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Integrating with different methods leads to different results?
The 2019 Stack Overflow Developer Survey Results Are InIntegrating with respect to different variablesTwo different results with contour integrationDifferent results when integrating 1/(x^2-9) with computer toolsVolume of truncated cone with different methods gives different results.Probability problem gives different answers with different methodsThe Sign of the Direction of Motion & Difficulty of the IntegralDo different methods of integration yields different results?Different answers using different methodsDoes integrating over different curves give different results?Compute $intfracxx+1dx$
$begingroup$
Acceleration of a particle is given as $a = 0.1(t-5)^2$. The particle is initially at rest.
Here is the way I initially attempted it. (I didn't use substitution as the question in the textbook appears before substitution is officially taught.)
$$a:=:0.1t^2-t+2.5$$
so
$$v = frac130t^3-0.5t^2+2.5t+c$$
$c$ evaluates to zero.
But my textbook uses a substitution and gets:
$$frac130left(t-5right)^3 +c$$
In this case, c has a value when t= 0 and so the end result is different. Wouldn't this lead to different values in general?
integration proof-verification
asked Apr 8 at 1:25
Gab N.Gab N.
534
$endgroup$
add a comment |
$begingroup$
Acceleration of a particle is given as $a = 0.1(t-5)^2$. The particle is initially at rest.
Here is the way I initially attempted it. (I didn't use substitution as the question in the textbook appears before substitution is officially taught.)
$$a:=:0.1t^2-t+2.5$$
so
$$v = frac130t^3-0.5t^2+2.5t+c$$
$c$ evaluates to zero.
But my textbook uses a substitution and gets:
$$frac130left(t-5right)^3 +c$$
In this case, c has a value when t= 0 and so the end result is different. Wouldn't this lead to different values in general?
integration proof-verification
asked Apr 8 at 1:25
Gab N.Gab N.
534
$endgroup$
2
$begingroup$
No it won't. The constants will always adjust accordingly to make the results match in the end.
$endgroup$
– ZeroXLR
Apr 8 at 1:33
1
$begingroup$
The values of your $c$ and the answer’s $c$ will be different, but that’s just because their answer also has a constant term in the polynomial expression. The total combined constant term in their answer ($tfrac130(-5)^3+c$) will be the same as yours, namely, $0$.
$endgroup$
– MPW
Apr 8 at 1:38
add a comment |
$begingroup$
Acceleration of a particle is given as $a = 0.1(t-5)^2$. The particle is initially at rest.
Here is the way I initially attempted it. (I didn't use substitution as the question in the textbook appears before substitution is officially taught.)
$$a:=:0.1t^2-t+2.5$$
so
$$v = frac130t^3-0.5t^2+2.5t+c$$
$c$ evaluates to zero.
But my textbook uses a substitution and gets:
$$frac130left(t-5right)^3 +c$$
In this case, c has a value when t= 0 and so the end result is different. Wouldn't this lead to different values in general?
integration proof-verification
asked Apr 8 at 1:25
Gab N.Gab N.
534
$endgroup$
Acceleration of a particle is given as $a = 0.1(t-5)^2$. The particle is initially at rest.
Here is the way I initially attempted it. (I didn't use substitution as the question in the textbook appears before substitution is officially taught.)
$$a:=:0.1t^2-t+2.5$$
so
$$v = frac130t^3-0.5t^2+2.5t+c$$
$c$ evaluates to zero.
But my textbook uses a substitution and gets:
$$frac130left(t-5right)^3 +c$$
In this case, c has a value when t= 0 and so the end result is different. Wouldn't this lead to different values in general?
integration proof-verification
integration proof-verification
asked Apr 8 at 1:25
Gab N.Gab N.
534
asked Apr 8 at 1:25
Gab N.Gab N.
534
asked Apr 8 at 1:25
Gab N.Gab N.
534
asked Apr 8 at 1:25
Gab N.Gab N.
534
asked Apr 8 at 1:25
Gab N.Gab N.
534
534
2
$begingroup$
No it won't. The constants will always adjust accordingly to make the results match in the end.
$endgroup$
– ZeroXLR
Apr 8 at 1:33
1
$begingroup$
The values of your $c$ and the answer’s $c$ will be different, but that’s just because their answer also has a constant term in the polynomial expression. The total combined constant term in their answer ($tfrac130(-5)^3+c$) will be the same as yours, namely, $0$.
$endgroup$
– MPW
Apr 8 at 1:38
add a comment |
2
$begingroup$
No it won't. The constants will always adjust accordingly to make the results match in the end.
$endgroup$
– ZeroXLR
Apr 8 at 1:33
1
$begingroup$
The values of your $c$ and the answer’s $c$ will be different, but that’s just because their answer also has a constant term in the polynomial expression. The total combined constant term in their answer ($tfrac130(-5)^3+c$) will be the same as yours, namely, $0$.
$endgroup$
– MPW
Apr 8 at 1:38
2
2
$begingroup$
No it won't. The constants will always adjust accordingly to make the results match in the end.
$endgroup$
– ZeroXLR
Apr 8 at 1:33
$begingroup$
No it won't. The constants will always adjust accordingly to make the results match in the end.
$endgroup$
– ZeroXLR
Apr 8 at 1:33
1
1
$begingroup$
The values of your $c$ and the answer’s $c$ will be different, but that’s just because their answer also has a constant term in the polynomial expression. The total combined constant term in their answer ($tfrac130(-5)^3+c$) will be the same as yours, namely, $0$.
$endgroup$
– MPW
Apr 8 at 1:38
$begingroup$
The values of your $c$ and the answer’s $c$ will be different, but that’s just because their answer also has a constant term in the polynomial expression. The total combined constant term in their answer ($tfrac130(-5)^3+c$) will be the same as yours, namely, $0$.
$endgroup$
– MPW
Apr 8 at 1:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that you ended up with
$$
v(t) = frac130 t^3 - frac12 t^2 + frac52 t
$$
But the book's answer must have $c = 5^3/30$, so the complete answer is
$$
beginsplit
v(t) &= frac130 (t-5)^3 + frac5^330\
&= frac(t-5)^3-5^330 \
&= fract^3 - 3cdot 5 t^2 + 3 cdot 5^2t - 5^3 + 5^330 \
&= fract^330 - fract^22 + frac5t2\
endsplit
$$
which is the same answer as the one you ended up with.
edited Apr 8 at 8:41
John Omielan
4,9362217
answered Apr 8 at 1:34
gt6989bgt6989b
35.8k22557
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that you ended up with
$$
v(t) = frac130 t^3 - frac12 t^2 + frac52 t
$$
But the book's answer must have $c = 5^3/30$, so the complete answer is
$$
beginsplit
v(t) &= frac130 (t-5)^3 + frac5^330\
&= frac(t-5)^3-5^330 \
&= fract^3 - 3cdot 5 t^2 + 3 cdot 5^2t - 5^3 + 5^330 \
&= fract^330 - fract^22 + frac5t2\
endsplit
$$
which is the same answer as the one you ended up with.
edited Apr 8 at 8:41
John Omielan
4,9362217
answered Apr 8 at 1:34
gt6989bgt6989b
35.8k22557
$endgroup$
add a comment |
$begingroup$
Note that you ended up with
$$
v(t) = frac130 t^3 - frac12 t^2 + frac52 t
$$
But the book's answer must have $c = 5^3/30$, so the complete answer is
$$
beginsplit
v(t) &= frac130 (t-5)^3 + frac5^330\
&= frac(t-5)^3-5^330 \
&= fract^3 - 3cdot 5 t^2 + 3 cdot 5^2t - 5^3 + 5^330 \
&= fract^330 - fract^22 + frac5t2\
endsplit
$$
which is the same answer as the one you ended up with.
edited Apr 8 at 8:41
John Omielan
4,9362217
answered Apr 8 at 1:34
gt6989bgt6989b
35.8k22557
$endgroup$
add a comment |
$begingroup$
Note that you ended up with
$$
v(t) = frac130 t^3 - frac12 t^2 + frac52 t
$$
But the book's answer must have $c = 5^3/30$, so the complete answer is
$$
beginsplit
v(t) &= frac130 (t-5)^3 + frac5^330\
&= frac(t-5)^3-5^330 \
&= fract^3 - 3cdot 5 t^2 + 3 cdot 5^2t - 5^3 + 5^330 \
&= fract^330 - fract^22 + frac5t2\
endsplit
$$
which is the same answer as the one you ended up with.
edited Apr 8 at 8:41
John Omielan
4,9362217
answered Apr 8 at 1:34
gt6989bgt6989b
35.8k22557
$endgroup$
Note that you ended up with
$$
v(t) = frac130 t^3 - frac12 t^2 + frac52 t
$$
But the book's answer must have $c = 5^3/30$, so the complete answer is
$$
beginsplit
v(t) &= frac130 (t-5)^3 + frac5^330\
&= frac(t-5)^3-5^330 \
&= fract^3 - 3cdot 5 t^2 + 3 cdot 5^2t - 5^3 + 5^330 \
&= fract^330 - fract^22 + frac5t2\
endsplit
$$
which is the same answer as the one you ended up with.
edited Apr 8 at 8:41
John Omielan
4,9362217
answered Apr 8 at 1:34
gt6989bgt6989b
35.8k22557
edited Apr 8 at 8:41
John Omielan
4,9362217
edited Apr 8 at 8:41
John Omielan
4,9362217
edited Apr 8 at 8:41
John Omielan
4,9362217
4,9362217
answered Apr 8 at 1:34
gt6989bgt6989b
35.8k22557
answered Apr 8 at 1:34
gt6989bgt6989b
35.8k22557
answered Apr 8 at 1:34
gt6989bgt6989b
35.8k22557
35.8k22557
add a comment |
add a comment |
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$begingroup$
No it won't. The constants will always adjust accordingly to make the results match in the end.
$endgroup$
– ZeroXLR
Apr 8 at 1:33
1
$begingroup$
The values of your $c$ and the answer’s $c$ will be different, but that’s just because their answer also has a constant term in the polynomial expression. The total combined constant term in their answer ($tfrac130(-5)^3+c$) will be the same as yours, namely, $0$.
$endgroup$
– MPW
Apr 8 at 1:38