Definite integrals : how do we approach in solving a problem
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While practicing definite integrals I came across a question and now I am stuck
Question:
let f be a continous satisfying $f(x+y) = f(x) + f(y) + f(x)cdot f(y)$ for all real $x$ and $y$ and $f'(0)= -1$.
Find the value of $int_0^1 f(x) , dx$.
I tried to solve by find the function but no success and i am confused.
Please tell me how to solve these kind of questions.
definite-integrals functional-calculus
edited Nov 14 at 13:31
callculus
17.6k31427
asked Nov 14 at 13:20
Shantanu Kaushik
223
add a comment |
up vote
0
down vote
favorite
While practicing definite integrals I came across a question and now I am stuck
Question:
let f be a continous satisfying $f(x+y) = f(x) + f(y) + f(x)cdot f(y)$ for all real $x$ and $y$ and $f'(0)= -1$.
Find the value of $int_0^1 f(x) , dx$.
I tried to solve by find the function but no success and i am confused.
Please tell me how to solve these kind of questions.
definite-integrals functional-calculus
edited Nov 14 at 13:31
callculus
17.6k31427
asked Nov 14 at 13:20
Shantanu Kaushik
223
$g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
– Raymond Manzoni
Nov 14 at 13:27
I guess you may take $lim yto 0$?
– Mythomorphic
Nov 14 at 13:44
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
While practicing definite integrals I came across a question and now I am stuck
Question:
let f be a continous satisfying $f(x+y) = f(x) + f(y) + f(x)cdot f(y)$ for all real $x$ and $y$ and $f'(0)= -1$.
Find the value of $int_0^1 f(x) , dx$.
I tried to solve by find the function but no success and i am confused.
Please tell me how to solve these kind of questions.
definite-integrals functional-calculus
edited Nov 14 at 13:31
callculus
17.6k31427
asked Nov 14 at 13:20
Shantanu Kaushik
223
While practicing definite integrals I came across a question and now I am stuck
Question:
let f be a continous satisfying $f(x+y) = f(x) + f(y) + f(x)cdot f(y)$ for all real $x$ and $y$ and $f'(0)= -1$.
Find the value of $int_0^1 f(x) , dx$.
I tried to solve by find the function but no success and i am confused.
Please tell me how to solve these kind of questions.
definite-integrals functional-calculus
definite-integrals functional-calculus
edited Nov 14 at 13:31
callculus
17.6k31427
asked Nov 14 at 13:20
Shantanu Kaushik
223
edited Nov 14 at 13:31
callculus
17.6k31427
asked Nov 14 at 13:20
Shantanu Kaushik
223
edited Nov 14 at 13:31
callculus
17.6k31427
edited Nov 14 at 13:31
callculus
17.6k31427
edited Nov 14 at 13:31
callculus
17.6k31427
17.6k31427
asked Nov 14 at 13:20
Shantanu Kaushik
223
asked Nov 14 at 13:20
Shantanu Kaushik
223
asked Nov 14 at 13:20
Shantanu Kaushik
223
223
$g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
– Raymond Manzoni
Nov 14 at 13:27
I guess you may take $lim yto 0$?
– Mythomorphic
Nov 14 at 13:44
add a comment |
$g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
– Raymond Manzoni
Nov 14 at 13:27
I guess you may take $lim yto 0$?
– Mythomorphic
Nov 14 at 13:44
$g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
– Raymond Manzoni
Nov 14 at 13:27
$g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
– Raymond Manzoni
Nov 14 at 13:27
I guess you may take $lim yto 0$?
– Mythomorphic
Nov 14 at 13:44
I guess you may take $lim yto 0$?
– Mythomorphic
Nov 14 at 13:44
add a comment |
1 Answer
1
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1
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accepted
Rearrange
$$f(x+y)-f(x)=f(y)[1+f(x)]$$
Divide both sides by $y$ and take the limit
$$lim_{yto0}frac{f(x+y)-f(x)}{y}=lim_{yto0}=frac{f(y)[1+f(x)]}{y}$$
$$f'(x)=[1+f(x)]cdotleft[lim_{yto0}frac{f(y)}{y}right]$$
Can you proceed from here?
Notice $f(0)=0$ and given $f'(0)=-1$
$$lim_{yto0}frac{f(y)}{y}=-1$$
So $$f'(x)=-[1+f(x)]$$ $$int^1_0f'(x)dx=int^1_0-1-f(x)dx$$
$$int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$
answered Nov 14 at 14:00
Mythomorphic
5,2941733
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Rearrange
$$f(x+y)-f(x)=f(y)[1+f(x)]$$
Divide both sides by $y$ and take the limit
$$lim_{yto0}frac{f(x+y)-f(x)}{y}=lim_{yto0}=frac{f(y)[1+f(x)]}{y}$$
$$f'(x)=[1+f(x)]cdotleft[lim_{yto0}frac{f(y)}{y}right]$$
Can you proceed from here?
Notice $f(0)=0$ and given $f'(0)=-1$
$$lim_{yto0}frac{f(y)}{y}=-1$$
So $$f'(x)=-[1+f(x)]$$ $$int^1_0f'(x)dx=int^1_0-1-f(x)dx$$
$$int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$
answered Nov 14 at 14:00
Mythomorphic
5,2941733
add a comment |
up vote
1
down vote
accepted
Rearrange
$$f(x+y)-f(x)=f(y)[1+f(x)]$$
Divide both sides by $y$ and take the limit
$$lim_{yto0}frac{f(x+y)-f(x)}{y}=lim_{yto0}=frac{f(y)[1+f(x)]}{y}$$
$$f'(x)=[1+f(x)]cdotleft[lim_{yto0}frac{f(y)}{y}right]$$
Can you proceed from here?
Notice $f(0)=0$ and given $f'(0)=-1$
$$lim_{yto0}frac{f(y)}{y}=-1$$
So $$f'(x)=-[1+f(x)]$$ $$int^1_0f'(x)dx=int^1_0-1-f(x)dx$$
$$int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$
answered Nov 14 at 14:00
Mythomorphic
5,2941733
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Rearrange
$$f(x+y)-f(x)=f(y)[1+f(x)]$$
Divide both sides by $y$ and take the limit
$$lim_{yto0}frac{f(x+y)-f(x)}{y}=lim_{yto0}=frac{f(y)[1+f(x)]}{y}$$
$$f'(x)=[1+f(x)]cdotleft[lim_{yto0}frac{f(y)}{y}right]$$
Can you proceed from here?
Notice $f(0)=0$ and given $f'(0)=-1$
$$lim_{yto0}frac{f(y)}{y}=-1$$
So $$f'(x)=-[1+f(x)]$$ $$int^1_0f'(x)dx=int^1_0-1-f(x)dx$$
$$int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$
answered Nov 14 at 14:00
Mythomorphic
5,2941733
Rearrange
$$f(x+y)-f(x)=f(y)[1+f(x)]$$
Divide both sides by $y$ and take the limit
$$lim_{yto0}frac{f(x+y)-f(x)}{y}=lim_{yto0}=frac{f(y)[1+f(x)]}{y}$$
$$f'(x)=[1+f(x)]cdotleft[lim_{yto0}frac{f(y)}{y}right]$$
Can you proceed from here?
Notice $f(0)=0$ and given $f'(0)=-1$
$$lim_{yto0}frac{f(y)}{y}=-1$$
So $$f'(x)=-[1+f(x)]$$ $$int^1_0f'(x)dx=int^1_0-1-f(x)dx$$
$$int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$
answered Nov 14 at 14:00
Mythomorphic
5,2941733
answered Nov 14 at 14:00
Mythomorphic
5,2941733
answered Nov 14 at 14:00
Mythomorphic
5,2941733
answered Nov 14 at 14:00
Mythomorphic
5,2941733
5,2941733
add a comment |
add a comment |
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$g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
– Raymond Manzoni
Nov 14 at 13:27
I guess you may take $lim yto 0$?
– Mythomorphic
Nov 14 at 13:44