Exercise 1. Consider the function f: dom(f) CRR defined as f(x) = x²-6x+8|ex-2 (e) Determine, if it exists, a real number a < 2 such that fe² f(x) dx = 2 e² f(x) dx. We have to solve the equation in the unknown a, given by (E) L(x²-6 6x+8)e dx == Integrating by parts, we get the following primitives Therefore (x²-6x (x2 - 6x+8)e* dx. 6r+8)e dr=8-6 re-edr+ze-2fre de dx = 8e-6xe + 6e + x²e - 2xe + 2e+c e [x28x+16]+c=e(x-4)²+c, CER. why not e-z & why a negative sign I understand this Integration by parts twice f(x) dr=4² - (0-4)² I do not understand how the limits [f(x) dr = 4². → were substituted Therefore equation (E) is equivalent to 4e2 - ea (a-4)² = 4e2, or -ea (a - 4)² = 0, whose only solution is a = 4. We cannot accept this solution, since we gave the condition a < 2. Therefore the equation has no solutions. Exercise 1. Consider the function f: dom(f) CRR defined as f(x) = x²-6x+8|ex-2. (e) Determine, if it exists, a real number a < 2 such that fe² f(x) da = 2 e² f(x) dx. We have to solve the equation in the unknown a, given by (E) L(x²-6 6r+8) dr == (r²-6x+8)e* dr. 12 why not e-z & why a negative sign Integrating by parts, we get the following primitives Therefore (x²-6x 6x+8)e dx = 8e6 -6-d]+-2/dr e dx + x2e-2 xe dr 8e6xe+ 6e + x²e - 2xe + 2e+c e [x2-8x+16]+c=e(x-4)²+c, CER. = I understand this Integration by parts twice f(r) dr-42-(0-4)² I do not understand how the limits f(x) dr=42 → were substituted Therefore equation (E) is equivalent to 4e2 - ea (a - 4)² = 4e², or -ea (a - 4)² = 0, whose only solution is a = 4. We cannot accept this solution, since we gave the condition a < 2. Therefore the equation has no solutions.
Exercise 1. Consider the function f: dom(f) CRR defined as f(x) = x²-6x+8|ex-2 (e) Determine, if it exists, a real number a < 2 such that fe² f(x) dx = 2 e² f(x) dx. We have to solve the equation in the unknown a, given by (E) L(x²-6 6x+8)e dx == Integrating by parts, we get the following primitives Therefore (x²-6x (x2 - 6x+8)e* dx. 6r+8)e dr=8-6 re-edr+ze-2fre de dx = 8e-6xe + 6e + x²e - 2xe + 2e+c e [x28x+16]+c=e(x-4)²+c, CER. why not e-z & why a negative sign I understand this Integration by parts twice f(x) dr=4² - (0-4)² I do not understand how the limits [f(x) dr = 4². → were substituted Therefore equation (E) is equivalent to 4e2 - ea (a-4)² = 4e2, or -ea (a - 4)² = 0, whose only solution is a = 4. We cannot accept this solution, since we gave the condition a < 2. Therefore the equation has no solutions. Exercise 1. Consider the function f: dom(f) CRR defined as f(x) = x²-6x+8|ex-2. (e) Determine, if it exists, a real number a < 2 such that fe² f(x) da = 2 e² f(x) dx. We have to solve the equation in the unknown a, given by (E) L(x²-6 6r+8) dr == (r²-6x+8)e* dr. 12 why not e-z & why a negative sign Integrating by parts, we get the following primitives Therefore (x²-6x 6x+8)e dx = 8e6 -6-d]+-2/dr e dx + x2e-2 xe dr 8e6xe+ 6e + x²e - 2xe + 2e+c e [x2-8x+16]+c=e(x-4)²+c, CER. = I understand this Integration by parts twice f(r) dr-42-(0-4)² I do not understand how the limits f(x) dr=42 → were substituted Therefore equation (E) is equivalent to 4e2 - ea (a - 4)² = 4e², or -ea (a - 4)² = 0, whose only solution is a = 4. We cannot accept this solution, since we gave the condition a < 2. Therefore the equation has no solutions.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Exercise 1. Consider the function f: dom(f) CRR defined as f(x) = x²-6x+8|ex-2
(e) Determine, if it exists, a real number a < 2 such that fe² f(x) dx = 2 e² f(x) dx.
We have to solve the equation in the unknown a, given by
(E)
L(x²-6
6x+8)e dx ==
Integrating by parts, we get the following primitives
Therefore
(x²-6x
(x2
- 6x+8)e* dx.
6r+8)e dr=8-6 re-edr+ze-2fre de
dx
= 8e-6xe + 6e + x²e - 2xe + 2e+c
e [x28x+16]+c=e(x-4)²+c, CER.
why not e-z
& why a negative sign
I understand this
Integration by parts twice
f(x) dr=4² - (0-4)² I do not understand how the limits
[f(x) dr = 4².
→
were substituted
Therefore equation (E) is equivalent to 4e2 - ea (a-4)² = 4e2, or -ea (a - 4)² = 0, whose
only solution is a = 4. We cannot accept this solution, since we gave the condition a < 2.
Therefore the equation has no solutions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2Fda75133b-aea2-4e0e-ba3b-c576f7ce1708%2Fq7eqdm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercise 1. Consider the function f: dom(f) CRR defined as f(x) = x²-6x+8|ex-2
(e) Determine, if it exists, a real number a < 2 such that fe² f(x) dx = 2 e² f(x) dx.
We have to solve the equation in the unknown a, given by
(E)
L(x²-6
6x+8)e dx ==
Integrating by parts, we get the following primitives
Therefore
(x²-6x
(x2
- 6x+8)e* dx.
6r+8)e dr=8-6 re-edr+ze-2fre de
dx
= 8e-6xe + 6e + x²e - 2xe + 2e+c
e [x28x+16]+c=e(x-4)²+c, CER.
why not e-z
& why a negative sign
I understand this
Integration by parts twice
f(x) dr=4² - (0-4)² I do not understand how the limits
[f(x) dr = 4².
→
were substituted
Therefore equation (E) is equivalent to 4e2 - ea (a-4)² = 4e2, or -ea (a - 4)² = 0, whose
only solution is a = 4. We cannot accept this solution, since we gave the condition a < 2.
Therefore the equation has no solutions.
![Exercise 1. Consider the function f: dom(f) CRR defined as f(x) = x²-6x+8|ex-2.
(e) Determine, if it exists, a real number a < 2 such that fe² f(x) da = 2 e² f(x) dx.
We have to solve the equation in the unknown a, given by
(E)
L(x²-6
6r+8) dr == (r²-6x+8)e* dr.
12
why not e-z
& why a negative sign
Integrating by parts, we get the following primitives
Therefore
(x²-6x
6x+8)e dx = 8e6
-6-d]+-2/dr
e dx + x2e-2 xe dr
8e6xe+ 6e + x²e - 2xe + 2e+c
e [x2-8x+16]+c=e(x-4)²+c, CER.
=
I understand this
Integration by parts twice
f(r) dr-42-(0-4)² I do not understand how the limits
f(x) dr=42
→
were substituted
Therefore equation (E) is equivalent to 4e2 - ea (a - 4)² = 4e², or -ea (a - 4)² = 0, whose
only solution is a = 4. We cannot accept this solution, since we gave the condition a < 2.
Therefore the equation has no solutions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2Fda75133b-aea2-4e0e-ba3b-c576f7ce1708%2Fg2dgqon_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercise 1. Consider the function f: dom(f) CRR defined as f(x) = x²-6x+8|ex-2.
(e) Determine, if it exists, a real number a < 2 such that fe² f(x) da = 2 e² f(x) dx.
We have to solve the equation in the unknown a, given by
(E)
L(x²-6
6r+8) dr == (r²-6x+8)e* dr.
12
why not e-z
& why a negative sign
Integrating by parts, we get the following primitives
Therefore
(x²-6x
6x+8)e dx = 8e6
-6-d]+-2/dr
e dx + x2e-2 xe dr
8e6xe+ 6e + x²e - 2xe + 2e+c
e [x2-8x+16]+c=e(x-4)²+c, CER.
=
I understand this
Integration by parts twice
f(r) dr-42-(0-4)² I do not understand how the limits
f(x) dr=42
→
were substituted
Therefore equation (E) is equivalent to 4e2 - ea (a - 4)² = 4e², or -ea (a - 4)² = 0, whose
only solution is a = 4. We cannot accept this solution, since we gave the condition a < 2.
Therefore the equation has no solutions.
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