Math 1131
Tuesday
November 14
d
(f ) means take the derivative of f .
dx
R
f dx means find the antiderivatives of f
Definition (Antiderivative)
A function F is an antiderivative of f on an interval I provided
F 0 (x) = f (x) for all x in I.
d 3
(x + sin(x)) = 3x2 + cos x
dx
x3 + sin(x), x3 + sin(x) − 8, x3 + sin(x) + 15 are all
antiderivatives of 3x2 + cos x.
Z
We write
3x2 + cos x dx = x3 + sin(x) + C
Tuesday November 14 ()
Math 1131
2/1
R
f dx is called the indefinite integral of f .
R
f dx = F (x) + C, when F 0 = f
This is the family of all functions whose derivative is f .
Tuesday November 14 ()
Math 1131
3/1
R
f dx is called the indefinite integral of f .
R
f dx = F (x) + C, when F 0 = f
This is the family of all functions whose derivative is f .
Examples:
R 3x
e − cos (2x) dx = 13 e3x − 12 sin (2x) + C
R 4
x + 3x − 9 dx = 15 x5 + 32 x2 − 9x + C
Theorem
Two functions with the same derivative differ only by a constant.
Hence, if F is one antiderivative of f , the all antiderivatives of f
have the form F (x) + C.
Tuesday November 14 ()
Math 1131
3/1
For each derivative rule there corresponds an antiderivative rule.
d x
(e ) = ex
dx
1
d
(
xn+1 ) = xn
dx n + 1
d
(sin x) = cos x
dx
Z
=⇒
Z
=⇒
ex dx = ex + C
xn dx =
1
xn+1 + C
n+1
Z
cos x dx = sin x + C
=⇒
d
1
=⇒
(arcsin x) = √
dx
1 − x2
Z
1
√
dx = sin−1 x + C
1 − x2
d
1
(ln |x|) =
dx
x
Z
1
dx = ln |x| + C
x
Tuesday November 14 ()
=⇒
Math 1131
4/1
More derivative rules that give us antiderivative rules.
Z
Z
Z
f (x) + g(x) dx = f (x) dx + g(x) dx
If F 0 = f, G0 = g then
d
(F (x) + G(x)) = F 0 (x) + G0 (x) = f (x) + g(x) =⇒
dx
Z
Z
Z
f (x) + g(x) dx = F (x) + G(x) + C = f (x) dx + g(x) dx
Tuesday November 14 ()
Math 1131
5/1
More derivative rules that give us antiderivative rules.
Z
Z
Z
f (x) + g(x) dx = f (x) dx + g(x) dx
If F 0 = f, G0 = g then
d
(F (x) + G(x)) = F 0 (x) + G0 (x) = f (x) + g(x) =⇒
dx
Z
Z
Z
f (x) + g(x) dx = F (x) + G(x) + C = f (x) dx + g(x) dx
Z
Z
kf (x) dx = k
f (x) dx
d
If F 0 = f then
(kF (x)) = kF 0 (x) = kf (x) =⇒
dx
Z
Z
kf (x) dx = kF (x) + C = k f (x) dx
Tuesday November 14 ()
Math 1131
5/1
How do we use these rules in practice?
Problem: Find the indefinite integral
R
9ex + 8x5 dx
Solution: We usually apply these rules mentally without actually
writing them down.
Tuesday November 14 ()
Math 1131
6/1
How do we use these rules in practice?
Problem: Find the indefinite integral
R
9ex + 8x5 dx
Solution: We usually apply these rules mentally without actually
writing them down.
1
R
9ex + 8x5 dx =
Tuesday November 14 ()
R
9ex dx +
R
8x5 dx
Math 1131
Sum Rule
6/1
How do we use these rules in practice?
Problem: Find the indefinite integral
R
9ex + 8x5 dx
Solution: We usually apply these rules mentally without actually
writing them down.
1
2
R
R
R
9ex + 8x5 dx = 9ex dx + 8x5 dx Sum Rule
R
R
= 9 ex dx + 8 x5 dx Constant Multiple Rule
Tuesday November 14 ()
Math 1131
6/1
How do we use these rules in practice?
Problem: Find the indefinite integral
R
9ex + 8x5 dx
Solution: We usually apply these rules mentally without actually
writing them down.
R
R
9ex + 8x5 dx = 9ex dx + 8x5 dx Sum Rule
R
R
2
= 9 ex dx + 8 x5 dx Constant Multiple Rule
x6
x
3
= 9e + 8( ) + C Rules from derivative tables
6
Z
x6
x
5
x
Answer:
9e + 8x dx = 9e + 8( ) + C
6
1
R
Tuesday November 14 ()
Math 1131
6/1
Clicker Question
Z
Evaluate the definite integral
e5x + x dx
1
(a) e5x + x2 + C
2
1
(b) 5e5x + x2 + C
2
1
1
(c) e5x + x2 + C
5
2
(d) e5x + x2 + C
(e) None of these
Tuesday November 14 ()
Math 1131
7/1
Clicker Question
Z
Evaluate the definite integral
e5x + x dx
1
(a) e5x + x2 + C
2
1
(b) 5e5x + x2 + C
2
1
1
(c) e5x + x2 + C
5
2
(d) e5x + x2 + C
(e) None of these
d 1 5x 1 2
( e + x ) = e5x + x
dx 5
2
Tuesday November 14 ()
Answer = c
Math 1131
7/1
Find an estimate of the area of the shaded green region between
the x-axis and the graph of f (x) = x2 + 1 between x = 0 and 2.
We estimate the area using four
rectangles whose base lies on
x-axis and right upper corner lies
on the graph.
Tuesday November 14 ()
Math 1131
8/1
Find an estimate of the area of the shaded green region between
the x-axis and the graph of f (x) = x2 + 1 between x = 0 and 2.
We estimate the area using four
rectangles whose base lies on
x-axis and right upper corner lies
on the graph.
1 1
Area of rectangle 1 = f ( )
2 2
1
Area of rectangle 2 = f (1)
2
1 3
Area of rectangle 3 = f ( )
2 2
1
Area of rectangle 4 = f (2)
2
Tuesday November 14 ()
Math 1131
8/1
We add the areas of these rectangles together to get an estimate
for the area under the graph of f (x) = x2 + 1.
1 1
1 2
1 3
1 4
A ≈ f( ) + f( ) + f( ) + f( )
2 2
2 2
2 2
2 2
1
1 13
1
1 5
A ≈ ( ) + (2) + ( ) + (5) = 5.75
2 4
2
2 4
2
This is an over estimate.
Tuesday November 14 ()
Math 1131
9/1
We add the areas of these rectangles together to get an estimate
for the area under the graph of f (x) = x2 + 1.
1 1
1 2
1 3
1 4
A ≈ f( ) + f( ) + f( ) + f( )
2 2
2 2
2 2
2 2
1
1 13
1
1 5
A ≈ ( ) + (2) + ( ) + (5) = 5.75
2 4
2
2 4
2
This is an over estimate.
Notation: This approximating sum is called a Riemann Sum.
The points 0, 12 , 1, 23 , 2 used to divide the interval [0,2] into four
smaller intervals are called grid points.
The length of each small interval is denoted by ∆x and ∆x = 0.5.
Tuesday November 14 ()
Math 1131
9/1
We do another approximation by taking the rectangle heights to
be the function value at the left endpoint of each small interval.
1
Area of rectangle 1 = f (0)
2
1 1
Area of rectangle 2 = f ( )
2 2
1
Area of rectangle 3 = f (1)
2
1 3
Area of rectangle 4 = f ( )
2 2
Tuesday November 14 ()
Math 1131
10 / 1
This approximating sum for the
area under f (x) = x2 + 1 is
called a Left Riemann Sum.
Same grid points points as
before, 0, 12 , 1, 23 , 2, dividing the
interval [0,2] into four smaller
intervals.
Interval length remains
∆x = 0.5.
Tuesday November 14 ()
Math 1131
11 / 1
This approximating sum for the
area under f (x) = x2 + 1 is
called a Left Riemann Sum.
Same grid points points as
before, 0, 12 , 1, 23 , 2, dividing the
interval [0,2] into four smaller
intervals.
Interval length remains
∆x = 0.5.
1
1 1
1 2
1 3
A ≈ f (0) + f ( ) + f ( ) + f ( )
2
2 2
2 2
2 2
1
1 5
1
1 13
A ≈ (1) + ( ) + (2) + ( ) = 3.75
2
2 4
2
2 4
Underestimate
Tuesday November 14 ()
Math 1131
11 / 1
Using the same grid points we approximate the area under
f (x) = x2 + 1 with a Midpoint Riemann Sum:
Instead of using the right or left
endpoints of each subinterval we
use the midpoint of each
subinterval to find the height of
each rectangle.
(Actual area = 4.6666666 . . .)
Tuesday November 14 ()
Math 1131
12 / 1
Using the same grid points we approximate the area under
f (x) = x2 + 1 with a Midpoint Riemann Sum:
Instead of using the right or left
endpoints of each subinterval we
use the midpoint of each
subinterval to find the height of
each rectangle.
(Actual area = 4.6666666 . . .)
1 1
1 3
1 5
1 7
A ≈ f( ) + f( ) + f( ) + f( )
2 4
2 4
2 4
2 4
1 17
1 25
1 41
1 65
A ≈ ( ) + ( ) + ( ) + ( ) = 4.625
2 16
2 16
2 16
2 16
Tuesday November 14 ()
Math 1131
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Clicker Question
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Use a Right Riemann Sum with
4 subintervals to approximate
the area between the graph of
f (x) = x2 + 2x, 1 ≤ x ≤ 3 and
the x-axis.
1 3/2 2 5/2 3
(a) 37
(b) 47/4
Tuesday November 14 ()
(c) 79/4
(d) 81/4
Math 1131
(e) 79/2
13 / 1
Clicker Question
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Use a Right Riemann Sum with
4 subintervals to approximate
the area between the graph of
f (x) = x2 + 2x, 1 ≤ x ≤ 3 and
the x-axis.
1 3/2 2 5/2 3
(a) 37
(b) 47/4
(c) 79/4
(d) 81/4
(e) 79/2
A ≈ f (3/2)∆x + f (2)∆x + f (5/2)∆x + f (3)∆x
A ≈ (f (3/2) + f (2) + f (5/2) + f (3))∆x
1
A ≈ ((3/2)2 + 2(3/2) + f (2) + f (5/2) + f (3))( ) = 79/4
2
Answer = c
Tuesday November 14 ()
Math 1131
13 / 1
Better Approximations
The easiest way to get a better approximation is to take more
rectangles. Here are the graphs showing eight rectangles and the
estimations for each of the three choices for rectangle heights that
we used above.
Rright = 5.1875
Rlef t = 4.1875
Rmidpoint = 4.65265
Right Riemann Sum
Midpt Riemann Sum
Left Riemann Sum
2−0
1
Eight intervals, nine grid points. ∆x =
=
8
4
Tuesday November 14 ()
Math 1131
14 / 1
Example
:
Estimate the area between f (x) = x3 − 5x2 + 6x + 5, 0 ≤ x ≤ 4
and the x-axis using n = 5 subintervals and all three cases above
for the heights of each rectangle.
Tuesday November 14 ()
Math 1131
15 / 1
Example
:
Estimate the area between f (x) = x3 − 5x2 + 6x + 5, 0 ≤ x ≤ 4
and the x-axis using n = 5 subintervals and all three cases above
for the heights of each rectangle.
Grid points: 0, 0.8, 1.6, 2.4, 3.2, 4
The width of each subinterval
will be
4−0
∆x =
= 0.8
5
We will approximate the area using a Left Riemann Sum, a Right
Riemann Sum, and a Midpoint Riemann Sum.
Tuesday November 14 ()
Math 1131
15 / 1
First approximate with a Right Riemann Sum, evaluating the
function using the right endpoints for the rectangle height.
∆x = 0.8
Unlike the first area we looked
at, the rectangles will give both
over and underestimates of the
area, depending on where we are
on the curve.
Area of 4-th rectangle is
f (3.2)∆x = [(3.2)3 − 5(3.2)2 + 6(3.2) + 5](0.8)
Tuesday November 14 ()
Math 1131
16 / 1
First approximate with a Right Riemann Sum, evaluating the
function using the right endpoints for the rectangle height.
∆x = 0.8
Unlike the first area we looked
at, the rectangles will give both
over and underestimates of the
area, depending on where we are
on the curve.
Area of 4-th rectangle is
f (3.2)∆x = [(3.2)3 − 5(3.2)2 + 6(3.2) + 5](0.8)
Adding up the area of all 5 rectangles gives us a Right Riemann
Sum that approximates the area under the curve.
A ≈ 0.8f (0.8) + 0.8f (1.6) + 0.8f (2.4) + 0.8f (3.2) + 0.8f (2) = 28.96
Tuesday November 14 ()
Math 1131
16 / 1
Now approximate by a Left Riemann Sum, using the left
endpoints to compute the heights of the rectangles.
∆x = 0.8
Area of 4-th rectangle is f (2.4)∆x
= [(2.4)3 − 5(2.4)2 + 6(2.4) + 5](0.8)
Tuesday November 14 ()
Math 1131
17 / 1
Now approximate by a Left Riemann Sum, using the left
endpoints to compute the heights of the rectangles.
∆x = 0.8
Area of 4-th rectangle is f (2.4)∆x
= [(2.4)3 − 5(2.4)2 + 6(2.4) + 5](0.8)
Adding up the area of all 5 rectangles gives us a Left Riemann
Sum that approximates the area under the curve.
A ≈ 0.8f (0) + 0.8f (0.8) + 0.8f (1.6) + 0.8f (2.4) + 0.8f (3.2) = 22.56
Tuesday November 14 ()
Math 1131
17 / 1
Now use the midpoints to compute the rectangle heights.
Area of 4-th rectangle is
f (2.8)∆x
[(2.8)3 − 5(2.8)2 + 6(2.8) + 5](0.8)
2.8 is the midpoint of the 4-th
interval.
Tuesday November 14 ()
Math 1131
18 / 1
Now use the midpoints to compute the rectangle heights.
Area of 4-th rectangle is
f (2.8)∆x
[(2.8)3 − 5(2.8)2 + 6(2.8) + 5](0.8)
2.8 is the midpoint of the 4-th
interval.
Adding up the area of all 5 rectangles gives us a Midpoint
Riemann Sum that approximates the area under the curve.
A ≈ 0.8f (.4) + 0.8f (1.2) + 0.8f (2) + 0.8f (2.8) + 0.8f (3.6) = 25.12
(The Exact area = 25.3333 . . . )
Tuesday November 14 ()
Math 1131
18 / 1
The general case.
To find the area from [a, b] between the graph of f (x) and the
x-axis when f (x) > 0, we divide the interval [a, b] into n equal
b−a
subintervals each of length ∆x =
. We will use a Right
n
Riemann Sum.
Tuesday November 14 ()
Math 1131
19 / 1
The general case.
To find the area from [a, b] between the graph of f (x) and the
x-axis when f (x) > 0, we divide the interval [a, b] into n equal
b−a
subintervals each of length ∆x =
. We will use a Right
n
Riemann Sum.
Grid points:
x0 = a, x1 = a + ∆x, x2 =
a + 2∆x, . . . + xn = a + n∆x = b
Area of i-th rectangle = f (xi )∆x
Tuesday November 14 ()
Math 1131
19 / 1
The general case.
To find the area from [a, b] between the graph of f (x) and the
x-axis when f (x) > 0, we divide the interval [a, b] into n equal
b−a
subintervals each of length ∆x =
. We will use a Right
n
Riemann Sum.
Grid points:
x0 = a, x1 = a + ∆x, x2 =
a + 2∆x, . . . + xn = a + n∆x = b
Area of i-th rectangle = f (xi )∆x
The Right Riemann Sum approximating the area is
A ≈ f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + · · · + f (xn )∆x
We need more efficient notation.
Tuesday November 14 ()
Math 1131
19 / 1
Sigma Notation
5
X
= 21 + 22 + 23 + 24 + 25
i=1
Tuesday November 14 ()
Math 1131
20 / 1
Sigma Notation
5
X
= 21 + 22 + 23 + 24 + 25
i=1
1
1 1
1 2
1 3
A ≈ f (0) + f ( ) + f ( ) + f ( )
2
2 2
2 2
2 2
Can be written A ≈
3
X
1
i=0
Tuesday November 14 ()
i
f( )
2 2
Math 1131
20 / 1
Sigma Notation
5
X
= 21 + 22 + 23 + 24 + 25
i=1
1
1 1
1 2
1 3
A ≈ f (0) + f ( ) + f ( ) + f ( )
2
2 2
2 2
2 2
Can be written A ≈
3
X
1
i=0
i
f( )
2 2
1
1 5
1
1 13
A ≈ (1) + ( ) + (2) + ( ) = 3.75
2
2 4
2
2 4
X 1
i
Can be written A ≈
35 (1 + ( )2 ) = 3.75
2
2
i=0
Tuesday November 14 ()
Math 1131
20 / 1
Using summation notation the area estimate by a Riemann Sum
can be written as
A≈
n
X
f (x̄i )∆x
i=1
(Here x̄i can be the left or right endpoint or the midpoint.)
Tuesday November 14 ()
Math 1131
21 / 1
Using summation notation the area estimate by a Riemann Sum
can be written as
A≈
n
X
f (x̄i )∆x
i=1
(Here x̄i can be the left or right endpoint or the midpoint.)
To get better estimations, we take n larger and larger. In fact, if
we let n go out to infinity we will get the exact area. In other
words,
Exact area = lim
n→∞
Tuesday November 14 ()
Math 1131
n
X
f (x̄i )∆x
i=1
21 / 1
What happens when we do this when f (x) < 0?
Example: f (x) = −4x2 is always negative on [0, 2]. We will use
the same ideas as before, using a Riemann Sum, to estimate the
area between the graph and the x-axis.
Tuesday November 14 ()
Math 1131
22 / 1
What happens when we do this when f (x) < 0?
Example: f (x) = −4x2 is always negative on [0, 2]. We will use
the same ideas as before, using a Riemann Sum, to estimate the
area between the graph and the x-axis.
Midpoint Riemann Sum.
We partition [0, 2] into 8 equal
intervals using grid points
0, 14 , 24 , 34 , 1, 45 , 64 , 74 , 2.
Interval length is
2−0
∆x =
= 0.25.
8
Tuesday November 14 ()
Math 1131
22 / 1
What happens when we do this when f (x) < 0?
Example: f (x) = −4x2 is always negative on [0, 2]. We will use
the same ideas as before, using a Riemann Sum, to estimate the
area between the graph and the x-axis.
Midpoint Riemann Sum.
We partition [0, 2] into 8 equal
intervals using grid points
0, 14 , 24 , 34 , 1, 45 , 64 , 74 , 2.
Interval length is
2−0
∆x =
= 0.25.
8
In the Midpoint Riemann Sum, for the first term we multiply the
1
1
width of the first rectangle, ∆x = 0.25, by f ( ), where is the
8
8
midpoint of the first interval.
Tuesday November 14 ()
Math 1131
22 / 1
Midpoint Riemann Sum.
First term = width of first rectangle
1
1
times f ( ), where is the midpoint of
8
8
the first interval.
1
1
31
f ( ) = ( )2 − 4 = −
(negative)
8
8
8
1
f ( )∆x = −(area of rectangle 1)
8
Tuesday November 14 ()
Math 1131
23 / 1
Midpoint Riemann Sum.
First term = width of first rectangle
1
1
times f ( ), where is the midpoint of
8
8
the first interval.
1
1
31
f ( ) = ( )2 − 4 = −
(negative)
8
8
8
1
f ( )∆x = −(area of rectangle 1)
8
The Midpoint Riemann Sum
1 1
1 5
1 7
1 9
1 11
f( ) + f( ) + f( ) + f( ) + f( )
4 8
4 8
4 8
4 8
4 8
1 13
1 15
+ f ( ) + f ( ) = −5.34375
4 8
4 8
n
X
Rmidpoint =
f (x̄i )∆x ≈ −(area between x-axis and curve)
i=1
Tuesday November 14 ()
Math 1131
23 / 1
Midpoint Riemann Sum.
When the rectangle is above the x-axis,
the summand f (x̄i )∆x equals the area
of the i-th rectangle.
When the rectangle is below the x-axis,
the summand f (x̄i )∆x equals −(area of
the i-th rectangle).
Tuesday November 14 ()
Math 1131
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Midpoint Riemann Sum.
When the rectangle is above the x-axis,
the summand f (x̄i )∆x equals the area
of the i-th rectangle.
When the rectangle is below the x-axis,
the summand f (x̄i )∆x equals −(area of
the i-th rectangle).
Rmidpoint =
n
X
f (xi )∆x ≈ − (area below x-axis)+(area above)
i=1
Tuesday November 14 ()
Math 1131
24 / 1
Midpoint Riemann Sum.
When the rectangle is above the x-axis,
the summand f (x̄i )∆x equals the area
of the i-th rectangle.
When the rectangle is below the x-axis,
the summand f (x̄i )∆x equals −(area of
the i-th rectangle).
Rmidpoint =
n
X
f (xi )∆x ≈ − (area below x-axis)+(area above)
i=1
Definition (Net Area)
The sum of the areas of the parts above the x-axis minus the
sum of the areas of the parts below the x-axis.
Rmidpoint =
n
X
f (xi )∆x ≈ Net Area
i=1
Tuesday November 14 ()
Math 1131
24 / 1
Clicker Question
1
3
5
7
9 11
Use a Right Riemann Sum with
5 subintervals to approximate
the net area between the graph
of f (x), 1 ≤ x ≤ 11 and the
x-axis.
y = f(x)
Each square is 1 unit
(a) 8
(b) 16
Tuesday November 14 ()
(c) 18
(d) 20
Math 1131
(e) 36
25 / 1
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11 − 1
=2
5
Grid points: 0, 3, 5, 7, 9, 11
∆x =
y = f(x)
Right Riemann Sum
A ≈ f (3)∆x + f (5)∆x + f (7)∆x + f (9)∆x + f (11)∆x
A ≈ 10 + 10 + 6 − 4 − 6 = 16
Answer = b
Tuesday November 14 ()
Math 1131
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