How can I find $h(y)$?
$begingroup$
Given the following function
$$ f(t) = begin{cases}
t &text{for }tleq2,,\
2 &text{for } tgeq 2,,\
end{cases}
$$
and $$h'(y)=f(y),,$$
how can I find $h(y)$ ?
calculus integration ordinary-differential-equations functions derivatives
edited Dec 12 '18 at 0:01
Batominovski
1
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1306
$endgroup$
add a comment |
$begingroup$
Given the following function
$$ f(t) = begin{cases}
t &text{for }tleq2,,\
2 &text{for } tgeq 2,,\
end{cases}
$$
and $$h'(y)=f(y),,$$
how can I find $h(y)$ ?
calculus integration ordinary-differential-equations functions derivatives
edited Dec 12 '18 at 0:01
Batominovski
1
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1306
$endgroup$
$begingroup$
You need to integrate $f(t)$
$endgroup$
– Patricio
Nov 11 '18 at 18:26
$begingroup$
How can I do it in this case?
$endgroup$
– Software_t
Nov 11 '18 at 18:30
$begingroup$
Do it case by case: $int t mathrm dt=frac12 t^2+c$ for $tle 2$ and $int 2 mathrm dt=2t+c$ for $t ge 2$
$endgroup$
– Mohammad Zuhair Khan
Nov 11 '18 at 18:35
add a comment |
$begingroup$
Given the following function
$$ f(t) = begin{cases}
t &text{for }tleq2,,\
2 &text{for } tgeq 2,,\
end{cases}
$$
and $$h'(y)=f(y),,$$
how can I find $h(y)$ ?
calculus integration ordinary-differential-equations functions derivatives
edited Dec 12 '18 at 0:01
Batominovski
1
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1306
$endgroup$
Given the following function
$$ f(t) = begin{cases}
t &text{for }tleq2,,\
2 &text{for } tgeq 2,,\
end{cases}
$$
and $$h'(y)=f(y),,$$
how can I find $h(y)$ ?
calculus integration ordinary-differential-equations functions derivatives
calculus integration ordinary-differential-equations functions derivatives
edited Dec 12 '18 at 0:01
Batominovski
1
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1306
edited Dec 12 '18 at 0:01
Batominovski
1
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1306
edited Dec 12 '18 at 0:01
Batominovski
1
edited Dec 12 '18 at 0:01
Batominovski
1
edited Dec 12 '18 at 0:01
Batominovski
1
1
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1306
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1306
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1306
1306
$begingroup$
You need to integrate $f(t)$
$endgroup$
– Patricio
Nov 11 '18 at 18:26
$begingroup$
How can I do it in this case?
$endgroup$
– Software_t
Nov 11 '18 at 18:30
$begingroup$
Do it case by case: $int t mathrm dt=frac12 t^2+c$ for $tle 2$ and $int 2 mathrm dt=2t+c$ for $t ge 2$
$endgroup$
– Mohammad Zuhair Khan
Nov 11 '18 at 18:35
add a comment |
$begingroup$
You need to integrate $f(t)$
$endgroup$
– Patricio
Nov 11 '18 at 18:26
$begingroup$
How can I do it in this case?
$endgroup$
– Software_t
Nov 11 '18 at 18:30
$begingroup$
Do it case by case: $int t mathrm dt=frac12 t^2+c$ for $tle 2$ and $int 2 mathrm dt=2t+c$ for $t ge 2$
$endgroup$
– Mohammad Zuhair Khan
Nov 11 '18 at 18:35
$begingroup$
You need to integrate $f(t)$
$endgroup$
– Patricio
Nov 11 '18 at 18:26
$begingroup$
You need to integrate $f(t)$
$endgroup$
– Patricio
Nov 11 '18 at 18:26
$begingroup$
How can I do it in this case?
$endgroup$
– Software_t
Nov 11 '18 at 18:30
$begingroup$
How can I do it in this case?
$endgroup$
– Software_t
Nov 11 '18 at 18:30
$begingroup$
Do it case by case: $int t mathrm dt=frac12 t^2+c$ for $tle 2$ and $int 2 mathrm dt=2t+c$ for $t ge 2$
$endgroup$
– Mohammad Zuhair Khan
Nov 11 '18 at 18:35
$begingroup$
Do it case by case: $int t mathrm dt=frac12 t^2+c$ for $tle 2$ and $int 2 mathrm dt=2t+c$ for $t ge 2$
$endgroup$
– Mohammad Zuhair Khan
Nov 11 '18 at 18:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An antiderivative is not unique. So $c=h(0)$ gives
$$h(y)-h(0)=int_0^yf(t)dt=int_0^ytdt =y^2/2$$
for $yle 2$, so
$$h(y)=y^2/2+c$$
when $yle 2$. But if $y>2$,
$$h(y)-h(2)=int_2^yf(t)dt=int_2^y2dt=2(y-2).$$
Id est,
$$h(y)=h(2)+2t-4=(2^2/2+c)+2y-4=2y-2+c$$
when $y>2$. (You should check btw that $h$ is differentiable, and it is.)
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An antiderivative is not unique. So $c=h(0)$ gives
$$h(y)-h(0)=int_0^yf(t)dt=int_0^ytdt =y^2/2$$
for $yle 2$, so
$$h(y)=y^2/2+c$$
when $yle 2$. But if $y>2$,
$$h(y)-h(2)=int_2^yf(t)dt=int_2^y2dt=2(y-2).$$
Id est,
$$h(y)=h(2)+2t-4=(2^2/2+c)+2y-4=2y-2+c$$
when $y>2$. (You should check btw that $h$ is differentiable, and it is.)
$endgroup$
add a comment |
$begingroup$
An antiderivative is not unique. So $c=h(0)$ gives
$$h(y)-h(0)=int_0^yf(t)dt=int_0^ytdt =y^2/2$$
for $yle 2$, so
$$h(y)=y^2/2+c$$
when $yle 2$. But if $y>2$,
$$h(y)-h(2)=int_2^yf(t)dt=int_2^y2dt=2(y-2).$$
Id est,
$$h(y)=h(2)+2t-4=(2^2/2+c)+2y-4=2y-2+c$$
when $y>2$. (You should check btw that $h$ is differentiable, and it is.)
$endgroup$
add a comment |
$begingroup$
An antiderivative is not unique. So $c=h(0)$ gives
$$h(y)-h(0)=int_0^yf(t)dt=int_0^ytdt =y^2/2$$
for $yle 2$, so
$$h(y)=y^2/2+c$$
when $yle 2$. But if $y>2$,
$$h(y)-h(2)=int_2^yf(t)dt=int_2^y2dt=2(y-2).$$
Id est,
$$h(y)=h(2)+2t-4=(2^2/2+c)+2y-4=2y-2+c$$
when $y>2$. (You should check btw that $h$ is differentiable, and it is.)
$endgroup$
An antiderivative is not unique. So $c=h(0)$ gives
$$h(y)-h(0)=int_0^yf(t)dt=int_0^ytdt =y^2/2$$
for $yle 2$, so
$$h(y)=y^2/2+c$$
when $yle 2$. But if $y>2$,
$$h(y)-h(2)=int_2^yf(t)dt=int_2^y2dt=2(y-2).$$
Id est,
$$h(y)=h(2)+2t-4=(2^2/2+c)+2y-4=2y-2+c$$
when $y>2$. (You should check btw that $h$ is differentiable, and it is.)
edited Nov 11 '18 at 18:49
answered Nov 11 '18 at 18:36
user614671
add a comment |
add a comment |
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$begingroup$
You need to integrate $f(t)$
$endgroup$
– Patricio
Nov 11 '18 at 18:26
$begingroup$
How can I do it in this case?
$endgroup$
– Software_t
Nov 11 '18 at 18:30
$begingroup$
Do it case by case: $int t mathrm dt=frac12 t^2+c$ for $tle 2$ and $int 2 mathrm dt=2t+c$ for $t ge 2$
$endgroup$
– Mohammad Zuhair Khan
Nov 11 '18 at 18:35