Find the value $int_0^1f(x) ,mathrm dx$ without words
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Let $f(x)=ax^3+bx^2+cx+d$ such that: $$f(0)=1,quad f(0.5)=5, quad f(1)=15$$
Find: $$int_0^1f(x) ,mathrm dx$$
It is said that it can be solved without words.
integration definite-integrals
edited Aug 13 at 8:33
Rodrigo de Azevedo
12.6k41751
asked Aug 13 at 8:14
communnites
1,204431
add a comment |Â
up vote
3
down vote
favorite
Let $f(x)=ax^3+bx^2+cx+d$ such that: $$f(0)=1,quad f(0.5)=5, quad f(1)=15$$
Find: $$int_0^1f(x) ,mathrm dx$$
It is said that it can be solved without words.
integration definite-integrals
edited Aug 13 at 8:33
Rodrigo de Azevedo
12.6k41751
asked Aug 13 at 8:14
communnites
1,204431
$Kavi No,It is right
– communnites
Aug 13 at 8:24
2
That depends what is your definition of a 'word'.
– uniquesolution
Aug 13 at 8:31
The answers that had a chance to make it show that a good hint would have been much better than playing the MSEthical correctness card.
– Christian Blatter
Aug 13 at 8:44
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $f(x)=ax^3+bx^2+cx+d$ such that: $$f(0)=1,quad f(0.5)=5, quad f(1)=15$$
Find: $$int_0^1f(x) ,mathrm dx$$
It is said that it can be solved without words.
integration definite-integrals
edited Aug 13 at 8:33
Rodrigo de Azevedo
12.6k41751
asked Aug 13 at 8:14
communnites
1,204431
Let $f(x)=ax^3+bx^2+cx+d$ such that: $$f(0)=1,quad f(0.5)=5, quad f(1)=15$$
Find: $$int_0^1f(x) ,mathrm dx$$
It is said that it can be solved without words.
integration definite-integrals
edited Aug 13 at 8:33
Rodrigo de Azevedo
12.6k41751
asked Aug 13 at 8:14
communnites
1,204431
edited Aug 13 at 8:33
Rodrigo de Azevedo
12.6k41751
edited Aug 13 at 8:33
Rodrigo de Azevedo
12.6k41751
edited Aug 13 at 8:33
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Aug 13 at 8:14
communnites
1,204431
asked Aug 13 at 8:14
communnites
1,204431
asked Aug 13 at 8:14
communnites
1,204431
1,204431
$Kavi No,It is right
– communnites
Aug 13 at 8:24
2
That depends what is your definition of a 'word'.
– uniquesolution
Aug 13 at 8:31
The answers that had a chance to make it show that a good hint would have been much better than playing the MSEthical correctness card.
– Christian Blatter
Aug 13 at 8:44
add a comment |Â
$Kavi No,It is right
– communnites
Aug 13 at 8:24
2
That depends what is your definition of a 'word'.
– uniquesolution
Aug 13 at 8:31
The answers that had a chance to make it show that a good hint would have been much better than playing the MSEthical correctness card.
– Christian Blatter
Aug 13 at 8:44
$Kavi No,It is right
– communnites
Aug 13 at 8:24
$Kavi No,It is right
– communnites
Aug 13 at 8:24
2
2
That depends what is your definition of a 'word'.
– uniquesolution
Aug 13 at 8:31
That depends what is your definition of a 'word'.
– uniquesolution
Aug 13 at 8:31
The answers that had a chance to make it show that a good hint would have been much better than playing the MSEthical correctness card.
– Christian Blatter
Aug 13 at 8:44
The answers that had a chance to make it show that a good hint would have been much better than playing the MSEthical correctness card.
– Christian Blatter
Aug 13 at 8:44
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
12
down vote
accepted
$$int_0^1 f(x) dx = dfrac1-06left(f(0)+4f(0.5)+f(1)right)=6$$
answered Aug 13 at 8:25
mathcounterexamples.net
25k21754
@JackD'Aurizio Simpson’s rule error is in $f^(4)(xi)$. Therefore exact for a degree $3$ polynomial.
– mathcounterexamples.net
Aug 13 at 8:31
1
You are right, my bad. All the cubic polynomials fulfilling $f(0)=A, f(1/2)=B, f(1)=C$ have the same integral over $(0,1)$ since they differ by a multiple of $x(x-1/2)(x-1)$ which has integral zero.
– Jack D'Aurizio♦
Aug 13 at 8:34
1
Nice! But since the solution is not complete without your explanatory comment, and that comment contained words, I don't think this counts as a solution “without words†(whatever that was supposed to mean to begin with).
– Hans Lundmark
Aug 13 at 11:12
@HansLundmark A solution without word can be a big debate. If you know Simpson’s rule, you don’t need words. Just like you don’t need words to explain why multiplication of reals is commutative. All depends on what you accept as given backgrounds....
– mathcounterexamples.net
Aug 13 at 22:30
Yes, I agree. My comment was more of a criticism towards the question than towards your answer.
– Hans Lundmark
Aug 14 at 8:00
 |Â
show 1 more comment
up vote
7
down vote
$f(0) = 1implies d = 1, f(0.5) = 5 implies dfraca8 +dfracb4 + dfracc2 = 4, f(1) = 15 implies 2a+2b+2c = 28implies displaystyle int_0^1 f(x)dx = dfraca4+dfracb3+dfracc2+d= dfrac3a+4b+6c12+1=dfrac32+2812+1 = 6$
answered Aug 13 at 8:28
DeepSea
69.1k54284
add a comment |Â
up vote
1
down vote
$$ f(x) = s + t(x-.5)^2 + u(x-.5) + v(x-.5)^3. tag1 $$
$$ f(.5) = 5 implies s=5. tag2 $$
$$ 16 = 1 + 15 = f(0) + f(1) = 2cdot 5 + t/2 + 0 + 0 implies , t = 12. tag3 $$
$$ int f(x) ,dx = 5x + 4(x-.5)^3 + u (x-.5)^2/2 + v(x-.5)^4/4. tag4 $$
$$ int_0^1 f(x) ,dx = 5 + 4(1/8+1/8) + u,0 + v,0 = 6. tag5 $$
answered Aug 13 at 15:17
Somos
11.7k11033
add a comment |Â
up vote
0
down vote
The answer is a definite real number.
To see that, you have to solve the system of equations which you are given and express all parameters in terms of $a$ (say). Then you can evaluate the integral and the $a$'s will cancel out.
I'm going to leave the work to you.
answered Aug 13 at 8:24
asdf
3,408519
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
$$int_0^1 f(x) dx = dfrac1-06left(f(0)+4f(0.5)+f(1)right)=6$$
answered Aug 13 at 8:25
mathcounterexamples.net
25k21754
@JackD'Aurizio Simpson’s rule error is in $f^(4)(xi)$. Therefore exact for a degree $3$ polynomial.
– mathcounterexamples.net
Aug 13 at 8:31
1
You are right, my bad. All the cubic polynomials fulfilling $f(0)=A, f(1/2)=B, f(1)=C$ have the same integral over $(0,1)$ since they differ by a multiple of $x(x-1/2)(x-1)$ which has integral zero.
– Jack D'Aurizio♦
Aug 13 at 8:34
1
Nice! But since the solution is not complete without your explanatory comment, and that comment contained words, I don't think this counts as a solution “without words†(whatever that was supposed to mean to begin with).
– Hans Lundmark
Aug 13 at 11:12
@HansLundmark A solution without word can be a big debate. If you know Simpson’s rule, you don’t need words. Just like you don’t need words to explain why multiplication of reals is commutative. All depends on what you accept as given backgrounds....
– mathcounterexamples.net
Aug 13 at 22:30
Yes, I agree. My comment was more of a criticism towards the question than towards your answer.
– Hans Lundmark
Aug 14 at 8:00
 |Â
show 1 more comment
up vote
12
down vote
accepted
$$int_0^1 f(x) dx = dfrac1-06left(f(0)+4f(0.5)+f(1)right)=6$$
answered Aug 13 at 8:25
mathcounterexamples.net
25k21754
@JackD'Aurizio Simpson’s rule error is in $f^(4)(xi)$. Therefore exact for a degree $3$ polynomial.
– mathcounterexamples.net
Aug 13 at 8:31
1
You are right, my bad. All the cubic polynomials fulfilling $f(0)=A, f(1/2)=B, f(1)=C$ have the same integral over $(0,1)$ since they differ by a multiple of $x(x-1/2)(x-1)$ which has integral zero.
– Jack D'Aurizio♦
Aug 13 at 8:34
1
Nice! But since the solution is not complete without your explanatory comment, and that comment contained words, I don't think this counts as a solution “without words†(whatever that was supposed to mean to begin with).
– Hans Lundmark
Aug 13 at 11:12
@HansLundmark A solution without word can be a big debate. If you know Simpson’s rule, you don’t need words. Just like you don’t need words to explain why multiplication of reals is commutative. All depends on what you accept as given backgrounds....
– mathcounterexamples.net
Aug 13 at 22:30
Yes, I agree. My comment was more of a criticism towards the question than towards your answer.
– Hans Lundmark
Aug 14 at 8:00
 |Â
show 1 more comment
up vote
12
down vote
accepted
up vote
12
down vote
accepted
$$int_0^1 f(x) dx = dfrac1-06left(f(0)+4f(0.5)+f(1)right)=6$$
answered Aug 13 at 8:25
mathcounterexamples.net
25k21754
$$int_0^1 f(x) dx = dfrac1-06left(f(0)+4f(0.5)+f(1)right)=6$$
answered Aug 13 at 8:25
mathcounterexamples.net
25k21754
answered Aug 13 at 8:25
mathcounterexamples.net
25k21754
answered Aug 13 at 8:25
mathcounterexamples.net
25k21754
answered Aug 13 at 8:25
mathcounterexamples.net
25k21754
25k21754
@JackD'Aurizio Simpson’s rule error is in $f^(4)(xi)$. Therefore exact for a degree $3$ polynomial.
– mathcounterexamples.net
Aug 13 at 8:31
1
You are right, my bad. All the cubic polynomials fulfilling $f(0)=A, f(1/2)=B, f(1)=C$ have the same integral over $(0,1)$ since they differ by a multiple of $x(x-1/2)(x-1)$ which has integral zero.
– Jack D'Aurizio♦
Aug 13 at 8:34
1
Nice! But since the solution is not complete without your explanatory comment, and that comment contained words, I don't think this counts as a solution “without words†(whatever that was supposed to mean to begin with).
– Hans Lundmark
Aug 13 at 11:12
@HansLundmark A solution without word can be a big debate. If you know Simpson’s rule, you don’t need words. Just like you don’t need words to explain why multiplication of reals is commutative. All depends on what you accept as given backgrounds....
– mathcounterexamples.net
Aug 13 at 22:30
Yes, I agree. My comment was more of a criticism towards the question than towards your answer.
– Hans Lundmark
Aug 14 at 8:00
 |Â
show 1 more comment
@JackD'Aurizio Simpson’s rule error is in $f^(4)(xi)$. Therefore exact for a degree $3$ polynomial.
– mathcounterexamples.net
Aug 13 at 8:31
1
You are right, my bad. All the cubic polynomials fulfilling $f(0)=A, f(1/2)=B, f(1)=C$ have the same integral over $(0,1)$ since they differ by a multiple of $x(x-1/2)(x-1)$ which has integral zero.
– Jack D'Aurizio♦
Aug 13 at 8:34
1
Nice! But since the solution is not complete without your explanatory comment, and that comment contained words, I don't think this counts as a solution “without words†(whatever that was supposed to mean to begin with).
– Hans Lundmark
Aug 13 at 11:12
@HansLundmark A solution without word can be a big debate. If you know Simpson’s rule, you don’t need words. Just like you don’t need words to explain why multiplication of reals is commutative. All depends on what you accept as given backgrounds....
– mathcounterexamples.net
Aug 13 at 22:30
Yes, I agree. My comment was more of a criticism towards the question than towards your answer.
– Hans Lundmark
Aug 14 at 8:00
@JackD'Aurizio Simpson’s rule error is in $f^(4)(xi)$. Therefore exact for a degree $3$ polynomial.
– mathcounterexamples.net
Aug 13 at 8:31
@JackD'Aurizio Simpson’s rule error is in $f^(4)(xi)$. Therefore exact for a degree $3$ polynomial.
– mathcounterexamples.net
Aug 13 at 8:31
1
1
You are right, my bad. All the cubic polynomials fulfilling $f(0)=A, f(1/2)=B, f(1)=C$ have the same integral over $(0,1)$ since they differ by a multiple of $x(x-1/2)(x-1)$ which has integral zero.
– Jack D'Aurizio♦
Aug 13 at 8:34
You are right, my bad. All the cubic polynomials fulfilling $f(0)=A, f(1/2)=B, f(1)=C$ have the same integral over $(0,1)$ since they differ by a multiple of $x(x-1/2)(x-1)$ which has integral zero.
– Jack D'Aurizio♦
Aug 13 at 8:34
1
1
Nice! But since the solution is not complete without your explanatory comment, and that comment contained words, I don't think this counts as a solution “without words†(whatever that was supposed to mean to begin with).
– Hans Lundmark
Aug 13 at 11:12
Nice! But since the solution is not complete without your explanatory comment, and that comment contained words, I don't think this counts as a solution “without words†(whatever that was supposed to mean to begin with).
– Hans Lundmark
Aug 13 at 11:12
@HansLundmark A solution without word can be a big debate. If you know Simpson’s rule, you don’t need words. Just like you don’t need words to explain why multiplication of reals is commutative. All depends on what you accept as given backgrounds....
– mathcounterexamples.net
Aug 13 at 22:30
@HansLundmark A solution without word can be a big debate. If you know Simpson’s rule, you don’t need words. Just like you don’t need words to explain why multiplication of reals is commutative. All depends on what you accept as given backgrounds....
– mathcounterexamples.net
Aug 13 at 22:30
Yes, I agree. My comment was more of a criticism towards the question than towards your answer.
– Hans Lundmark
Aug 14 at 8:00
Yes, I agree. My comment was more of a criticism towards the question than towards your answer.
– Hans Lundmark
Aug 14 at 8:00
 |Â
show 1 more comment
up vote
7
down vote
$f(0) = 1implies d = 1, f(0.5) = 5 implies dfraca8 +dfracb4 + dfracc2 = 4, f(1) = 15 implies 2a+2b+2c = 28implies displaystyle int_0^1 f(x)dx = dfraca4+dfracb3+dfracc2+d= dfrac3a+4b+6c12+1=dfrac32+2812+1 = 6$
answered Aug 13 at 8:28
DeepSea
69.1k54284
add a comment |Â
up vote
7
down vote
$f(0) = 1implies d = 1, f(0.5) = 5 implies dfraca8 +dfracb4 + dfracc2 = 4, f(1) = 15 implies 2a+2b+2c = 28implies displaystyle int_0^1 f(x)dx = dfraca4+dfracb3+dfracc2+d= dfrac3a+4b+6c12+1=dfrac32+2812+1 = 6$
answered Aug 13 at 8:28
DeepSea
69.1k54284
add a comment |Â
up vote
7
down vote
up vote
7
down vote
$f(0) = 1implies d = 1, f(0.5) = 5 implies dfraca8 +dfracb4 + dfracc2 = 4, f(1) = 15 implies 2a+2b+2c = 28implies displaystyle int_0^1 f(x)dx = dfraca4+dfracb3+dfracc2+d= dfrac3a+4b+6c12+1=dfrac32+2812+1 = 6$
answered Aug 13 at 8:28
DeepSea
69.1k54284
$f(0) = 1implies d = 1, f(0.5) = 5 implies dfraca8 +dfracb4 + dfracc2 = 4, f(1) = 15 implies 2a+2b+2c = 28implies displaystyle int_0^1 f(x)dx = dfraca4+dfracb3+dfracc2+d= dfrac3a+4b+6c12+1=dfrac32+2812+1 = 6$
answered Aug 13 at 8:28
DeepSea
69.1k54284
answered Aug 13 at 8:28
DeepSea
69.1k54284
answered Aug 13 at 8:28
DeepSea
69.1k54284
answered Aug 13 at 8:28
DeepSea
69.1k54284
69.1k54284
add a comment |Â
add a comment |Â
up vote
1
down vote
$$ f(x) = s + t(x-.5)^2 + u(x-.5) + v(x-.5)^3. tag1 $$
$$ f(.5) = 5 implies s=5. tag2 $$
$$ 16 = 1 + 15 = f(0) + f(1) = 2cdot 5 + t/2 + 0 + 0 implies , t = 12. tag3 $$
$$ int f(x) ,dx = 5x + 4(x-.5)^3 + u (x-.5)^2/2 + v(x-.5)^4/4. tag4 $$
$$ int_0^1 f(x) ,dx = 5 + 4(1/8+1/8) + u,0 + v,0 = 6. tag5 $$
answered Aug 13 at 15:17
Somos
11.7k11033
add a comment |Â
up vote
1
down vote
$$ f(x) = s + t(x-.5)^2 + u(x-.5) + v(x-.5)^3. tag1 $$
$$ f(.5) = 5 implies s=5. tag2 $$
$$ 16 = 1 + 15 = f(0) + f(1) = 2cdot 5 + t/2 + 0 + 0 implies , t = 12. tag3 $$
$$ int f(x) ,dx = 5x + 4(x-.5)^3 + u (x-.5)^2/2 + v(x-.5)^4/4. tag4 $$
$$ int_0^1 f(x) ,dx = 5 + 4(1/8+1/8) + u,0 + v,0 = 6. tag5 $$
answered Aug 13 at 15:17
Somos
11.7k11033
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$ f(x) = s + t(x-.5)^2 + u(x-.5) + v(x-.5)^3. tag1 $$
$$ f(.5) = 5 implies s=5. tag2 $$
$$ 16 = 1 + 15 = f(0) + f(1) = 2cdot 5 + t/2 + 0 + 0 implies , t = 12. tag3 $$
$$ int f(x) ,dx = 5x + 4(x-.5)^3 + u (x-.5)^2/2 + v(x-.5)^4/4. tag4 $$
$$ int_0^1 f(x) ,dx = 5 + 4(1/8+1/8) + u,0 + v,0 = 6. tag5 $$
answered Aug 13 at 15:17
Somos
11.7k11033
$$ f(x) = s + t(x-.5)^2 + u(x-.5) + v(x-.5)^3. tag1 $$
$$ f(.5) = 5 implies s=5. tag2 $$
$$ 16 = 1 + 15 = f(0) + f(1) = 2cdot 5 + t/2 + 0 + 0 implies , t = 12. tag3 $$
$$ int f(x) ,dx = 5x + 4(x-.5)^3 + u (x-.5)^2/2 + v(x-.5)^4/4. tag4 $$
$$ int_0^1 f(x) ,dx = 5 + 4(1/8+1/8) + u,0 + v,0 = 6. tag5 $$
answered Aug 13 at 15:17
Somos
11.7k11033
edited Aug 13 at 15:32
answered Aug 13 at 15:17
Somos
11.7k11033
answered Aug 13 at 15:17
Somos
11.7k11033
answered Aug 13 at 15:17
Somos
11.7k11033
11.7k11033
add a comment |Â
add a comment |Â
up vote
0
down vote
The answer is a definite real number.
To see that, you have to solve the system of equations which you are given and express all parameters in terms of $a$ (say). Then you can evaluate the integral and the $a$'s will cancel out.
I'm going to leave the work to you.
answered Aug 13 at 8:24
asdf
3,408519
add a comment |Â
up vote
0
down vote
The answer is a definite real number.
To see that, you have to solve the system of equations which you are given and express all parameters in terms of $a$ (say). Then you can evaluate the integral and the $a$'s will cancel out.
I'm going to leave the work to you.
answered Aug 13 at 8:24
asdf
3,408519
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The answer is a definite real number.
To see that, you have to solve the system of equations which you are given and express all parameters in terms of $a$ (say). Then you can evaluate the integral and the $a$'s will cancel out.
I'm going to leave the work to you.
answered Aug 13 at 8:24
asdf
3,408519
The answer is a definite real number.
To see that, you have to solve the system of equations which you are given and express all parameters in terms of $a$ (say). Then you can evaluate the integral and the $a$'s will cancel out.
I'm going to leave the work to you.
answered Aug 13 at 8:24
asdf
3,408519
edited Aug 13 at 8:30
answered Aug 13 at 8:24
asdf
3,408519
answered Aug 13 at 8:24
asdf
3,408519
answered Aug 13 at 8:24
asdf
3,408519
3,408519
add a comment |Â
add a comment |Â
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$Kavi No,It is right
– communnites
Aug 13 at 8:24
2
That depends what is your definition of a 'word'.
– uniquesolution
Aug 13 at 8:31
The answers that had a chance to make it show that a good hint would have been much better than playing the MSEthical correctness card.
– Christian Blatter
Aug 13 at 8:44