How can I find $h(y)$?












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 0:01









Batominovski

1




1










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


















  • $begingroup$
    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










1 Answer
1






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oldest

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0












$begingroup$

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.)






share|cite|improve this answer











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    $begingroup$

    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.)






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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.)






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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.)






        share|cite|improve this answer











        $endgroup$



        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.)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 11 '18 at 18:49

























        answered Nov 11 '18 at 18:36







        user614671





































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