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
33.1k33293
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1276
$endgroup$
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
33.1k33293
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1276
$endgroup$
$begingroup$
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
33.1k33293
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1276
$endgroup$
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
33.1k33293
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1276
edited Dec 12 '18 at 0:01
Batominovski
33.1k33293
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1276
edited Dec 12 '18 at 0:01
Batominovski
33.1k33293
edited Dec 12 '18 at 0:01
Batominovski
33.1k33293
edited Dec 12 '18 at 0:01
Batominovski
33.1k33293
33.1k33293
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1276
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1276
asked Nov 11 '18 at 18:24
Software_tSoftware_t
1276
1276
$begingroup$
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 |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2994214%2fhow-can-i-find-hy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2994214%2fhow-can-i-find-hy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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