The antiderivative of f₁(y) is Hence F(y) = 2√ dy = 2√
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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answer for step 6 please
![Step 5
Assume g₁(y) = 36- y². Therefore, g₁'(y)
Define f₁(y)
=
1
√y
Step 6
Therefore, f₁(g₁(y)) =
Multiply and divide the given integrand by -2.
-2y
2
1 ² = ² = √36 -
The antiderivative of
Hence F(y) = 2√
dy
=
2
=[/²
=
= -2
-1
클[Fg1(y)] 2
0
whereF(g₁(y)) is the antiderivative off₁(g₁(y)).
36 - Y
(36-12)-1/回
f₁ (v) is √ √ √ √ dy = 2√
1
-2 y.
2
[² f₁(9₁1 (1) ₁1 (1) dy
y dy]
-2 y dy](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32c5117c-6cba-4a4e-a2e4-0e889cc62fe2%2F582cb047-3da6-44d8-91f4-0b0e7bada51c%2F5cdoinj_processed.png&w=3840&q=75)
Transcribed Image Text:Step 5
Assume g₁(y) = 36- y². Therefore, g₁'(y)
Define f₁(y)
=
1
√y
Step 6
Therefore, f₁(g₁(y)) =
Multiply and divide the given integrand by -2.
-2y
2
1 ² = ² = √36 -
The antiderivative of
Hence F(y) = 2√
dy
=
2
=[/²
=
= -2
-1
클[Fg1(y)] 2
0
whereF(g₁(y)) is the antiderivative off₁(g₁(y)).
36 - Y
(36-12)-1/回
f₁ (v) is √ √ √ √ dy = 2√
1
-2 y.
2
[² f₁(9₁1 (1) ₁1 (1) dy
y dy]
-2 y dy
Transcribed Image Text:Consider the following equations.
f(y)
g(y)
=
= 0
y = 2
Sketch the region bounded by the graphs of the functions. Find the area of the region.
у
36 - y²
Step 1
Write the original algebraic functions:
f(y)
and g(y)
=
=
0
36
0
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