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Integral problem. Unsure of the approach.

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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 31 mins ago

Eevee Trainer

6,54811237

asked 40 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 31 mins ago

Eevee Trainer

6,54811237

asked 40 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 31 mins ago

Eevee Trainer

6,54811237

asked 40 mins ago

Jwan622Jwan622

2,16211530

$endgroup$

share|cite|improve this question

edited 31 mins ago

Eevee Trainer

6,54811237

asked 40 mins ago

Jwan622Jwan622

2,16211530

share|cite|improve this question

edited 31 mins ago

Eevee Trainer

6,54811237

asked 40 mins ago

Jwan622Jwan622

2,16211530

edited 31 mins ago

Eevee Trainer

6,54811237

edited 31 mins ago

Eevee Trainer

6,54811237

asked 40 mins ago

Jwan622Jwan622

2,16211530

asked 40 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 21 mins ago

Jyrki Lahtonen

109k13169374

answered 37 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 32 mins ago

Eevee TrainerEevee Trainer

6,54811237

$endgroup$

add a comment | 

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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 21 mins ago

Jyrki Lahtonen

109k13169374

answered 37 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 21 mins ago

Jyrki Lahtonen

109k13169374

answered 37 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 21 mins ago

Jyrki Lahtonen

109k13169374

answered 37 mins ago

Gareth MaGareth Ma

416213

$endgroup$

share|cite|improve this answer

edited 21 mins ago

Jyrki Lahtonen

109k13169374

answered 37 mins ago

Gareth MaGareth Ma

416213

edited 21 mins ago

Jyrki Lahtonen

109k13169374

edited 21 mins ago

Jyrki Lahtonen

109k13169374

answered 37 mins ago

Gareth MaGareth Ma

416213

answered 37 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 32 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 32 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 32 mins ago

Eevee TrainerEevee Trainer

6,54811237

$endgroup$

share|cite|improve this answer

answered 32 mins ago

Eevee TrainerEevee Trainer

6,54811237

answered 32 mins ago

Eevee TrainerEevee Trainer

6,54811237

answered 32 mins ago

Eevee TrainerEevee Trainer

6,54811237