a) The blue shaded region A₁ of the unit circle is a sector with central angle y. The tan shaded region A₂ is a right triangle with base measuring x and acute angles a, ß. The elementary geometric area formulas give 1 A₁ = r2 and b) Since B,y are [ CIRCLE ONE: vertical / alternate / corresponding] angles, we conclude that B = 2 c) We know that so that we can write sin ß = y = y(x) = A₂ = 2 h. d) The vertex of the triangle corresponding with ß is B(x, h). Use the fact that B lies on the unit circle centered at the origin to write h = h(x) = e) With the above we are able to write the shaded areas as functions of x: and A₁ = A₁(x) = A₂ = A₂(x) = f) Write the sum of A₁ and A₂ as a single definite integral ·b(x) f(t) dt. A₁ + A₂ =
a) The blue shaded region A₁ of the unit circle is a sector with central angle y. The tan shaded region A₂ is a right triangle with base measuring x and acute angles a, ß. The elementary geometric area formulas give 1 A₁ = r2 and b) Since B,y are [ CIRCLE ONE: vertical / alternate / corresponding] angles, we conclude that B = 2 c) We know that so that we can write sin ß = y = y(x) = A₂ = 2 h. d) The vertex of the triangle corresponding with ß is B(x, h). Use the fact that B lies on the unit circle centered at the origin to write h = h(x) = e) With the above we are able to write the shaded areas as functions of x: and A₁ = A₁(x) = A₂ = A₂(x) = f) Write the sum of A₁ and A₂ as a single definite integral ·b(x) f(t) dt. A₁ + A₂ =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
The second picture is the graph and the x changest from 0 to 1
Transcribed Image Text:~
=
B
(₁,√1-2²)
0 SIS X
7
X = 0.5
ⒸVI-X²³XSYS√1-2² (0sxsx)
√1-X² x{0≤x≤X}
X
0
0≤x≤
(x,0)
✓Label: X
X
X
1
X
X
X
X
-0.2
1.2+
0.8-
0.6+
0.4+
0.2+
0
-0.2+
-0.4+
0.2
0.4 X 0.6
0.8
1
1.2
1.4
1.6
1.8
2
![A₁
Va
a) The blue shaded region A₁ of the unit circle is a sector with central
angle y. The tan shaded region A₂ is a right triangle with base
measuring x and acute angles a, ß. The elementary geometric
area formulas give
=
2
√1-x²
c) We know that
1
so that we can write
X
r2 and A₂ =
b) Since B,y are [ CIRCLE ONE: vertical / alternate / corresponding]
angles, we conclude that
B =
sin ß =
=1/1²
y = y(x) =
h.
d) The vertex of the triangle corresponding with ß is B(x, h). Use the
fact that B lies on the unit circle centered at the origin to write
h = h(x) =
e) With the above we are able to write the shaded areas as functions
of x:
and
A₁ = A₁(x) =
A₂ = A₂(x) =
f) Write the sum of A₁ and A₂ as a single definite integral
pb(x)
S
f(t) dt.
A₁ + A₂ =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa1c93266-1293-41fb-a3ab-5f626105c986%2Ff1099be4-f63b-4ea2-8d47-f6cde0d0d577%2Fz1l3ys4k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A₁
Va
a) The blue shaded region A₁ of the unit circle is a sector with central
angle y. The tan shaded region A₂ is a right triangle with base
measuring x and acute angles a, ß. The elementary geometric
area formulas give
=
2
√1-x²
c) We know that
1
so that we can write
X
r2 and A₂ =
b) Since B,y are [ CIRCLE ONE: vertical / alternate / corresponding]
angles, we conclude that
B =
sin ß =
=1/1²
y = y(x) =
h.
d) The vertex of the triangle corresponding with ß is B(x, h). Use the
fact that B lies on the unit circle centered at the origin to write
h = h(x) =
e) With the above we are able to write the shaded areas as functions
of x:
and
A₁ = A₁(x) =
A₂ = A₂(x) =
f) Write the sum of A₁ and A₂ as a single definite integral
pb(x)
S
f(t) dt.
A₁ + A₂ =
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