Post on 11-Dec-2020
5.2 Sigma Notation and Limits of FiniteSums
Trig Example
Example
Estimate the area under the curve f (x) = sin x between x = 0and x = π with n = 3.
What do we need to do first? find ∆x
∆x = π−03 = π
3
Our intervals, therefore, will be [0, π3 ], [
π3 , 2π
3 ], and [ 2π3 , π].
Trig Example
Example
Estimate the area under the curve f (x) = sin x between x = 0and x = π with n = 3.
What do we need to do first?
find ∆x
∆x = π−03 = π
3
Our intervals, therefore, will be [0, π3 ], [
π3 , 2π
3 ], and [ 2π3 , π].
Trig Example
Example
Estimate the area under the curve f (x) = sin x between x = 0and x = π with n = 3.
What do we need to do first? find ∆x
∆x = π−03 = π
3
Our intervals, therefore, will be [0, π3 ], [
π3 , 2π
3 ], and [ 2π3 , π].
Trig Example
Example
Estimate the area under the curve f (x) = sin x between x = 0and x = π with n = 3.
What do we need to do first? find ∆x
∆x = π−03 = π
3
Our intervals, therefore, will be [0, π3 ], [
π3 , 2π
3 ], and [ 2π3 , π].
Trig Example
Example
Estimate the area under the curve f (x) = sin x between x = 0and x = π with n = 3.
What do we need to do first? find ∆x
∆x = π−03 = π
3
Our intervals, therefore, will be [0, π3 ], [
π3 , 2π
3 ], and [ 2π3 , π].
Trig Example
LHS:
π
3
[sin(0) + sin
(π
3
)+ sin
(2π
3
)]
=π
3
[0 +
√3
2+
√3
2
]=
π√
33
RHS:
π
3
[sin(π
3
)+ sin
(2π
3+ sin(π)
)]=
π
3
[√3
2+
√3
2+ 0
]=
π√
33
Trig Example
LHS:
π
3
[sin(0) + sin
(π
3
)+ sin
(2π
3
)]=
π
3
[0 +
√3
2+
√3
2
]=
π√
33
RHS:
π
3
[sin(π
3
)+ sin
(2π
3+ sin(π)
)]
=π
3
[√3
2+
√3
2+ 0
]=
π√
33
Trig Example
LHS:
π
3
[sin(0) + sin
(π
3
)+ sin
(2π
3
)]=
π
3
[0 +
√3
2+
√3
2
]=
π√
33
RHS:
π
3
[sin(π
3
)+ sin
(2π
3+ sin(π)
)]=
π
3
[√3
2+
√3
2+ 0
]=
π√
33
Trig Example
So, since the LHS and RHS sum are the same, we are prettyconfident that this is the exact area.
But, we would be wrong ... why?
This is because of the small n. Even though the estimates areequal, that is not enough to be guaranteed that we have theexact value.
∫ π
0sin(x)dx = −cos(x)
∣∣∣∣π0= −(−1− 1) = 2
Trig Example
So, since the LHS and RHS sum are the same, we are prettyconfident that this is the exact area.
But, we would be wrong ... why?
This is because of the small n. Even though the estimates areequal, that is not enough to be guaranteed that we have theexact value.
∫ π
0sin(x)dx = −cos(x)
∣∣∣∣π0= −(−1− 1) = 2
Trig Example
So, since the LHS and RHS sum are the same, we are prettyconfident that this is the exact area.
But, we would be wrong ... why?
This is because of the small n. Even though the estimates areequal, that is not enough to be guaranteed that we have theexact value.
∫ π
0sin(x)dx = −cos(x)
∣∣∣∣π0= −(−1− 1) = 2
Trig Example
So, since the LHS and RHS sum are the same, we are prettyconfident that this is the exact area.
But, we would be wrong ... why?
This is because of the small n. Even though the estimates areequal, that is not enough to be guaranteed that we have theexact value.
∫ π
0sin(x)dx = −cos(x)
∣∣∣∣π0= −(−1− 1) = 2
Sigma Notation
We have a better way to express sums then to write everythingout. This is called Sigma notation.
n
∑k=1
ak
where ak is the general term and k is the index.
So, in the last example,
LHS =2
∑k=0
sin(
kπ
3
)RHS =
3
∑k=1
sin(
kπ
3
)
Sigma Notation
We have a better way to express sums then to write everythingout. This is called Sigma notation.
n
∑k=1
ak
where ak is the general term and k is the index.
So, in the last example,
LHS =
2
∑k=0
sin(
kπ
3
)RHS =
3
∑k=1
sin(
kπ
3
)
Sigma Notation
We have a better way to express sums then to write everythingout. This is called Sigma notation.
n
∑k=1
ak
where ak is the general term and k is the index.
So, in the last example,
LHS =2
∑k=0
sin(
kπ
3
)
RHS =3
∑k=1
sin(
kπ
3
)
Sigma Notation
We have a better way to express sums then to write everythingout. This is called Sigma notation.
n
∑k=1
ak
where ak is the general term and k is the index.
So, in the last example,
LHS =2
∑k=0
sin(
kπ
3
)RHS =
3
∑k=1
sin(
kπ
3
)
Sigma Notation
We have a better way to express sums then to write everythingout. This is called Sigma notation.
n
∑k=1
ak
where ak is the general term and k is the index.
So, in the last example,
LHS =2
∑k=0
sin(
kπ
3
)RHS =
3
∑k=1
sin(
kπ
3
)
Sigma Notation Examples
Example
List the terms of4
∑k=1
k2
4
∑k=1
k2 = 1 + 4 + 9 + 16
Sigma Notation Examples
Example
List the terms of4
∑k=1
k2
4
∑k=1
k2 = 1 + 4 + 9 + 16
Sigma Notation Examples
Example
List the terms of3
∑k=1
1k + 2
3
∑k=1
1k + 2
=13+
14+
15
Sigma Notation Examples
Example
List the terms of3
∑k=1
1k + 2
3
∑k=1
1k + 2
=13+
14+
15
Rules for Summations
1 Sums∑(an + bn) = ∑ an + ∑ bn
2 Differences∑(an − bn) = ∑ an −∑ bn
3 Constant multiples
∑ kan = k ∑ an
for k 6= 04 Constant value
n
∑k=1
c = cn
Rules for Summations
1 Sums∑(an + bn) = ∑ an + ∑ bn
2 Differences∑(an − bn) = ∑ an −∑ bn
3 Constant multiples
∑ kan = k ∑ an
for k 6= 04 Constant value
n
∑k=1
c = cn
Rules for Summations
1 Sums∑(an + bn) = ∑ an + ∑ bn
2 Differences∑(an − bn) = ∑ an −∑ bn
3 Constant multiples
∑ kan = k ∑ an
for k 6= 0
4 Constant valuen
∑k=1
c = cn
Rules for Summations
1 Sums∑(an + bn) = ∑ an + ∑ bn
2 Differences∑(an − bn) = ∑ an −∑ bn
3 Constant multiples
∑ kan = k ∑ an
for k 6= 04 Constant value
n
∑k=1
c = cn
Important Summations
n
∑k=1
k =n(n + 1)
2
n
∑k=1
k2 =n(n + 1)(2n + 1)
6
n
∑k=1
k3 =
[n(n + 1)
2
]2
Important Summations
n
∑k=1
k =n(n + 1)
2
n
∑k=1
k2 =n(n + 1)(2n + 1)
6
n
∑k=1
k3 =
[n(n + 1)
2
]2
Important Summations
n
∑k=1
k =n(n + 1)
2
n
∑k=1
k2 =n(n + 1)(2n + 1)
6
n
∑k=1
k3 =
[n(n + 1)
2
]2
Using these rules and summations
Example
10
∑k=1
(2k2 − k
)
= 210
∑k=1
k2 +10
∑k=1
k
= 2(
10(11)(21)6
)− 10(11)
2= 715
Using these rules and summations
Example
10
∑k=1
(2k2 − k
)= 2
10
∑k=1
k2 +10
∑k=1
k
= 2(
10(11)(21)6
)− 10(11)
2= 715
Using these rules and summations
Example
10
∑k=1
(2k2 − k
)= 2
10
∑k=1
k2 +10
∑k=1
k
= 2(
10(11)(21)6
)− 10(11)
2
= 715
Using these rules and summations
Example
10
∑k=1
(2k2 − k
)= 2
10
∑k=1
k2 +10
∑k=1
k
= 2(
10(11)(21)6
)− 10(11)
2= 715
Using these rules and summations
Example
20
∑k=4
k2
=20
∑k=1
k2 −3
∑k=1
k2
=20(21)(41)
6− 3(4)(7)
6= 2870− 14
= 2856
Using these rules and summations
Example
20
∑k=4
k2
=20
∑k=1
k2 −3
∑k=1
k2
=20(21)(41)
6− 3(4)(7)
6= 2870− 14
= 2856
Using these rules and summations
Example
20
∑k=4
k2
=20
∑k=1
k2 −3
∑k=1
k2
=20(21)(41)
6− 3(4)(7)
6
= 2870− 14= 2856
Using these rules and summations
Example
20
∑k=4
k2
=20
∑k=1
k2 −3
∑k=1
k2
=20(21)(41)
6− 3(4)(7)
6= 2870− 14
= 2856
Using these rules and summations
Example
7
∑k=3
(3k3 + 2k
)
7
∑k=1
(3k3 + 2k
)−
2
∑k=1
(3k3 + 2k
)3
7
∑k=1
k3 + 27
∑k=1
k− 32
∑k=1
k3 − 22
∑k=1
k
= 3
[(7(8)
2
)2]+ 2
(7(8)
2
)− 3
[(2(3)
2
)2]− 2
(2(3)
2
)= 2375
Using these rules and summations
Example
7
∑k=3
(3k3 + 2k
)7
∑k=1
(3k3 + 2k
)−
2
∑k=1
(3k3 + 2k
)
37
∑k=1
k3 + 27
∑k=1
k− 32
∑k=1
k3 − 22
∑k=1
k
= 3
[(7(8)
2
)2]+ 2
(7(8)
2
)− 3
[(2(3)
2
)2]− 2
(2(3)
2
)= 2375
Using these rules and summations
Example
7
∑k=3
(3k3 + 2k
)7
∑k=1
(3k3 + 2k
)−
2
∑k=1
(3k3 + 2k
)3
7
∑k=1
k3 + 27
∑k=1
k− 32
∑k=1
k3 − 22
∑k=1
k
= 3
[(7(8)
2
)2]+ 2
(7(8)
2
)− 3
[(2(3)
2
)2]− 2
(2(3)
2
)= 2375
Using these rules and summations
Example
7
∑k=3
(3k3 + 2k
)7
∑k=1
(3k3 + 2k
)−
2
∑k=1
(3k3 + 2k
)3
7
∑k=1
k3 + 27
∑k=1
k− 32
∑k=1
k3 − 22
∑k=1
k
= 3
[(7(8)
2
)2]+ 2
(7(8)
2
)− 3
[(2(3)
2
)2]− 2
(2(3)
2
)
= 2375
Using these rules and summations
Example
7
∑k=3
(3k3 + 2k
)7
∑k=1
(3k3 + 2k
)−
2
∑k=1
(3k3 + 2k
)3
7
∑k=1
k3 + 27
∑k=1
k− 32
∑k=1
k3 − 22
∑k=1
k
= 3
[(7(8)
2
)2]+ 2
(7(8)
2
)− 3
[(2(3)
2
)2]− 2
(2(3)
2
)= 2375
More Properties
Theorem
limn→∞
1n
n
∑k=1
1 = 1
limn→∞
1n2
n
∑k=1
k =12
limn→∞
1n3
n
∑k=1
k2 =13
limn→∞
1n4
n
∑k=1
k3 =14
More Properties
Theorem
limn→∞
1n
n
∑k=1
1 = 1
limn→∞
1n2
n
∑k=1
k =12
limn→∞
1n3
n
∑k=1
k2 =13
limn→∞
1n4
n
∑k=1
k3 =14
More Properties
Theorem
limn→∞
1n
n
∑k=1
1 = 1
limn→∞
1n2
n
∑k=1
k =12
limn→∞
1n3
n
∑k=1
k2 =13
limn→∞
1n4
n
∑k=1
k3 =14
More Properties
Theorem
limn→∞
1n
n
∑k=1
1 = 1
limn→∞
1n2
n
∑k=1
k =12
limn→∞
1n3
n
∑k=1
k2 =13
limn→∞
1n4
n
∑k=1
k3 =14
Riemann Sums
To find the approximate area under a curve f (x), we break theregion into rectangular regions (since we can easily find thearea of a rectangle) and we add their areas together.
x0 x1 x2 x3 x4 x5
Riemann Sums
We partition the interval [a, b] into n rectangles with endpoints
a = x0, x1, . . . , xn−1, xn = b
This gives the subintervals
∆x1 = x1 − x0, ∆x2 = x2 − x1, . . . , ∆xn = xn − xn−1
Note: Nothing says we have to partition the interval intosubintervals of the same size. When we do, ∆xi = ∆x = b−a
nand we say we have a regular partition.
Riemann Sums
We partition the interval [a, b] into n rectangles with endpoints
a = x0, x1, . . . , xn−1, xn = b
This gives the subintervals
∆x1 = x1 − x0, ∆x2 = x2 − x1, . . . , ∆xn = xn − xn−1
Note: Nothing says we have to partition the interval intosubintervals of the same size. When we do, ∆xi = ∆x = b−a
nand we say we have a regular partition.
Riemann Sums
We partition the interval [a, b] into n rectangles with endpoints
a = x0, x1, . . . , xn−1, xn = b
This gives the subintervals
∆x1 = x1 − x0, ∆x2 = x2 − x1, . . . , ∆xn = xn − xn−1
Note: Nothing says we have to partition the interval intosubintervals of the same size. When we do, ∆xi = ∆x = b−a
nand we say we have a regular partition.
Riemann Sums
If we did use a consistant ∆x, we would have something like
x0 x1 x2 x3 x4 x5 x6 x7 x8
Riemann Sums
In general, we have
RHS = f (x1)∆x + · · ·+ f (xn)∆x =n
∑k=1
f (xk)∆x
LHS = f (x0)∆x + · · ·+ f (xn−1)∆x =n−1
∑k=0
f (xk)∆x
where each [xk−1, xk] is the kth subinterval. We are dividinga ≤ x ≤ b into n subintervals, each of size 1
n .
Riemann Sums
If we have an infinite number of subintervals, then we get theoverestimates and the underestimates to be the same, we getthe over and under estimates to become the same.
limn→∞
n−1
∑k=0
f (xk)∆x = limn→∞
n
∑k=1
f (xk)∆x =∫ b
af (x)dx
We call a, b the limits of integration and we call f (x) theintegrand.
Riemann Sums
If we have an infinite number of subintervals, then we get theoverestimates and the underestimates to be the same, we getthe over and under estimates to become the same.
limn→∞
n−1
∑k=0
f (xk)∆x = limn→∞
n
∑k=1
f (xk)∆x =∫ b
af (x)dx
We call a, b the limits of integration and we call f (x) theintegrand.
Formal Definition of the Integral
Let f (x) be defined on [a, b]. The number I is the definiteintegral of f over [a, b] and I is the limit of the Riemann sumprovided:
Given ε > 0, there exists δ > 0 such that for all partitionsp = {x0, x1, . . . , xn} of [a, b] with ||ρ|| < δ and any choiceof ck in {xk−1, xk}, we have∣∣∣∣∣ n
∑k=1
f (xk)∆xk − I
∣∣∣∣∣ < ε
||ρ|| is the norm of the partition, defined as the largest of thesubinterval lengths.
Formal Definition of the Integral
Let f (x) be defined on [a, b]. The number I is the definiteintegral of f over [a, b] and I is the limit of the Riemann sumprovided:
Given ε > 0, there exists δ > 0 such that for all partitionsp = {x0, x1, . . . , xn} of [a, b] with ||ρ|| < δ and any choiceof ck in {xk−1, xk}, we have∣∣∣∣∣ n
∑k=1
f (xk)∆xk − I
∣∣∣∣∣ < ε
||ρ|| is the norm of the partition, defined as the largest of thesubinterval lengths.
Riemann Sums
That is,
lim||ρ||→∞
n
∑k=1
f (xk)∆xk
is the same as ∫ b
af (x)dx
Sums and Integrals
Example
For f (x) = cos(x2) on [π, 3π], the area bound by the curve andthe x-axis could be expressed as
limn→∞
n
∑k=1
cos(xk)2∆xk
or ∫ 3π
πcos(x2)dx
Using n Subdivisions to Find a Sum
Example
Find the RHS for f (x) = x2 − 2 by dividing [0, 1] into n equalsubintervals. Then, take the limit as n→ ∞ to calculate the areabetween the x-axis and the curve.
Since we want n equal subintervals, ∆x = 1n .
Our partition would be [0, 1n ], [
1n , 2
n ], . . . , [n−1n , 1]
Using n Subdivisions to Find a Sum
Example
Find the RHS for f (x) = x2 − 2 by dividing [0, 1] into n equalsubintervals. Then, take the limit as n→ ∞ to calculate the areabetween the x-axis and the curve.
Since we want n equal subintervals, ∆x = 1n .
Our partition would be [0, 1n ], [
1n , 2
n ], . . . , [n−1n , 1]
Using n Subdivisions to Find a Sum
Example
Find the RHS for f (x) = x2 − 2 by dividing [0, 1] into n equalsubintervals. Then, take the limit as n→ ∞ to calculate the areabetween the x-axis and the curve.
Since we want n equal subintervals, ∆x = 1n .
Our partition would be [0, 1n ], [
1n , 2
n ], . . . , [n−1n , 1]
Using n Subdivisions to Find a Sum
Then we get the following:
RHS = f(
1n
)1n+ f
(2n
)1n+ · · ·+ f
(nn
) 1n
=n
∑k−1
f(
kn
)1n
=n
∑k−1
((kn
)2
− 2
)1n
=1n3
n
∑k=1
k2 − 1n
n
∑k=1
2
= limn→∞
n(n + 1)(2n + 1)6n3 − 2
= limn→∞
2n3 + 3n2 + n6n3 − 2
Using n Subdivisions to Find a Sum
Then we get the following:
RHS = f(
1n
)1n+ f
(2n
)1n+ · · ·+ f
(nn
) 1n
=n
∑k−1
f(
kn
)1n
=n
∑k−1
((kn
)2
− 2
)1n
=1n3
n
∑k=1
k2 − 1n
n
∑k=1
2
= limn→∞
n(n + 1)(2n + 1)6n3 − 2
= limn→∞
2n3 + 3n2 + n6n3 − 2
Using n Subdivisions to Find a Sum
Then we get the following:
RHS = f(
1n
)1n+ f
(2n
)1n+ · · ·+ f
(nn
) 1n
=n
∑k−1
f(
kn
)1n
=n
∑k−1
((kn
)2
− 2
)1n
=1n3
n
∑k=1
k2 − 1n
n
∑k=1
2
= limn→∞
n(n + 1)(2n + 1)6n3 − 2
= limn→∞
2n3 + 3n2 + n6n3 − 2
Using n Subdivisions to Find a Sum
Then we get the following:
RHS = f(
1n
)1n+ f
(2n
)1n+ · · ·+ f
(nn
) 1n
=n
∑k−1
f(
kn
)1n
=n
∑k−1
((kn
)2
− 2
)1n
=1n3
n
∑k=1
k2 − 1n
n
∑k=1
2
= limn→∞
n(n + 1)(2n + 1)6n3 − 2
= limn→∞
2n3 + 3n2 + n6n3 − 2
Using n Subdivisions to Find a Sum
Then we get the following:
RHS = f(
1n
)1n+ f
(2n
)1n+ · · ·+ f
(nn
) 1n
=n
∑k−1
f(
kn
)1n
=n
∑k−1
((kn
)2
− 2
)1n
=1n3
n
∑k=1
k2 − 1n
n
∑k=1
2
= limn→∞
n(n + 1)(2n + 1)6n3 − 2
= limn→∞
2n3 + 3n2 + n6n3 − 2
Using n Subdivisions to Find a Sum
Then we get the following:
RHS = f(
1n
)1n+ f
(2n
)1n+ · · ·+ f
(nn
) 1n
=n
∑k−1
f(
kn
)1n
=n
∑k−1
((kn
)2
− 2
)1n
=1n3
n
∑k=1
k2 − 1n
n
∑k=1
2
= limn→∞
n(n + 1)(2n + 1)6n3 − 2
= limn→∞
2n3 + 3n2 + n6n3 − 2
Using n Subdivisions to Find a Sum
When we total this, we get RHS = 13 − 2 = − 5
3 .
Notice that the estimate of the sum is negative. What does thismean?
It means that the region bound by the curve and the x-axis isbelow the x-axis. Also note that we were asked to find the RHSand not to find the area of the region ...
Using n Subdivisions to Find a Sum
When we total this, we get RHS = 13 − 2 = − 5
3 .
Notice that the estimate of the sum is negative. What does thismean?
It means that the region bound by the curve and the x-axis isbelow the x-axis. Also note that we were asked to find the RHSand not to find the area of the region ...
Using n Subdivisions to Find a Sum
When we total this, we get RHS = 13 − 2 = − 5
3 .
Notice that the estimate of the sum is negative. What does thismean?
It means that the region bound by the curve and the x-axis isbelow the x-axis. Also note that we were asked to find the RHSand not to find the area of the region ...
Example
Write the Riemann sum for f (x) = 2x over the interval from[1, 4] with n equal subdivisions.
n
∑k=1
f(
1 +3kn
)3n
=n
∑k=1
2(
1 +3kn
)3n
Example
Write the Riemann sum for f (x) = 2x over the interval from[1, 4] with n equal subdivisions.
n
∑k=1
f(
1 +3kn
)3n
=n
∑k=1
2(
1 +3kn
)3n
Example
Write the Riemann sum for f (x) = 2x over the interval from[1, 4] with n equal subdivisions.
n
∑k=1
f(
1 +3kn
)3n
=n
∑k=1
2(
1 +3kn
)3n
Using n Subdivisions to Find a Sum
If we wanted to continue so that we could solve, we wouldhave
limn→∞
3n
n
∑k=1
(2 +
6kn
)
= limn→∞
[3n
n
∑k=1
2 +3n
n
∑k=1
(6kn
)]
= 3(2) + limn→∞
18n2
n
∑k=1
k
= 3(2) + limn→∞
18n2
(n(n + 1)
2
)
= 6 + 18(
12
)= 15
Using n Subdivisions to Find a Sum
If we wanted to continue so that we could solve, we wouldhave
limn→∞
3n
n
∑k=1
(2 +
6kn
)
= limn→∞
[3n
n
∑k=1
2 +3n
n
∑k=1
(6kn
)]
= 3(2) + limn→∞
18n2
n
∑k=1
k
= 3(2) + limn→∞
18n2
(n(n + 1)
2
)
= 6 + 18(
12
)= 15
Using n Subdivisions to Find a Sum
If we wanted to continue so that we could solve, we wouldhave
limn→∞
3n
n
∑k=1
(2 +
6kn
)
= limn→∞
[3n
n
∑k=1
2 +3n
n
∑k=1
(6kn
)]
= 3(2) + limn→∞
18n2
n
∑k=1
k
= 3(2) + limn→∞
18n2
(n(n + 1)
2
)
= 6 + 18(
12
)= 15
Using n Subdivisions to Find a Sum
If we wanted to continue so that we could solve, we wouldhave
limn→∞
3n
n
∑k=1
(2 +
6kn
)
= limn→∞
[3n
n
∑k=1
2 +3n
n
∑k=1
(6kn
)]
= 3(2) + limn→∞
18n2
n
∑k=1
k
= 3(2) + limn→∞
18n2
(n(n + 1)
2
)
= 6 + 18(
12
)= 15
Using n Subdivisions to Find a Sum
If we wanted to continue so that we could solve, we wouldhave
limn→∞
3n
n
∑k=1
(2 +
6kn
)
= limn→∞
[3n
n
∑k=1
2 +3n
n
∑k=1
(6kn
)]
= 3(2) + limn→∞
18n2
n
∑k=1
k
= 3(2) + limn→∞
18n2
(n(n + 1)
2
)
= 6 + 18(
12
)= 15
Solving with Geometry
Example ∫ 1
−1
√1− x2 dx
We can solve this, even though we don’t know what theantiderivative is, by using geometry. What does this integrandrepresent?
Solving with Geometry
Example ∫ 1
−1
√1− x2 dx
We can solve this, even though we don’t know what theantiderivative is, by using geometry. What does this integrandrepresent?
Solving with Geometry
Example ∫ 1
−1
√1− x2 dx
We can solve this, even though we don’t know what theantiderivative is, by using geometry. What does this integrandrepresent?
Solving with Geometry
We need to note that this is the top half of the unit circle.
What is the area of a circle? A(r) = πr2
The circle has a radius of 1 and we only need half the circle,so ... ∫ 1
−1
√1− x2 dx =
π
2
Solving with Geometry
We need to note that this is the top half of the unit circle.
What is the area of a circle?
A(r) = πr2
The circle has a radius of 1 and we only need half the circle,so ... ∫ 1
−1
√1− x2 dx =
π
2
Solving with Geometry
We need to note that this is the top half of the unit circle.
What is the area of a circle? A(r) = πr2
The circle has a radius of 1 and we only need half the circle,so ... ∫ 1
−1
√1− x2 dx =
π
2
Solving with Geometry
We need to note that this is the top half of the unit circle.
What is the area of a circle? A(r) = πr2
The circle has a radius of 1 and we only need half the circle,so ...
∫ 1
−1
√1− x2 dx =
π
2
Solving with Geometry
We need to note that this is the top half of the unit circle.
What is the area of a circle? A(r) = πr2
The circle has a radius of 1 and we only need half the circle,so ... ∫ 1
−1
√1− x2 dx =
π
2