4.1 Analysis of functions I: Increase, decrease and
concavity
Definition
Let f be defined on an interval and let x1 and x2 denote points in that
interval.
a) f is said to be increasing on the interval if
f (x1 ) < f (x2 ) whenever x1 < x2
b) f is said to be decreasing on the interval if
f (x1 ) > f (x2 ) whenever x1 < x2
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Theorem
Let f be a function that is continuous on a closed interval [a, b] and
differentiable on the open interval (a, b)
a) If f 0 (x) > 0 for every value of x in (a, b), then f is increasing on
[a, b].
b) If f 0 (x) < 0 for every value of x in (a, b), then f is decreasing on
[a, b]
c) If f 0 (x) = 0 for every value of x in (a, b), then f is constant on [a, b]
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Example: Find where f (x) = x 2 − 5x + 6 is increasing, decreasing or
constant.
f 0 (x) = 2x − 5
x
f 0 (x)
f(x)
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2
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Example: Find where f (x) = x 2 − 5x + 6 is increasing, decreasing or
constant.
f 0 (x) = 2x − 5
x
f 0 (x)
f(x)
&
0
-
5
2
&
0
+
%
f (x) = x 2 − 5x + 6 is decreasing on (−∞, 52 ] and increasing on
[ 25 , +∞). It is nowhere constant!
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Concavity
Concave up: The slope of tangent lines is increasing. Equivalently, the
derivative function f 0 (x) is increasing, that is
(f 0 )0 (x) = f 00 (x) > 0
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Concavity
Concave up: The slope of tangent lines is increasing. Equivalently, the
derivative function f 0 (x) is increasing, that is
(f 0 )0 (x) = f 00 (x) > 0
Concave down: The slope of tangent lines is decreasing. Equivalently,
the derivative function f 0 (x) is decreasing, that is
(f 0 )0 (x) = f 00 (x) < 0
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Theorem
a) If f 00 (x) > 0 on an open interval (a, b), then the graph of f is
concave up on (a, b).
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Theorem
a) If f 00 (x) > 0 on an open interval (a, b), then the graph of f is
concave up on (a, b).
b) If f 00 (x) < 0 on an open interval (a, b), then the graph of f is
concave down on (a, b).
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Example: Find where f (x) = 3x 3 − 4x + 3 is concave up or concave
down.
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Example: Find where f (x) = 3x 3 − 4x + 3 is concave up or concave
down.
f 0 (x) = 9x 2 − 4
f 00 (x) = 18x
x
f 00 (x)
f(x)
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a
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The graph of f is concave up on (0, +∞). It is concave down on
(−∞, 0).
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The graph of f is concave up on (0, +∞). It is concave down on
(−∞, 0).
Definition
If f is continuous on an open interval containing x0 and if the graph of f
changes the direction of its concavity at x0 , then the point (x0 , f (x0 )) on
the graph of f is called an inflection point of f .
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Example: Find the largest intervals on which f (x) = 3x 4 − 4x 3 is
increasing, decreasing; find the largest open intervals on which f is
concave up, concave down and find the x coordinates of all inflection
points.
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f 0 (x) = 12x 2 (x − 1)
f 00 (x) = 12x(3x − 2)
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f 0 (x) = 12x 2 (x − 1)
f 00 (x) = 12x(3x − 2)
x
12x 2
x-1
f 0 (x)
12x
3x-2
f 00 (x)
f(x)
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2
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The function f is increasing on [1, +∞),
decreasing on (−∞, 1].
Its graph is concave up on (−∞, 0] and [ 23 , +∞)
concave down on (0, 32 ). The graph has an inflection point at x = 0 and
x = 32 .
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4.2 Analysis of functions II: Relative extrema, graphing
polynomials
A critical point for a function f is any value of x in the domain of f at
which f 0 (x) = 0 or at which f is not differentiable. Recall an earlier
statement that relative extrema of a function f can occur only at critical
points.
If f has a relative extremum at x0 , then either f 0 (x0 ) = 0 or f is not
differentiable at x0 ; in other words, x0 is a critical point.
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Example 1.
f (x) = x 2 − 5x + 6. Since f 0 (x) = 2x − 5 is defined everywhere and
2x − 5 = 0 if and only if x = 52 , we conclude that f can have a relative
extremum only at x = 52 and nowhere else!
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Example 1.
f (x) = x 2 − 5x + 6. Since f 0 (x) = 2x − 5 is defined everywhere and
2x − 5 = 0 if and only if x = 52 , we conclude that f can have a relative
extremum only at x = 52 and nowhere else!
Example 2.
f (x) =
0
(
x
f (x) =
if x ≥ 0
−x
if x < 0
(
1
if x>0
−1
if x<0
f 0 (x) is undefined at x = 0 and it is nowhere zero, so f can have a
relative extremum only at x = 0.
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First derivative test
Suppose f is continuous at a critical point x0 .
a) If f 0 (x) > 0 on an open interval extending left from x0 and
f 0 (x) < 0 on an open interval extending right from x0 , then f has a
relative maximum at x0 .
b) If f 0 (x) < 0 on an open interval extending left from x0 and
f 0 (x) > 0 on an open interval extending right from x0 , then f has a
relative minimum at x0 .
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Example
f (x) = 4x 2 + 2x − 5
f 0 (x) = 8x + 2
f 0 (x) = 0 ⇔ x = − 14
x
f 0 (x)
f(x)
&
- 14
0
0
+
%
+
f (x) has a relative minimum at x = − 14 .
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Example: f (x) = x 3 + 3x 2 − 9x + 1
f 0 (x) = 3(x − 1)(x + 3)
f 0 (x) = 0 ⇔ x = 1 or x = −3.
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x
f 0 (x)
f(x)
+
%
-3
0
0
&
&
1
0
+
%
f (x) has a relative maximum at x = −3, a relative minimum at x = 1.
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Example
f (x) = x 3
f 0 (x) = 3x 2
f 0 (x) = 0 ⇔ x = 0
x
f 0 (x)
f(x)
+
%
0
0
+
%
f (x) has no relative extremum.
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Second derivative test
At a relative maximum, the graph of the function f is concave down, so
as seen before, the second derivative should be negative at this point.
Similarly, at a relative minimum, the graph is concave up, therefore the
second derivative should be positive at such a point.
Theorem
Suppose f is twice differentiable at a stationary point x0 (that is
f 0 (x0 ) = 0. Then:
a) If f 00 (x0 ) > 0, then f has a relative minimum at x0 .
b) If f 00 (x0 ) < 0, then f has a relative maximum at x0 .
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Example
f (x) = 2x 3 − 3x 2 − 3
f 0 (x) = 6x 2 − 6x = 6x(x − 1)
f 0 (x) = 0 ⇔ x = 0 or x = 1
x = 0 and x = 1 are stationary points.
f 00 (x) = 12x − 6
Since f 00 (0) = −6 < 0, f (x) has a relative maximum at x = 0; and
since f 00 (1) = 6 > 0, f (x) has a relative minimum at x = 1.
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Simple curve sketching.
Information to be included on the graphs
X and Y intercepts.
Critical points, relative extrema.
Increasing, decreasing behavior.
Concavity, inflection points.
Behavior at ±∞
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Example
Graph f (x) = x(x 2 − 1)2 .
X intercepts: Solve 0 = x(x 2 − 1)2
x=0, x=±1
Y intercept: Evaluate f (0) = 0. So y = 0.
lim f (x) = +∞
x→+∞
lim f (x) = −∞
x→−∞
f 0 (x) = (x 2 − 1)(5x 2 − 1)
The critical points are x = ±1, x = ± √15
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f 00 (x) = 4x(5x 2 − 3)
f 00 (x)
q
= 0 ⇔ x = 0, or x = ± 35 .
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f 00 (x) = 4x(5x 2 − 3)
f 00 (x)
x
f0
f 00
f
q
= 0 ⇔ x = 0, or x = ± 35 .
+
%
a
-1
0
0
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a
0
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0
1
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( 15 ) 2
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1
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limx→−∞ f (x) = −∞, no global minimum.
limx→+∞ f (x) = +∞, no global maximum.
q
Local minima are f (− 15 ), and f (1).
q
Local maxima are f (−1) and f ( 15 ).
q
q
q
q
Inflection points are (− 35 , f (− 35 )), (0, f (0)) and ( 35 , f ( 35 )).
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4.3 Analysis of functions III: Rational functions, cusps
and vertical tangents
Vertical asymptotes:
Example: y =
1
.
(x+2)3
lim
x→−2−
1
= −∞
(x + 2)3
lim +
1
= +∞
(x + 2)3
x→−2
The graph of y =
x = −2.
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1
(x+2)3
will show a (two-sided) vertical asymptote at
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x =a
is a vertical asymptote for the graph of y = f (x) if
lim f (x) = ±∞
x→a
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x =a
is a vertical asymptote for the graph of y = f (x) if
lim f (x) = ±∞
x→a
Horizontal asymptotes:
Example: y =
1
.
x2
lim
1
=0
x2
lim
1
=0
x2
x→+∞
x→−∞
The graph will show a (two-sided) horizontal asymptote y = 0
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If
lim f (x) = d,
x→±∞
then
y =d
is a horizontal asymptote for the graph of y = f (x).
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If
lim f (x) = d,
x→±∞
then
y =d
is a horizontal asymptote for the graph of y = f (x).
Slant asymptotes: y = mx + b
m = lim f 0 (x)
x→±∞
b = lim [f (x) − mx]
x→±∞
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Example: Find slant asymptotes for the graph of f (x) =
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x 2 +x−1
x−1 .
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Example: Find slant asymptotes for the graph of f (x) =
f 0 (x) =
x 2 +x−1
x−1 .
x 2 − 2x
(x − 1)2
m = lim f 0 (x) = 1
x→±∞
b = lim [
x→±∞
2x − 1
x2 + x − 1
− x] = lim
=2
x→±∞
x −1
x −1
y = x + 2 is a (two-sided) slant asymptote.
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Complete graph
Graph f (x) =
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x3
x 2 −5
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Complete graph
Graph f (x) =
x3
x 2 −5
x-intercept: x = 0
y -intercept: y = 0
limx→−∞ f (x) = −∞
limx→+∞ f (x) = +∞
f 0 (x) =
x 2 (x 2 −15)
.
(x 2 −5)2
f 00 (x) =
10x(x 2 +15)
3
(x 2 −5)
√
√
Critical points are x = 0, x = ± 15, f 0 (x) undefined at x = ± 5.
√
f 00 (x) = 0 at x = 0, undefined at x = ± 5
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√
Vertical asymptote at x = ± 5
No horizontal asymptote.
limx→±∞ f 0 (x) = 1
3
limx→±∞ [ x 2x−5 − x] = 0. So
y =x
is a two-sided slant asymptote.
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x
f 0 (x)
f 00 (x)
f(x)
+
%
a
√
− 15
0
&
a
√
− 5
?
?
?
?
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0
0
0
0
√
&
a
5
?
?
?
?
√
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15
0
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`
√
Local minimum value f ( 15)
√
Local maximum value f (− 15)
Inflection point at x = 0
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Vertical tangent lines and cusps.
Examples:
1
4
f (x) = 6x 3 + 3x 3 .
f 0 (x) = 2x −2/3 + 4x 1/3
Observe that f (0) is defined and
lim f 0 (x) = +∞
x→0
The graph will have a vertical tangent line at x = 0
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Another example:
f (x) = (x − 4)2/3 .
f 0 (x) =
1
2
(x − 4)− 3
3
Observe that f (4) is defined and
lim f 0 (x) = +∞
x→4+
lim f 0 (x) = −∞
x→4−
The graph will show a cusp at x = 4.
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Graph f (x) = x 1/3 (2 − x)2/3 .
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Graph f (x) = x 1/3 (2 − x)2/3 .
x intercepts: x = 0, x = 2
y intercept: y = 0
Computations show f 0 (x) = 13 x 2/32−3x
. Critical points are
(2−x)1/3
0
x = 2/3, f (x) undefined at x = 0 and x = 2.
Vertical tangent at x = 0, cusp at x = 2.
1
Computations show f 00 (x) = − 98 x 5/3 (2−x)
4/3 . Undefined at x = 0
and x = 2
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x
f 0 (x)
f 00 (x)
f(x)
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+
%
`
0
?
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0
+
%
a
2/3
0
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a
2
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+
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a
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x
f 0 (x)
f 00 (x)
f(x)
+
+
%
`
0
?
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0
+
%
a
2/3
0
&
a
2
?
?
+
%
a
A local minimum value is f (2). A local maximum value is f (2/3)
Inflection point at x = 0
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4.4 Absolute maxima and minima
Recall: If f (x) is continuous on [a, b], then there exists numbers c and
d in [a, b] such that f (c) is the minimum value and f (d) is the
maximum value of f (x) on [a, b]. Moreover each of c and d is either
critical points of f (x) or one of the endpoints a or b.
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Program for looking for extremal values of a continuous function
f on a closed interval [a, b].
1. Find all critical points of f (x) in [a, b]
2. Evaluate f (x) at those points and at the endpoints a and b
3. Identify the maximum and minimum values by comparison.
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Examples:
Find the absolute extrema for:
f (x) = √ 3x2
4x +1
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on the interval [−1, 1]
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Extrema values on open or unbounded intervals
Examples State whether the given function attains a maximum value
or a minimum value (or both).
1) f (x) =
2) f (x) =
1
on (−∞, +∞)
x 2 +1
1
x(1−x) on (0, 1).
Evaluation at endpoints is replaced by right and left limits as
appropriate.
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4.5 Applied maximum and minimum problems
Example A rectangle has its two lower corners on the X -axis and its
two upper corners on the curve y = 16 − x 2 . For all such rectangles,
what are the dimensions of the one with largest area?
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Strategy in solving applied maximum and minimum problems
1. Drawing a picture and labeling quantities.
2. Find a formula for the quantity to be maximized or minimized.
3. Using information from the text, express the quantity to be
optimized as a function of one variable.
4. Find the interval of possible values for this variable.
5. Use the previous procedure to find the absolute extrema on the
closed interval from 4.
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More applied extrema problems: Open or unbounded
intervals
Example: A closed rectangular container with a square base is to have
a volume of 2000cm3 . It costs twice as much per square centimeter for
the top and bottom as it does for the sides. Find the dimensions of the
container of least cost.
The procedure is similar, except evaluation at endpoints is replaced by
right and left limits.
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Example 2
A lamp is suspended above the center of a round table of radius r .
How high above the table should the lamp be placed to achieve
maximum illumination at the edge of the table?
Assuming that the illumination I is directly proportional to the cosine of
the angle of incidence φ of the light rays and inversely proportional to
the square of the distance l from the light source.
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Observation: A unique local extremum on an open interval is
necessarily an absolute extremum!
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4.8 Rolle’s Theorem; Mean Value Theorem
Recall: (Rolle’s Theorem) Let f be differentiable on (a, b) and
continuous on [a, b]. If f (a) = 0 = f (b), then there is at least one point
c in (a, b) where
f 0 (c) = 0.
Recall: (Mean Value Theorem) Let f (x) be differentiable on (a, b) and
continuous on [a, b]. Then there is at least one point c in (a, b) where
f 0 (c) =
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f (b) − f (a)
.
b−a
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Independent proof of Mean Value Theorem
The secant line joining (a, f (a)) and (b, f (b)) has equation
y − f (a) =
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f (b) − f (a)
(x − a)
b−a
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Independent proof of Mean Value Theorem
The secant line joining (a, f (a)) and (b, f (b)) has equation
y − f (a) =
f (b) − f (a)
(x − a)
b−a
The function
v (x) = f (x) − f (a) −
f (b) − f (a)
(x − a)
b−a
satisfies the hypotheses in Rolle’s Theorem.
Indeed, v (x) is differentiable on (a, b), continuous on [a, b] and
v (a) = 0 = v (b).
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So, by Rolle’s Theorem, there is at least one c in (a, b) such that
v 0 (c) = 0.
But v 0 (c) = 0 is equivalent to
f 0 (c) −
f (b) − f (a)
=0
b−a
That is
f 0 (c) =
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f (b) − f (a)
.
b−a
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Exercise: Does f (x) = x + x1 satisfy the Mean Value Theorem’s
hypotheses on [3, 4]. If it does, find all values c in ]3, 4[ that satisfy the
conclusion of the Theorem.
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5.2 The indefinite integral
The area Problem
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A(x) will denote the shaded area under the graph of y = f (x) and
between fixed a and variable x.
Claim: A0 (x) = f (x).
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Proof:
A(x + h) − A(x)
h
h→0
A0 (x) = lim
assuming h > 0 for now:
hf (x) ≤ A(x + h) − A(x) ≤ hf (x + h)
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f (x) ≤
lim f (x) ≤
h→0
f (x) ≤
A(x+h)−A(x)
h
A(x+h)−A(x)
limh→0
h
A0 (x)
≤ f (x + h)
≤ lim f (x + h)
h→0
≤ f (x)
The same inequalities will be obtained assuming h < 0. So
A0 (x) = f (x)
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Example: Find the area A(x) of the region under the graph of y = x 2 ,
between x = 1 and x to the right of 1.
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Example: Find the area A(x) of the region under the graph of y = x 2 ,
between x = 1 and x to the right of 1.
3
A0 (x) = x 2 . So A(x) = x3 + C where C could be any constant. But
since A(1) = 0, it follows that
0=
1
+C
3
and hence C = − 13 .
A(x) =
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x3 1
− .
3
3
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Remark: To find an area to the left of 1, a negative sign needs to be
introduced.
For example the area between 1 and
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1
2
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1
is −A( 12 ) = −( 24
− 13 ) =
7
24 .
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A function F (x) such that F 0 (x) = f (x) is called an antiderivative of
f (x).
If F (x) is an antiderivative of f (x), then for any constant C, F (x) + C is
another antiderivative of f (x). It is true also that any two antiderivatives
of f (x) differ by a (locally) constant function.
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The process of finding antiderivatives is called integration. We shall
use the notation
Z
f (x)dx
to represent all antiderivatives of f (x). That is
Z
f (x)dx = F (x) + C
where F (x) is one of the antiderivatives of f (x).
the indefinite integral of f (x).
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R
f (x)dx is also called
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Exercises: Find the integrals:
R
1) x −6 dx
R
2) x 2/3 dx
R
3) 2x1 3 dx =
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Z
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(u 3 − 2u + 7)du =
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Z
(u 3 − 2u + 7)du =
u4
− u 2 + 7u + C
4
We have used the following:
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Properties of the indefinite integral
R
R
Cf (x)dx = C
R
f (x)dx
R
R
(f (x) + g(x))dx = f (x)dx + g(x)dx
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R
(1 + x 2 )(2 − x)dx =
R
(t −2 − cos t)dt =
R
R
R
sin 2x
cos x dx =
x 2 sin x+2 sin x
dx
2+x 2
2
=
cot xdx =
Find f (x) such that f 0 (x) + sin x = 0 and f (0) = 2.
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Some integration formulas
d x r +1
dx ( r +1 )
= x r , for r 6= −1, we deduce that
R
d
2t ⇒
tan
t
=
sec
sec2 tdt = tan t + C
dt
R
d
sin xdx = − cos x + C
dx (− cos x) = sin x ⇒
R
d
cos xdx = sin x + C
dx sin x = cos x ⇒
Since
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R
x r dx =
x r +1
r +1
+ C.
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5.3 Integration by substitution
Basic integration by substitution formula:
If f (x) = g(u) du
dx , then
Z
Z
Z
du
f (x)dx = g(u) dx = g(u)du
dx
R
Example: Evaluate (3x − 1)5 dx
Put u(x) = 3x − 1, then
Letting g(u) =
u5
3 ,
du
dx
=3
we see that
(3x − 1)5 = g(u)
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du
dx
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Therefore
Z
5
(3x − 1) dx
Z
g(u)
=
du
dx
dx
Z
=
=
=
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Z
g(u)du =
u5
du
3
1 u6
+C
3 6
1
(3x − 1)6 + C
18
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Exercises
1)
R
sec2 (4x + 1)dx
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Exercises
1)
R
sec2 (4x + 1)dx
2)
R
sin (3x)dx
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Exercises
1)
R
sec2 (4x + 1)dx
2)
R
sin (3x)dx
3)
R
x cos (3x 2 )dx
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Exercises
1)
R
sec2 (4x + 1)dx
2)
R
sin (3x)dx
3)
R
x cos (3x 2 )dx
4)
R
tan3 (5x) sec2 (5x)dx
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More exercises
1)
R
√ 3x
4x 2 +5
dx
R
2) (4x 2 + 12x + 9)2/3 dx
√
R
3) cos 4θ 2 − sin 4θdθ
R
4) sin x(5/x)
dx
2
R 2
5) x sec2 (x 3 )dx
R
6) cos2 2t sin 2tdt
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10.1 Parametric equations; Tangent lines for
parametric curves
A parametric curve in the x, y plane is a pair of functions:
x = f (t),
y = g(t),
t ∈ [a, b]
x = cos t,
y = sin t,
0 ≤ t ≤ 2π
Example 1:
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The graph is the unit circle as we can see by eliminating t and
obtaining
x2 + y2 = 1
Example 2:
x=
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1 − t2
,
1 + t2
y=
2t
,
1 + t2
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−∞ < t < +∞
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Eliminating t, we get
x2 + y2 = 1
the unit circle again!
Exercises:
Eliminate the parameter and then sketch the curve:
a) x = t 2 + 3t,
y =t −2
b) x = et ,
y = 4e2t
c) x = et ,
y = e−t
d) x = 5 cos t,
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y = 5 sin t
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First eliminate the parameter and sketch the curve. Then describe the
motion of the point (x(t), y (t)) as t varies in the given interval:
1) x = sin2 πt,
2) x = cos t,
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y = cos2 πt;
y = sin2 t;
0 ≤ t ≤ 2π
−π ≤ t ≤ π
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Tangent lines to parametric curves
Given parametric equations of a curve:
x = f (t),
We wish to find
When
dx
dt
dy
dx
y = g(t)
at a given t = t0 .
6= 0, one has:
dy
=
dx
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dy
dt
dx
dt
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Example:
x = t sin t,
y = t cos t
find the equation of the tangent line at t = π2 .
dy
cos t − t sin t
=
dx
sin t + t cos t
At t = π2 ,
dy
dx
= − π2 .
The point corresponding to t =
equation:
π
2
is ( π2 , 0). So the tangent line has
π
π
y = − (x − )
2
2
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Exercises:
1. First write the equation of the tangent line to the given parametric
curve at the point that corresponds to the given value of t, then
2
calculate ddxy2 to determine whether the curve is concave upward or
concave downward at this point.
a) x = 2t 2 + 1,
b) x = t sin t,
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y = 3t 3 + 2;
y = t cos t;
t =1
t=
π
2
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Polar coordinates
(r , θ) :
0 ≤ r < +∞,
x = r cos θ,
0 ≤ θ < 2π
y = r sin θ
Conversion formulas:
r2 = x2 + y2
y
tan θ =
for x 6= 0
x
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Exercises:
Express the given rectangular equation in polar form.
1) xy = 1
2) x 2 − y 2 = 1
3) x + y = 4
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Express the given polar equation in rectangular form.
1) r = 3
2) θ =
3π
4
3) r = sin 2θ
4) r = 1 − cos 2θ
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Graphing polar curves
Polar curves are usually given by equations of the form
r = f (θ)
Example: Graph the circle r = 2 sin θ
The period T = 2π will be divided into 4 quarters,
θ = T4 , θ = 2 T4 , θ = 3 T4 , θ = 4 T4 . We will include any additional
reference points θ satisfying r = 0, that is, 2 sin θ = 0 in this exemple.
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θ
r
0
0
π
2
2
π
0
3 π2
-2
2π
0
Observe that the point r = −2, θ = 3 π2 will be plotted as
r = 2, θ = 3 π2 + π, that is, a reflection through the pole is applied.
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Cardioids are given by equations like:
r = a(1 ± sin θ)
r = a(1 ± cos θ)
Example: Graph r = 2 + 2 sin θ
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Rose curves are given by equations of the form:
r = a cos nθ
r = a sin nθ
When n is even, the rose has 2n petals. When n is odd, the rose will
have n petals.
Examples: Graph r = 3 sin 3θ.
Graph r = cos 2θ
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Polar curves as parametric curves
From a polar curve r = f (θ) we get a parametric curve
x = f (θ) cos θ,
y = f (θ) sin θ
In this case, the slope of tangent lines will be given by:
dr
r cos θ + sin θ dθ
dy
=
dr
dx
−r sin θ + cos θ dθ
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Example: Find the slope of the tangent line to the circle r = 4 cos θ at
the point where θ = π6 .
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