Math 1142
Solutions for Spring 2005 Final Exam
1) The “limit at infinity” for the rational function of polynomials,
lim
x→∞
x + 1
, can
2 x2 + 4
be calculated in a couple of different ways. Early on in the course, we learned a method in
which we divide the numerator and the denominator by the highest power of x that
1
= 0
appears in the denominator function; we then apply the special
Limit Law, lim
€
x→ ∞ xp
for p > 0 :
x + 1
2
x + 1
x + 1
1 x2
x
x2
lim
=
lim
⋅
=
lim
x → ∞ 2 x2 + 4
x → ∞ 2 x2 + 4
x →€∞ 2 x 2
1 x2
+ 42
2
x
x
1 + 1
x
x 2 = lim 0 + 0 = 0 .
= lim
x→∞ 2 + 4
x→∞ 2 + 0
2
x
(A)
€
We can also use L’Hôpital’s Rule to resolve this limit, since simple application of
€
the Limit Laws alone produces the indeterminate ratio
lim
x→∞
∞
∞
:
x + 1
( x + 1) '
1
= lim
= lim
= 0 .
x → ∞ (2 x2 + 4 )'
x→ ∞ 4x
2 x2 + 4
€
2) In€
order to evaluate the first derivative of the rational function
f (x) =
2x + 3
,
x2 + 1
we will need to apply the Quotient Rule to determine the derivative function first:
f '(x) =
=
€
( x 2 + 1) ⋅ ( 2 x + 3) ' − ( 2 x + 3) ⋅ ( x 2 + 1) '
( x 2 + 1) ⋅ ( 2) − ( 2 x + 3) ⋅ ( 2 x )
=
€
( x 2 + 1) 2
( x 2 + 1) 2
( 2 x 2 + 2) − ( 4 x 2 + 6 x )
2 − 6 x − 2 x2
=
( x 2 + 1) 2
( x 2 + 1) 2
.
Since we just want to find the value of the derivative function, it isn’t necessary to tidy
up the algebra further. We find
€
f ' (1) =
2 − 6 x − 2 x2
( x 2 + 1) 2
=
x=1
2 − 6 ⋅ 1 − 2 ⋅ 12
2 − 6 − 2
−6
3
=
=
= −
.
2
2
4
4
2
(1 + 1)
(D)
€
2
( 2 x + 5)
3) What we are calculating here is the derivative of the function y =
which we are unable to determine directly. The problem statement is setting up a
3
,
dy
dy du
=
⋅
dx
du dx , with this composite function
€
→ y = u = u1/ 2 , u = ( 2 x 2 + 5) 3 . Up to this
differentiation by the Chain Rule, using
analyzed as y =
( 2 x 2 + 5) 3
14243
u
€
dy
d
d
=
( u1/ 2 ) ⋅
[ ( 2 x 2 + 5) 3 ] . However,
dx
du
dx
point, the derivative can be written as
2
3
and differentiate it term-by-term, we
x 2 + 5 ) 3 = v 3 . The
should apply the Chain Rule once more, declaring ( 2
1
424
3
€
v
differentiation becomes €
unless
€ we intend to multiply out ( 2 x + 5)
dy
dy du
dv
d
d
d
=
⋅
⋅
=
( u1/ 2 ) ⋅
(v3 ) ⋅
( 2 x 2 + 5)
€ dv
dx
du
dv
dx
du
dx
€
3 ⋅ 4 ⋅ v2 ⋅ x
1
= ( u −1/ 2 ) ⋅ ( 3v 2 ) ⋅ ( 4 x ) =
2
2 ⋅ u1/ 2
€
( 2 x 2 + 5) 2 ⋅ x
( 2 x 2 + 5) 2
=
6
x
⋅
[ ( 2 x 2 + 5) 3 ] 1/ 2
( 2 x 2 + 5) 3 / 2
= 6 ⋅
= 6 x ( 2 x 2 + 5)1/ 2 or 6 x
2 x 2 + 5 . (C) €
4) The slope of the tangent line at x = a to the curve described by f ( x ) is given by
the value
€ of f ’ ( a ) ; so the work for this problem is actually similar to that for
Problem 2 . However, the function is expressed here as a product of factors, so we will
need to differentiate it by applying the Product Rule:
1
1
f ' ( x ) = 2 ⋅ [ ( x + 1) ' ⋅ ( x 2 − 5 x + 3) + ( x + 1) ⋅ ( x 2 − 5 x + 3) ' ]
= 2 ⋅ [ (−
€
1
1
+ 0 ) ⋅ ( x 2 − 5 x + 3) + ( x + 1) ⋅ ( 2 x − 5) ] x2
( x − 1 ) ’ = −x
−2
The slope of the tangent line to this curve at x = 1 is then
€
f ' (1) = 2 ⋅ [ (−
€
€
1
1
) ⋅ (12 − 5 ⋅ 1 + 3) + ( + 1) ⋅ ( 2 ⋅ 1 − 5) ] 1
12
= 2 ⋅ [ (− 1) ⋅ (− 1) + ( 2 ) ⋅ (− 3) ] = 2 (1 − 6 ) = − 10 .
(E)
5) When we are working with a piecewise-defined function, we must take care to look at
the correct “branch” of the definition when determining a limit. Since we want to find
x2 − 9
lim− f ( x ) , we must use f ( x ) =
x − 3 , for x < 3 .
x→3
However, we cannot simply apply the Limit Laws here: doing so presents us with
0
the indeterminate ratio
. We can calculate the limit we are interested in either by
€
0
removing a “cancelling factor”:
€
x2 − 9
lim f€( x ) = lim x − 3 = lim
x → 3−
x → 3−
x → 3−
( x + 3) ( x − 3)
=
x − 3
lim x + 3 = 6
x → 3−
or by using L’Hôpital’s Rule:
x2 − 9
€
( x 2 − 9) '
2x
lim x − 3 = lim ( x − 3) ' = lim 1 = lim 2 x = 6 .
x → 3−
x → 3−
x → 3−
x → 3−
(A)
2
€
6) The definite integral
∫
x ( x 2 − 1) 4 dx could be calculated by multiplying out the
1
factor ( x − 1) , multiplying that in turn by x , and then integrating the resulting
2
4
2
ninth-degree polynomial term-by-term. However, because the expression x − 1 has
the derivative 2
€x , which is a simple multiple of the factor x , we can use a
2
€ “u-substitution” here, with u = x − 1 , du = 2 x dx ⇒ x dx = ½ du , and the
limits of integration changed from x = 1 to 2 to u = 12 − 1 = 0 to 22 − 1 = 3 :
€
2
∫
3
x ( x 2 − 1) 4 dx →
1
∫
0
1
1
u 4 ( du ) =
2
2
3
∫
u 4 du =
0
3
1 ⎛ u 5 ⎞
⎜ ⎟
2 ⎝ 5 ⎠
0
⎛ 5 ⎞ ⎛ 5 ⎞
243
= ⎜ 3 ⎟ − ⎜ 0 ⎟ =
.
10
⎝10 ⎠ ⎝ 10 ⎠
€
(E)
7) The average value of a function f ( x ) over an interval [ a , b ] is given by
∫ ab f ( x ) dx €
f ( x ) [a, b ] =
. So the average of f ( x ) = x ( x – 2 ) on [ 0 , 2 ] is
b−a
2
∫
x ( x − 2)
€
[0, 2]
=
2
∫
x ( x − 2) dx
0
2−0
= (
=
x 2 − 2 x dx
0
2
(
=
x3
− x2 )
3
2
2
0
= (
x3
x2
−
)
6
2
8 − 12
23
22
03
02
8
4
4
2
−
) − (
−
) =
−
=
= − or −
.
6
2
6
2
6
2
6
6
3
(C)
€
This function is negative in the interval ( 0 , 2 ) , so it is entirely reasonable that
€ on the interval is found to be negative.
the average value
2
0
8) Owing to the nature of the choices offered in this problem, it will be useful to
x 3 − 3x
calculate the first derivative of the function f ( x ) = e
that derivative function. Using the Chain Rule, we obtain
and find the properties of
u48
6
47
3
3
du
d
d u
d
f '(x) =
(e x − 3x ) =
(e ) ⋅€
= eu ⋅
( x 3 − 3x ) = e x − 3x ⋅ ( 3 x 2 − 3 ) .
dx
du
dx
dx
€
Since an exponential function can only be positive, the sign of f ’ ( x ) is determined
entirely by the sign of the factor ( 3 x 2 − 3 ) = 3 ( x 2 − 1 ) . We have x 2 − 1 > 0
⇒ x 2 > 1 ⇒ | x | > √ 1 = 1 , so f ’ ( x ) > 0 for x < −1 and x > +1 ; by a similar
argument, f ’ ( x ) < 0 for −1 < x < +1 . This tells us that f ( x ) is increasing on the
intervals ( − ∞ , −1 ) and ( 1 , ∞ ) and decreasing on the interval ( −1 , 1 ) .
(C)
9) Since we need to describe the concavity properties of the curve described by
f ( x ) = x 4 − 4 x 3 + 10 , we should compute the first two derivatives of this function:
f’(x) = 4x
3
− 12 x
2
= 4 x 2 · ( x – 3 ) , f ’’ ( x ) = 12 x
2
− 24 x = 12 x · ( x – 2 ) .
We can see that f ’’ ( x ) = 0 for x = 0 and x = 2 ; by checking the signs of the
factors x and x – 2 , and thus of the product x ( x – 2 ) , we find that
sign of f ’’ ( x )
x < 0
0 < x < 2
x > 2
−·− = +
+·− = −
+·+ = +
The second derivative is positive over ( − ∞ , 0 ) and ( 2 , ∞ ) , so the curve for f ( x ) is
concave upward there; f ’’ ( x ) < 0 on ( 0 , 2 ) , so the curve is concave downward on
that interval.
(D)
10) While partial differentiation does involve investigating the change in the value of a
function of more than one variable due to making a change in only one of the variables,
the basic rules of differentiation still apply. So we can apply the Chain Rule in the usual
way in calculating the requested partial derivative:
⎡
⎤
∂u
∂
∂ ⎢
d
1
∂
2
2
f (x, y) =
ln ( x + y + 1 )⎥ =
( ln u ) ⋅
= u ⋅
( x 2 + y 2 + 1)
∂x
∂ x ⎢ 14243 ⎥
∂x
∂x
du
⎣
⎦
u
2x
1
= u ⋅ 2x = 2
.
x + y2 + 1
€
At the point ( 1, 0 ) , the value of this partial derivative is then
€
€
∂
f (x, y)
∂x
or
(1, 0)
f x (1, 0) =
12
2 ⋅1
2
=
= 1 .
2
2
+ 0 + 1
(B)
11) For functions of a single variable, we find relative (or local) extrema by determining
what values of x cause the first derivative of the function to become zero. We do
something similar for functions of more than one variable, except that we must find the
critical points where the first partial derivatives with respect to each of the variables are
zero.
For the given function f ( x, y ) = x
2
+ 2y
2
− x y + 6 , this leads to
fx =
∂
( x 2 + 2 y 2 − x y + 6) = 2 x − y = 0 ,
∂x
fy =
∂
( x 2 + 2 y 2 − x y + 6 ) = 4 y − x = 0 . ∂y
€
So the equations 2 x = y and 4 y = x must be satisfied simultaneously. We find by
substitution the first equation into the second one that 4 ( 2 x ) = x ⇒ 8 x = x ,
€ which has only the one solution x = 0 . Thus y also equals zero, meaning that our
function has just one critical point at ( 0, 0 ) .
In order to classify the critical points of a function of two variables, we compute
an index, D = fxx fyy − ( fxy )2 , which tells us whether the point represents a local
maximum for both variables, a local minimum for both, or a local maximum for one
variable but a local minimum for the other (a “saddle point”). To carry out the necessary
computation, we need the second partial derivatives of the function:
f xx =
∂2
∂
( x 2 + 2 y 2 − x y + 6) =
(2x − y ) = 2 ,
2
∂
x
∂x
f xy =
∂2
∂
( x2 + 2 y2 − x y + 6) =
(2 x − y ) = −1 ,
∂y ∂x
∂y
f yy =
∂2
∂
( x2 + 2 y2 − x y + 6) =
( 4 y − x ) = 4 . ∂y
∂ y2
€
€
In this case, the second partial derivatives have no dependence on either of the variables,
so the index simply gives us the constant D = 2 · 4 − ( −1 )2 = 7 > 0 . A positive
€ value for D tells us that the critical point is a local extremum for both variables, but not
which kind. To decide the matter, we also look at the sign of the second partial
derivative fxx : since it has a positive value, the critical point ( 0, 0 ) is a local minimum
(just as a value of x would locate a relative minimum if a function of one variable had
df
d2 f
= 0 and
> 0 there).
dx
d x2
€
(E)
12) The equation for some curves may be expressed in a form in which the variables x
and y cannot be separated algebraically to produce an equation y = f ( x ) .
Nevertheless, the equation still describes some function(s) implicitly, so the derivative
dy
can still be calculated. We apply the Chain Rule in order to differentiate the given
dx
curve equation implicitly with respect to x :
€
d
d
( x2 − x y + 2 y2 ) =
( 2) ⇒
dx
dx
d
d
d
( x2 ) −
(xy) +
(2 y2 ) = 0
dx
dx
dx
dy ⎤
⎡ d
⎤ ⎡ d
d
⇒ 2 x − ⎢
(x) ⋅ y + x ⋅
( y )⎥ + ⎢
(2 y2 ) ⋅
= 0 dx
dx ⎥⎦
⎣ dx
⎦ ⎣ dy
€
⇒ 2 x − (1 ⋅ y + x ⋅
€ ⇒ (4 y − x ) ⋅
€
dy
dy
dy
) + 4y⋅
= 0 ⇒ ( 2 x − y ) + (− x + 4 y ) ⋅
= 0 dx
dx
dx
dy
= y − 2x ⇒
dx
dy
y − 2x
=
dx
4y − x
. (A)
13) To calculate the area between two curves, we need to know where and how the curves
€ meet. The intersection points of the curves are found by setting the two functions which
describe the curves equal to one another and solving for the values of x :
x
2
− 2x = −x
2
⇒
2x
2
− 2x = 0
⇒
2x (x–1) = 0
⇒ either x = 0 or ( x – 1 ) = 0 ⇒ x = 1 .
So the area enclosed by the these two parabolas extends from x = 0 to x = 1 : these
will be the limits of integration for the area.
We also need to find which of these two lies “above” the other, since we integrate
the area from the “lower curve” to the “upper curve”. Since we have located the
intersection points, we know that whichever curve is the upper one remains so over the
interval [ 0 , 1 ] . A reasonably quick way to decide the question is to find the values of
the two functions at some value of x in this interval, preferably one where it is easy to
make the calculations:
for x = ½ :
y = x
2
y = −x
− 2 x --
y = ( ½ )2 − 2 ( ½ ) = ¼ − 1 = − ¾ ,
2
y = − ( ½ )2 = − ¼
--
So the parabola y = − x 2 lies above the parabola y = x
which we shall integrate.
.
2
− 2 x in the interval over
Our area integral is then
1
∫
1
2
(− x ) − ( x − 2 x ) dx =
0
∫
0
= (−
€
2
2x 3
(− 2x + 2 x ) dx = (−
+ x2 )
3
1
2
0
2 ⋅ 13
2 ⋅ 03
2
1
+ 12 ) − (−
+ 0 2 ) = (− + 1) − ( 0 + 0) =
.
3
3
3
3
While areas above or below the x-axis for a function f ( x ) may be positive or negative
(a “signed area”), the area between two curves is always taken to be positive.
€
14) One method of approximating the value of a definite integral follow’s Simpson’s
Rule, which “fits” parabolic arcs to portions of the curve described by y = f ( x ) and
applies the formula for the area under a parabola. Because three points at a time are
used along the curve, the area calculation uses two subdivisions of the interval from
x = a to x = b at a time; thus, Simpson’s Rule can only be used when the number of
subdivisions, n , of the interval is an even integer.
The definite integral in this Problem runs from x = −2 to x = +2 , so the
n = 4 subdivisions of this interval each have a width of Δ x =
b−a
2 − (−2 )
= 1 .
n =
4
The points at which we will then evaluate the function are x = −2 , −1 , 0 , 1 , and 2 .
The estimated value of the given integral, by Simpson’s Rule, is then
2
∫
−2
4
dx
x2 + 4
≈
⎡
⎞
⎛
⎞
⎛ 4 ⎞
⎛ 4 ⎞
⎛ 4 ⎞ ⎤
4
4
1 ⎢ ⎛
⋅ ⎜
+
4
+
2
+
4
+
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜ 2
⎟ ⎥
2
2
2
3 ⎢⎣ ⎝ (− 2) 2 + 4 ⎠
⎝ (− 1) + 4 ⎠
⎝ 0 + 4 ⎠
⎝ 1 + 4 ⎠
⎝ 2 + 4 ⎠ ⎥⎦
=
4 ⎡ 1
4
2
4
1 ⎤
4 ⎡ 1
4
2
4
1 ⎤
⋅ ⎢
+
+
+
+
= ⋅ ⎢ +
+
+
+ ⎥ 3 ⎣ 4 + 4
1+ 4
4
1+ 4
4 + 4 ⎥⎦
3 ⎣ 8
5
4
5
8 ⎦
=
94
47
=
≈ 3.133 . 30
15
€
€
€
€
We do not deal with trigonometric functions in this course, but the value of this integral
can be computed exactly as 2 · 2 arctan ( 2/2 ) = π ≈ 3.14159…
€
15) Once the meaning of the negative exponent in the expression on the right-hand side
of this differential equation is applied, it becomes clear that the equation is separable,
and so can be integrated to solve this “initial-value problem”:
dy
= x e −2y
dx
⇒
⇒
dy
x
= 2y
dx
e
⇒ e 2y dy = x dx ⇒
∫e
2y
dy =
∫ x dx
1 2y
1
e = x2 + C .
2
2
€ It is often more convenient at this stage to determine the value of the constant C for the
initial-value problem, rather than to find the function y = f ( x ) first (when possible)
and then “back-figure” for C . We are told that f ( 0 ) = 0 , thus
1 2 ⋅€0
1
e
=
⋅ 02 + C ⇒
2
2
1
1
⋅ 1 = 0 + C ⇒ C = . The specific function which
2
2
solve this initial-value problem is then
1 2y
1
1
e = x2 +
⇒ e 2y = x 2 + 1 ⇒ ln ( e 2y ) = ln ( x 2 + 1)
2
2
2
€
⇒ 2y = ln ( x 2 + 1) ⇒ y =
€
We can also write this as y =
1
ln ( x 2 + 1) . 2
1
ln x 2 + 1 , but it is not necessary, since x 2 + 1 is
2
never zero or negative, its value is always in the domain of the natural logarithm function.
€
€
16) The Extreme Value Theorem tells us that a function which is continuous on an
interval [ a , b ] must have an absolute maximum and an absolute minimum value in
that interval. While the (First) Derivative Test aids us in the location of local (or relative)
extrema for the function, these are not necessarily the largest or smallest values the
function has in the interval. So we must calculate the values of the function at the
endpoints of the intervals, as well as finding the local maximum or minimum, and
compare all of these values.
8
f ( x ) = 2 x + x on the interval [ 1 , 3 ] , its value at the
18 + 8
8
8
26
= 10 and f ( 3) = 2 ⋅ 3 +
=
=
endpoints are f (1) = 2 ⋅ 1 +
1
3
3
3 .
d
−1
−2
The first derivative
€ is f ' ( x ) = dx ( 2 x + 8 x ) = 2 − 8 x ; setting this to zero
8
8
= 2 ⇒ x 2 = 4 ⇒ x = ± 2 . The extremum at
gives €
us 2 − 2 = 0 ⇒
x
x2
For the function
x = − 2 is not in the interval we are concerned with, so we will only examine x = 2 .
€
The second derivative of our function is
f '' ( x ) =
d
16
( 2 − 8 x −2 ) = 16 x −3 or
,
dx
x3
€
which is positive at x = 2 , so this is the location of a local minimum. The value of the
8
= 8 . We find that f ( 2 ) < f ( 3 ) < f ( 1 ) , so the
function there is f ( 2) = 2 ⋅ 2 +
2
point ( 2, 8 ) is both the local and
€ the absolute minimum of f ( x ) on the interval, and
( 1, 10 ) is the absolute maximum.
€
17) We are asked to find two numbers x and y such that the sum x + 2y is 36 and
the product xy is made as large as possible. One approach we can take is to use the
constraining equation x + 2y = 36 to reduce the function of two variable f ( x , y ) = xy
to a function of a single variable, which we can maximize by looking for its local maximum:
eliminating x : x = 36 − 2y
⇒ f ( x , y ) = xy
 f ( y ) = ( 36 − 2y ) · y = 36 y − 2 y
2
⇒ f ’ ( y ) = 36 − 4 y = 0 ⇒ 4 y = 36 ⇒ y = 9
⇒ x = 36 − 2 · 9 = 18 .
This makes the largest possible product under the constraint xy = 9 · 18 = 162 .
We can make this determination a bit more directly by using the method of
“Lagrange multiplers”. We take the function of two variables f ( x , y ) = xy and use the
constraint to define a second function g ( x , y ) = x + 2y − 36 . The extrema for
f ( x , y ) are found where fx = λ · gx and fy = λ · gy , where λ is an as yet
undetermined constant, or “multiplier”. The partial derivatives in these equations are
€
€
fx =
∂
(x y ) = y ,
∂x
gx =
∂
∂
( x + 2 y − 36 ) = 1 , and g y =
( x + 2 y − 36 ) = 2 . ∂x
∂y
fy =
∂
(x y) = x ,
∂y
The Lagrange multiplier equations are then y = λ · 1 and x = λ · 2 ⇒ x = 2 y .
Inserting this result into the constraint equation gives us x + 2y = ( 2y ) + 2y = 36
⇒ 4y = 36 ⇒ y = 9 ⇒ x = 2 · 9 = 18 , as we found above. (In this particular
problem, we do not obtain a specific value for λ , but we could do so in other applications.)
18)
a) Here, we will calculate the indefinite integral
∫
100 + 25 x 2 dx
using an
already-determined anti-derivative, such as we would find in a table of integrals,
∫
a 2 + u 2 du =
u
2
a2 + u2 +
a2
ln u +
2
€
We will need to use a 2 = 100 ⇒ a = 10 and u
Placing these into the equation produces
€
∫
10 2 + ( 5x ) 2 ( 5 d x ) =
⇒ 5
∫
5x
2
100 + 25 x 2 dx =
€
⇒
∫
100 + 25 x 2 dx =
€
or
€
x
⋅
2
x
2
10 2 + ( 5x ) 2 +
5x
2
2
a2 + u2
= 25 x
2
+ C .
⇒ u = 5x , du = 5 dx .
10 2
ln ( 5x ) +
2
10 2 + ( 5x ) 2
100 + 25 x 2 + 50 ln ( 5x ) +
100 + 25 x 2 + 10 ln ( 5x ) +
25 ⋅
4 + x 2 + 10 ln ( 5x ) +
5x
⋅
2
4 + x 2 + 10 ln 5 ⋅
5x
⋅
2
4 + x 2 + 10 ln x +
5x
⋅
2
4 + x 2 + 10 ln x +
25 ⋅
+ C
100 + 25 x 2
100 + 25 x 2
4 + x2
+ C
+ C
+ C =
€
=
€ €
=
{x +
4 + x2
4 + x2
}
+ C
+ 10 ln 5 + C using properties of logarithms 4 + x2
+ C . Notice that we leave the arbitrary constant as C throughout this calculation. Properly
speaking, it is involved in the arithmetic and should be written as a different constant at
various points along the way, but because it has no specific value, no one really worries
about that…€
b) When we have to integrate a logarithmic function multiplied by a polynomial, we
will need to use integration-by-parts. We can make the work a little easier by using
properties of logarithms to write ln ( x 3 ) = 3 ln x . The indefinite integral is then
∫
x ln ( x 3 ) dx = 3
∫
⎡
⎢
1 2
x ln x dx = 3 ⎢ ln
x −
{x ⋅ {
2
⎢⎣ u
v
⎡ 1
1
= 3 ⎢ x 2 ln x −
2
⎣ 2
€
∫
∫
⎛
⎞ ⎤
1 2 ⎜ 1 ⎟ ⎥
x x dx ⎟ ⎥
2 ⎜⎜ {
{
⎟
v ⎝ du ⎠ ⎦⎥
⎤
⎡ 1
1 1 ⎤
x dx ⎥ = 3 ⎢ x 2 ln x − ⋅ x 2 ⎥ + C 2 2 ⎦
⎦
⎣ 2
=
€
3 2
3
x ln x − x 2 + C 2
4
or
1 2
3
1
3
x ( 3 ln x ) − x 2 + C = x 2 ln ( x 3 ) − x 2 + C .
2
4
2
4
€
Here also, the arbitrary constant is simply written as C in every step, even though all of
our arithmetic actually does apply to it as well.
€
G. Ruffa – 7/12