Mass of a wire positioned along the interval $[a, b]$
Consider a wire positioned along the interval $[a, b]$ on the $x$-axis and having a nonuniform distribution of mass. The mass of the segment lying along the interval $[a, x]$ is a function $F(x)$. Find the mass of the rod.
My answer:
$m=F(b)$
My textbook's answer:
$dm=F(x+dx)-F(x)=dF$
$m=int_a^b dF(x)=F(b)-F(a)$
Who is right and why?
integration
asked Nov 25 '18 at 21:54
Thomas
718416
add a comment |
Consider a wire positioned along the interval $[a, b]$ on the $x$-axis and having a nonuniform distribution of mass. The mass of the segment lying along the interval $[a, x]$ is a function $F(x)$. Find the mass of the rod.
My answer:
$m=F(b)$
My textbook's answer:
$dm=F(x+dx)-F(x)=dF$
$m=int_a^b dF(x)=F(b)-F(a)$
Who is right and why?
integration
asked Nov 25 '18 at 21:54
Thomas
718416
add a comment |
Consider a wire positioned along the interval $[a, b]$ on the $x$-axis and having a nonuniform distribution of mass. The mass of the segment lying along the interval $[a, x]$ is a function $F(x)$. Find the mass of the rod.
My answer:
$m=F(b)$
My textbook's answer:
$dm=F(x+dx)-F(x)=dF$
$m=int_a^b dF(x)=F(b)-F(a)$
Who is right and why?
integration
asked Nov 25 '18 at 21:54
Thomas
718416
Consider a wire positioned along the interval $[a, b]$ on the $x$-axis and having a nonuniform distribution of mass. The mass of the segment lying along the interval $[a, x]$ is a function $F(x)$. Find the mass of the rod.
My answer:
$m=F(b)$
My textbook's answer:
$dm=F(x+dx)-F(x)=dF$
$m=int_a^b dF(x)=F(b)-F(a)$
Who is right and why?
integration
integration
asked Nov 25 '18 at 21:54
Thomas
718416
asked Nov 25 '18 at 21:54
Thomas
718416
asked Nov 25 '18 at 21:54
Thomas
718416
asked Nov 25 '18 at 21:54
Thomas
718416
asked Nov 25 '18 at 21:54
Thomas
718416
718416
add a comment |
add a comment |
1 Answer
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Why not both? Since $F(a)$ is the mass along the segment $[a,a]$, which is just one point, we can asume $F(a)=0$ (and since it is also true that both answers are correct, this has to be the case.)
answered Nov 25 '18 at 22:03
Alejandro Nasif Salum
4,399118
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Why not both? Since $F(a)$ is the mass along the segment $[a,a]$, which is just one point, we can asume $F(a)=0$ (and since it is also true that both answers are correct, this has to be the case.)
answered Nov 25 '18 at 22:03
Alejandro Nasif Salum
4,399118
add a comment |
Why not both? Since $F(a)$ is the mass along the segment $[a,a]$, which is just one point, we can asume $F(a)=0$ (and since it is also true that both answers are correct, this has to be the case.)
answered Nov 25 '18 at 22:03
Alejandro Nasif Salum
4,399118
add a comment |
Why not both? Since $F(a)$ is the mass along the segment $[a,a]$, which is just one point, we can asume $F(a)=0$ (and since it is also true that both answers are correct, this has to be the case.)
answered Nov 25 '18 at 22:03
Alejandro Nasif Salum
4,399118
Why not both? Since $F(a)$ is the mass along the segment $[a,a]$, which is just one point, we can asume $F(a)=0$ (and since it is also true that both answers are correct, this has to be the case.)
answered Nov 25 '18 at 22:03
Alejandro Nasif Salum
4,399118
answered Nov 25 '18 at 22:03
Alejandro Nasif Salum
4,399118
answered Nov 25 '18 at 22:03
Alejandro Nasif Salum
4,399118
answered Nov 25 '18 at 22:03
Alejandro Nasif Salum
4,399118
4,399118
add a comment |
add a comment |
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