Sfrgttk

Aligning Systems of Equations

Did the characters in Moving Pictures not know about cameras like Twoflower's?

Can I do anything else with aspersions other than cast them?

In the Lost in Space intro why was Dr. Smith actor listed as a special guest star?

Can I legally make a website about boycotting a certain company?

How can a kingdom keep the secret of a missing monarch from the public?

What is an explicit bijection in combinatorics?

How to typeset a small black square as a binary operator?

Is it Safe to Plug an Extension Cord Into a Power Strip?

Was the Soviet N1 really capable of sending 9.6 GB/s of telemetry?

Why Doesn't It Completely Uninstall?

What does it mean when an external ID field follows a DML Statement?

Taking an academic pseudonym?

Found a major flaw in paper from home university – to which I would like to return

What does an unprocessed RAW file look like?

How do I avoid the "chosen hero" feeling?

Dot product with a constant

Was Opportunity's last message to Earth "My battery is low and it's getting dark"?

How bad is a Computer Science course that doesn't teach Design Patterns?

Is Screenshot Time-tracking Common?

How Create a list of the first 10,000 digits of Pi and sum it?

Exploding Numbers

Isn't a semicolon (';') needed after a function declaration in C++?

SQL Server 2017 crashes when backing up because filepath is wrong

Integral problem. Unsure of the approach.

Calculate the integral $int_0^{x} (lfloor t+1 rfloor)^3dt$Cauchy integral formula problemWhy is are these expressions in Leibniz notation not equivalent?Double integral problemIs it neccessary to split the integral?Evaluate the following line integral.$f_n(x) = frac{x^n}{1 + x^n}$ does not converge uniformly, yet we can change the order of integralBizarre Definite IntegralWhat's the integral of $(df/dx)^2$?Problem with a Hypergeometric function integral

3

$begingroup$

I have this integral:

$$int_0^1 frac{1 + 12t}{1 + 3t}dt$$

I can split this up into:

$$int_0^1 frac{1}{1+3t} dt + int_0^1frac{12t}{1+3t}dt$$

The left side:

$u = 1+3t$ and $du = 3dt$ and $frac{du}{3} = dt$

so $$frac{1}{3} int_0^1 frac{1}{u} du = frac{1}{3} ln |u| + C$$

But what about the right?

calculus integration

share|cite|improve this question

edited 34 mins ago

Eevee Trainer

6,54811237

asked 43 mins ago

Jwan622Jwan622

2,16211530

$endgroup$

add a comment | 

3

$begingroup$

I have this integral:

$$int_0^1 frac{1 + 12t}{1 + 3t}dt$$

I can split this up into:

$$int_0^1 frac{1}{1+3t} dt + int_0^1frac{12t}{1+3t}dt$$

The left side:

$u = 1+3t$ and $du = 3dt$ and $frac{du}{3} = dt$

so $$frac{1}{3} int_0^1 frac{1}{u} du = frac{1}{3} ln |u| + C$$

But what about the right?

calculus integration

share|cite|improve this question

edited 34 mins ago

Eevee Trainer

6,54811237

asked 43 mins ago

Jwan622Jwan622

2,16211530

$endgroup$

add a comment | 

$begingroup$

I have this integral:

$$int_0^1 frac{1 + 12t}{1 + 3t}dt$$

I can split this up into:

$$int_0^1 frac{1}{1+3t} dt + int_0^1frac{12t}{1+3t}dt$$

The left side:

$u = 1+3t$ and $du = 3dt$ and $frac{du}{3} = dt$

so $$frac{1}{3} int_0^1 frac{1}{u} du = frac{1}{3} ln |u| + C$$

But what about the right?

calculus integration

share|cite|improve this question

edited 34 mins ago

Eevee Trainer

6,54811237

asked 43 mins ago

Jwan622Jwan622

2,16211530

$endgroup$

share|cite|improve this question

edited 34 mins ago

Eevee Trainer

6,54811237

asked 43 mins ago

Jwan622Jwan622

2,16211530

share|cite|improve this question

edited 34 mins ago

Eevee Trainer

6,54811237

asked 43 mins ago

Jwan622Jwan622

2,16211530

edited 34 mins ago

Eevee Trainer

6,54811237

edited 34 mins ago

Eevee Trainer

6,54811237

asked 43 mins ago

Jwan622Jwan622

2,16211530

asked 43 mins ago

Jwan622Jwan622

2,16211530

2 Answers
2

active

oldest

votes

6

$begingroup$

Hint:

$$frac{12t}{1+3t} = 4 - frac{4}{1+3t}.$$ Now use the same approach as you did for the first part.
In fact you should’ve simplified the fraction before splitting into parts.

Remember to evaluate the integral at t=0 and t=1! ;)

share|cite|improve this answer

edited 23 mins ago

Jyrki Lahtonen

109k13169374

answered 39 mins ago

Gareth MaGareth Ma

416213

$endgroup$

add a comment | 

1

$begingroup$

At least on an initial glance, I'd make the substitution

$$u = 1+3t implies du = 3dt implies dt = frac{du}{3}$$

Note that $t = (u-1)/3$ and thus $12t = 4(u-1)$.

Then the integral becomes

$$int_0^1 frac{12t}{1 + 3t} dt = int_ast^ast frac{4(u-1)}{u} frac{dt}{3} = frac 4 3 int_ast^ast frac{u-1}{u}du = frac 4 3 int_ast^ast 1 - frac{1}{u}du$$

But this seems a bit excessive and unnecessary, even if it's the first thing to come to mind. (That's why I left the bounds as $ast$'s: in cases like these, overcomplication suggests an alternative approach.)

---

Another trick is to "add and subtract the same thing from the top"; this can be handy in a lot of math classes for manipulations if you add and subtract the right thing. In this case, we notice:

$$12t = 12t + 4 - 4 = 4(3t+1) - 4$$

Thus,

$$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$

share|cite|improve this answer

answered 35 mins ago

Eevee TrainerEevee Trainer

6,54811237

$endgroup$

add a comment | 

Your Answer

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3122196%2fintegral-problem-unsure-of-the-approach%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

Hint:

$$frac{12t}{1+3t} = 4 - frac{4}{1+3t}.$$ Now use the same approach as you did for the first part.
In fact you should’ve simplified the fraction before splitting into parts.

Remember to evaluate the integral at t=0 and t=1! ;)

share|cite|improve this answer

edited 23 mins ago

Jyrki Lahtonen

109k13169374

answered 39 mins ago

Gareth MaGareth Ma

416213

$endgroup$

add a comment | 

6

$begingroup$

Hint:

$$frac{12t}{1+3t} = 4 - frac{4}{1+3t}.$$ Now use the same approach as you did for the first part.
In fact you should’ve simplified the fraction before splitting into parts.

Remember to evaluate the integral at t=0 and t=1! ;)

share|cite|improve this answer

edited 23 mins ago

Jyrki Lahtonen

109k13169374

answered 39 mins ago

Gareth MaGareth Ma

416213

$endgroup$

add a comment | 

$begingroup$

Hint:

$$frac{12t}{1+3t} = 4 - frac{4}{1+3t}.$$ Now use the same approach as you did for the first part.
In fact you should’ve simplified the fraction before splitting into parts.

Remember to evaluate the integral at t=0 and t=1! ;)

share|cite|improve this answer

edited 23 mins ago

Jyrki Lahtonen

109k13169374

answered 39 mins ago

Gareth MaGareth Ma

416213

$endgroup$

share|cite|improve this answer

edited 23 mins ago

Jyrki Lahtonen

109k13169374

answered 39 mins ago

Gareth MaGareth Ma

416213

edited 23 mins ago

Jyrki Lahtonen

109k13169374

edited 23 mins ago

Jyrki Lahtonen

109k13169374

answered 39 mins ago

Gareth MaGareth Ma

416213

answered 39 mins ago

Gareth MaGareth Ma

416213

1

$begingroup$

At least on an initial glance, I'd make the substitution

$$u = 1+3t implies du = 3dt implies dt = frac{du}{3}$$

Note that $t = (u-1)/3$ and thus $12t = 4(u-1)$.

Then the integral becomes

$$int_0^1 frac{12t}{1 + 3t} dt = int_ast^ast frac{4(u-1)}{u} frac{dt}{3} = frac 4 3 int_ast^ast frac{u-1}{u}du = frac 4 3 int_ast^ast 1 - frac{1}{u}du$$

But this seems a bit excessive and unnecessary, even if it's the first thing to come to mind. (That's why I left the bounds as $ast$'s: in cases like these, overcomplication suggests an alternative approach.)

---

Another trick is to "add and subtract the same thing from the top"; this can be handy in a lot of math classes for manipulations if you add and subtract the right thing. In this case, we notice:

$$12t = 12t + 4 - 4 = 4(3t+1) - 4$$

Thus,

$$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$

share|cite|improve this answer

answered 35 mins ago

Eevee TrainerEevee Trainer

6,54811237

$endgroup$

add a comment | 

1

$begingroup$

At least on an initial glance, I'd make the substitution

$$u = 1+3t implies du = 3dt implies dt = frac{du}{3}$$

Note that $t = (u-1)/3$ and thus $12t = 4(u-1)$.

Then the integral becomes

$$int_0^1 frac{12t}{1 + 3t} dt = int_ast^ast frac{4(u-1)}{u} frac{dt}{3} = frac 4 3 int_ast^ast frac{u-1}{u}du = frac 4 3 int_ast^ast 1 - frac{1}{u}du$$

But this seems a bit excessive and unnecessary, even if it's the first thing to come to mind. (That's why I left the bounds as $ast$'s: in cases like these, overcomplication suggests an alternative approach.)

---

Another trick is to "add and subtract the same thing from the top"; this can be handy in a lot of math classes for manipulations if you add and subtract the right thing. In this case, we notice:

$$12t = 12t + 4 - 4 = 4(3t+1) - 4$$

Thus,

$$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$

share|cite|improve this answer

answered 35 mins ago

Eevee TrainerEevee Trainer

6,54811237

$endgroup$

add a comment | 

$begingroup$

At least on an initial glance, I'd make the substitution

$$u = 1+3t implies du = 3dt implies dt = frac{du}{3}$$

Note that $t = (u-1)/3$ and thus $12t = 4(u-1)$.

Then the integral becomes

$$int_0^1 frac{12t}{1 + 3t} dt = int_ast^ast frac{4(u-1)}{u} frac{dt}{3} = frac 4 3 int_ast^ast frac{u-1}{u}du = frac 4 3 int_ast^ast 1 - frac{1}{u}du$$

But this seems a bit excessive and unnecessary, even if it's the first thing to come to mind. (That's why I left the bounds as $ast$'s: in cases like these, overcomplication suggests an alternative approach.)

---

Another trick is to "add and subtract the same thing from the top"; this can be handy in a lot of math classes for manipulations if you add and subtract the right thing. In this case, we notice:

$$12t = 12t + 4 - 4 = 4(3t+1) - 4$$

Thus,

$$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$

share|cite|improve this answer

answered 35 mins ago

Eevee TrainerEevee Trainer

6,54811237

$endgroup$

share|cite|improve this answer

answered 35 mins ago

Eevee TrainerEevee Trainer

6,54811237

answered 35 mins ago

Eevee TrainerEevee Trainer

6,54811237

answered 35 mins ago

Eevee TrainerEevee Trainer

6,54811237