan1e )
[
2 12 sec tan ln sec tan Or use Formula 21 in the Table of Integrals.
4
]
(by Example 8 in Section 6.2)
[
]
sec tan lnsec tan s2 ln(s2 1)
Since tan e, we have sec 2 1 tan 2 1 e 2 and
S [es1 e 2 ln(e s1 e 2 ) s2 ln(s2 1)]
EXERCISES
A Click here for answers.
S
17–20
Use Simpson’s Rule with n 10 to approximate the area
of the surface obtained by rotating the curve about the x-axis. Compare your answer with the value of the integral produced by your
calculator.
Click here for solutions.
1–4
Set up, but do not evaluate, an integral for the area of the
surface obtained by rotating the curve about the given axis.
1. y ln x, 1 x 3;
2. y sin 2x, 0 x 2;
x-axis
3. y sec x, 0 x 4;
y-axis
4. y e x, 1 y 2;
■
■
■
■
■
17. y ln x,
x-axis
18. y x sx,
19. y sec x,
■
■
■
■
■
■
■
CAS
Find the area of the surface obtained by rotating the curve
about the x-axis.
0x2
6. 9x y 2 18,
7. y sx,
0x1
10. y x3
1
,
6
2x
11. x 3 y 2 232,
12. x 1 2y 2,
■
■
■
CAS
■
■
■
■
■
■
■
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0x1
■
■
■
a y a
■
■
■
■
■
■
■
■
■
■
■
■
0x3
■
■
■
■
■
■
■
■
■
■
0y1
■
■
■
0x1
■
■
■
■
■
■
■
■
26. A group of engineers is building a parabolic satellite dish
0 y a2
16. x a cosh ya,
■
■
the loop of the curve 3ay 2 xa x2 about the x-axis.
(b) Find the surface area if the loop is rotated about the
y-axis.
1y2
15. x sa 2 y 2,
■
25. (a) If a 0, find the area of the surface generated by rotating
The given curve is rotated about the y-axis. Find the area
of the resulting surface.
14. y 1 x 2,
■
■
13–16
3
x,
13. y s
■
23–24
Use a CAS to find the exact area of the surface obtained
by rotating the curve about the y-axis. If your CAS has trouble
evaluating the integral, express the surface area as an integral in the
other variable.
■
■
■
24. y lnx 1,
1y2
■
■
23. y x 3,
1y2
1
■
1x2
22. y sx 1,
x1
1
2
■
Use either a CAS or a table of integrals to find the exact
area of the surface obtained by rotating the given curve about the
x-axis.
■
9. y cosh x,
0x1
21–22
2
4x9
0 x 6
■
21. y 1x,
2x6
8. y cos 2x,
■
■
5–12
5. y x 3,
■
1x2
0 x 3
20. y s1 e x,
y-axis
■
1x3
whose shape will be formed by rotating the curve y ax 2
about the y-axis. If the dish is to have a 10-ft diameter and a
maximum depth of 2 ft, find the value of a and the surface area
of the dish.
■
6 ■ AREA OF A SURFACE OF REVOLUTION
27. The ellipse
32. Show that the surface area of a zone of a sphere that lies
2
2
x
y
2 1
a2
b
a
b
is rotated about the x-axis to form a surface called an ellipsoid.
Find the surface area of this ellipsoid.
28. Find the surface area of the torus in Exercise 41 in Section 7.2.
29. If the curve y f x, a x b, is rotated about the horizon-
tal line y c, where f x c, find a formula for the area of
the resulting surface.
CAS
30. Use the result of Exercise 29 to set up an integral to find the
area of the surface generated by rotating the curve y sx,
0 x 4, about the line y 4. Then use a CAS to evaluate
the integral.
31. Find the area of the surface obtained by rotating the circle
Thomson Brooks-Cole copyright 2007
x 2 y 2 r 2 about the line y r.
between two parallel planes is S dh, where d is the diameter of the sphere and h is the distance between the planes.
(Notice that S depends only on the distance between the planes
and not on their location, provided that both planes intersect
the sphere.)
33. Formula 4 is valid only when f x 0. Show that when f x
is not necessarily positive, the formula for surface area becomes
b
S y 2 f x s1 f x 2 dx
a
34. Let L be the length of the curve y f x, a x b, where
f is positive and has a continuous derivative. Let S f be the surface area generated by rotating the curve about the x-axis. If c
is a positive constant, define tx f x c and let St be the
corresponding surface area generated by the curve y tx,
a x b. Express St in terms of S f and L.
AREA OF A SURFACE OF REVOLUTION ■ 7
ANSWERS
Click here for solutions.
S
13. (145 s145 10 s10 )27
17. 9.023754
3
1.
y
3.
y
1
2 ln x s1 1x dx
4
0
2 x s1 sec x tan x2 dx
9. [1 4 e 2 e2 ]
Thomson Brooks-Cole copyright 2007
1
19. 13.527296
21. 4[4 ln(s17 4) 4 ln(s2 1) s17 4 s2 ]
2
5. (145 s145 1)27
15. a 2
7. (37 s37 17 s17 )6
11. 212
23. 6[ln(s10 3) 3 s10 ]
25. (a) a 23
(b) 56 s3 a 245
27. 2 [b 2 a 2b sin1(sa 2 b 2a)sa 2 b 2 ]
29.
xab 2 c f x s1 f x 2 dx
31. 4 2r 2
8 ■ AREA OF A SURFACE OF REVOLUTION
SOLUTIONS
1. y = ln x ⇒ ds =
1 + (dy/dx)2 dx =
3. y = sec x ⇒ ds =
S=
π/4
0
5. y = x3
7. y =
1 + (1/x)2 dx ⇒ S =
1 + (dy/dx)2 dx =
3
1
1 + (1/x)2 dx [by (7)]
2π(ln x)
1 + (sec x tan x)2 dx ⇒
1 + (sec x tan x)2 dx [by (8)]
2πx
⇒ y 0 = 3x2 . So
S=
2
0
=
2π
36
2
0
1 + (y 0 )2 dx = 2π
2πy
√
u du =
145
1
145
2 3/2
u
3
π
18
√
x3 1 + 9x4 dx
=
1
π
27
145
[u = 1 + 9x4 , du = 36x3 dx]
√
145 − 1
√
√ 2
x ⇒ 1 + (dy/dx)2 = 1 + [1/(2 x )] = 1 + 1/(4x). So
9
S=
2πy
1+
4
2
3
= 2π
dy
dx
9
1 3/2
4
4
x+
2
9
dx =
2π
√
x
1+
4
4π
3
=
1
(4x
8
+ 1)3/2
9
π
6
=
4
9
1
dx = 2π
4x
37
x+
4
1
4
dx
√
√
37 − 17 17
9. y = cosh x ⇒ 1 + (dy/dx)2 = 1 + sinh2 x = cosh2 x. So
1
0
S = 2π
=π 1+
11. x =
y2 + 2
1
3
3/2
1 1
(1
0 2
cosh x cosh x dx = 2π
1
2
or π 1 +
sinh 2
⇒ dx/dy =
1
2
y2 + 2
1/2
1
4
1
2
+ cosh 2x) dx = π x +
sinh 2x
1
0
e2 − e−2
y2 + 2 ⇒
(2y) = y
2
1 + (dx/dy)2 = 1 + y 2 y 2 + 2 = y 2 + 1 . So
S = 2π
13. y =
√
3
x ⇒ x = y3
S = 2π
=
15. x =
a2 − y 2
π
18
Thomson Brooks-Cole copyright 2007
0
1
4
1 4
4y
+ 12 y 2
2
1
= 2π 4 + 2 −
2
1
x
2
3
1 + 9y 4
a2
1
4
−
1
2
=
21π
2
⇒ 1 + (dx/dy)2 = 1 + 9y 4 . So
1 + (dx/dy)2 dy = 2π
3/2 2
1
=
π
27
145
2
1
y3
1 + 9y 4 dy =
a2 − y 2
a
dy = 2π
− y2
a2
a/2
a dy = 2πa y
0
2π
36
2
1
1 + 9y 4 36y 3 dy
√
√
145 − 10 10
y2
a2 − y 2
y2
a2
= 2
+ 2
= 2
2
2
2
−y
a −y
a −y
a − y2
a/2
2π
y y 2 + 1 dy = 2π
⇒ dx/dy = 12 (a2 − y 2 )−1/2 (−2y) = −y/
1 + (dx/dy)2 = 1 +
S=
2
1
a2 − y 2
⇒
⇒
a/2
0
= 2πa
a
− 0 = πa2 . Note that this is
2
the surface area of a sphere of radius a, and the length of the interval y = 0 to y = a/2 is
interval y = −a to y = a.
1
4
the length of the
AREA OF A SURFACE OF REVOLUTION ■ 9
17. y = ln x ⇒ dy/dx = 1/x ⇒ 1 + (dy/dx)2 = 1 + 1/x2
Let f(x) = ln x
S ≈ S10 = 2π ·
1 + 1/x2 . Since n = 10, ∆x =
1/5
3
3−1
10
=
1
.
5
3
1
⇒ S=
2π ln x
1 + 1/x2 dx.
Then
[f (1) + 4f (1.2) + 2f (1.4) + · · · + 2f (2.6) + 4f (2.8) + f(3)] ≈ 9.023754.
The value of the integral produced by a calculator is 9.024262 (to six decimal places).
19. y = sec x ⇒ dy/dx = sec x tan x ⇒ 1 + (dy/dx)2 = 1 + sec2 x tan2 x ⇒
S=
π/3
0
2π sec x
√
√
1 + sec2 x tan2 x dx. Let f (x) = sec x 1 + sec2 x tan2 x.
π
π/3 − 0
=
. Then
10
30
Since n = 10, ∆x =
S ≈ S10 = 2π ·
π/30
π
2π
f (0) + 4f
+ 2f
3
30
30
8π
30
+ · · · + 2f
9π
30
+ 4f
+f
π
3
≈ 13.527296.
The value of the integral produced by a calculator is 13.516987 (to six decimal places).
21. y = 1/x ⇒ ds =
2
2π ·
S=
1
4
=π
1
=π −
1
x
1 + (dy/dx)2 dx =
1+
2
1
dx = 2π
x4
1
1 + (−1/x2 )2 dx =
√
x4 + 1
dx = 2π
x3
√
√
1 + u2
1 + u2
24
+ ln u +
du
=
π
−
2
u
u
√
17
4
+ ln 4 +
√
17
+
√
2
1
− ln 1 +
4
1
1 + 1/x4 dx ⇒
√
u2 + 1
u2
1
2
[u = x2 , du = 2x dx]
du
4
1 + u2
1
√
2
√
2−
=π
√
17
4
+ ln
√
4 + √17
1+ 2
23. y = x3 and 0 ≤ y ≤ 1 ⇒ y 0 = 3x2 and 0 ≤ x ≤ 1.
S=
=
=
1
0
2πx
1 + (3x2 )2 dx = 2π
√
21
1 + u2 du =
π
3
3
0
π
3
√
3
10 +
2
3
0
√
1 + u2 16 du
[or use CAS]
√
1
ln 3 + 10
2
=
π
6
π
3
1
2u
[u = 3x2 , du = 6x dx]
√
1 + u2 +
1
2
ln u +
√
1 + u2
3
0
√
√
3 10 + ln 3 + 10
25. Since a > 0, the curve 3ay 2 = x(a − x)2 only has points with
x ≥ 0. (3ay 2 ≥ 0 ⇒ x(a − x)2 ≥ 0 ⇒ x ≥ 0.) The
curve is symmetric about the x-axis (since the equation is
unchanged when y is replaced by −y). y = 0 when x = 0 or a,
so the curve’s loop extends from x = 0 to x = a.
Thomson Brooks-Cole copyright 2007
d
d
dy
(3ay 2 ) =
[x(a − x)2 ] ⇒ 6ay
= x · 2(a − x)(−1) + (a − x)2
dx
dx
dx
⇒
dy
dx
2
=
(a − x)2 (a − 3x)2
(a − x)2 (a − 3x)2
3a
=
·
2
2
36a y
36a2
x(a − x)2
⇒
(a − x)[−2x + a − x]
dy
=
dx
6ay
the last fraction
is 1/y 2
=
(a − 3x)2
12ax
⇒
10 ■ AREA OF A SURFACE OF REVOLUTION
1+
2
dy
dx
=1+
(a + 3x)2
12ax
a2 − 6ax + 9x2
a2 + 6ax + 9x2
a2 − 6ax + 9x2
=
+
=
=
for x 6= 0.
12ax
12ax
12ax
12ax
12ax
a
a
(a) S =
2πy ds = 2π
x=0
0
a
π
=
3a
0
√
x(a − x) a + 3x
√
·√
dx = 2π
3a
12ax
π 2
(a + 2ax − 3x ) dx =
a x + ax2 − x3
3a
2
a
0
a
2
(a − x)(a + 3x)
dx
6a
0
π 3
π
πa2
(a + a3 − a3 ) =
· a3 =
.
3a
3a
3
=
Note that we have rotated the top half of the loop about the x-axis. This generates the full surface.
(b) We must rotate the full loop about the y-axis, so we get double the area obtained by rotating the top half of
the loop:
a
S = 2 · 2π
x=0
a
2π
= √
3a
=
27.
a
x ds = 4π
0
0
√
6 2
2π 3
2
+
a =
3
5
3
y2
x2
+
=1 ⇒
a2
b2
x1/2 (a + 3x) dx
0
=1+
=
a
=
0
√
2π 3
√
3 a
2 5/2 6 5/2
a
+ a
3
5
√
56π 3 a2
28 2
a =
15
45
y (dy/dx)
x
=− 2
b2
a
2
dy
dx
a
2π 2 3/2 6 5/2
ax
(ax1/2 + 3x3/2 ) dx = √
+ x
5
3a 3
√
2π 3
3
1+
a + 3x
4π
x√
dx = √
12ax
2 3a
b2 x
dy
=− 2
dx
a y
⇒
⇒
b4 x2 + a4 b2 1 − x2/a2
a4 b2 + b4 x2 − a2 b2 x2
b4 x2
b4 x2 + a4 y 2
=
=
=
a4 y 2
a4 y 2
a4 b2 (1 − x2/a2 )
a4 b2 − a2 b2 x2
a4 − a2 − b2 x2
a4 + b2 x2 − a2 x2
=
a4 − a2 x2
a2 (a2 − x2 )
The ellipsoid’s surface area is twice the area generated by rotating the first quadrant portion of the ellipse about the
x-axis. Thus,
a
2πy
S =2
1+
0
4πb
= 2
a
Thomson Brooks-Cole copyright 2007
2
a
dx = 4π
0
a
a4
0
4πb
30
= √
2
a a2 − b2
=
dy
dx
4πb
√
a2 a2 − b2
−
u
2
(a2
−
b2 )x2
b
a
4πb
dx = 2
a
√
a2 −b2
a
du
a4 − u2 √
a2 − b2
0
u
a4
sin−1 2
a4 − u2 +
2
a
√
a a2 − b2
2
a4 − (a2 − b2 )x2
√
dx
a a2 − x2
a2 − x2
[u =
√
a
a2 −b2
0
4
a4 − a2 (a2 − b2 ) +
a
sin−1
2
√
a2 − b2
a

√
a2 − b2
2
−1
a
b
sin
 2
a
√
= 2π
b +
a2 − b2
29. The analogue of f (x∗i ) in the derivation of (4) is now c − f(x∗i ), so
n
S = lim
n→∞ i=1
2π[c − f (x∗i )]
a2 − b2 x]
b
1 + [f 0 (x∗i )]2 ∆x =
a
2π[c − f (x)]
1 + [f 0 (x)]2 dx.




AREA OF A SURFACE OF REVOLUTION ■ 11
31. For the upper semicircle, f(x) =
√
√
r2 − x2 , f 0 (x) = −x/ r2 − x2 . The surface area generated is
r
S1 =
−r
2π r −
r
= 4π
0
r2 − x2
1+
r2
x2
dx = 4π
− x2
r
0
r−
r
√
dx
r2 − x2
r2 − x2
r2
√
− r dx
r 2 − x2
√
x
, so S2 = 4π
For the lower semicircle, f (x) = − r2 − x2 and f 0 (x) = √
r2 − x2
r
Thus, the total area is S = S1 + S2 = 8π
0
r2
√
2
r − x2
dx = 8π r2 sin−1
r
0
x
r
33. In the derivation of (4), we computed a typical contribution to the surface area to be 2π
r
0
r2
√
+ r dx.
2
r − x2
= 8πr2
π
2
= 4π 2 r2 .
yi−1 + yi
|Pi−1 Pi |, the area
2
of a frustum of a cone. When f (x) is not necessarily positive, the approximations yi = f(xi ) ≈ f (x∗i ) and
yi−1 = f (xi−1 ) ≈ f (x∗i ) must be replaced by yi = |f (xi )| ≈ |f (x∗i )| and yi−1 = |f (xi−1 )| ≈ |f (x∗i )|. Thus,
2π
yi−1 + yi
|Pi−1 Pi | ≈ 2π |f (x∗i )|
2
Thomson Brooks-Cole copyright 2007
obtain S =
b
a
2π |f(x)|
1 + [f 0 (x∗i )]2 ∆x. Continuing with the rest of the derivation as before, we
1 + [f 0 (x)]2 dx.