Integral problem. Unsure of the approach.
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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
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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
$endgroup$
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
$endgroup$
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
calculus integration
edited 1 hour ago
Eevee Trainer
6,55811237
6,55811237
asked 1 hour ago
Jwan622Jwan622
2,16211530
2,16211530
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add a comment |
2 Answers
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$begingroup$
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! ;)
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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,
$$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$
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2 Answers
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2 Answers
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$begingroup$
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! ;)
$endgroup$
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! ;)
$endgroup$
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! ;)
$endgroup$
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! ;)
edited 1 hour ago
Jyrki Lahtonen
109k13169374
109k13169374
answered 1 hour ago
Gareth MaGareth Ma
426213
426213
add a comment |
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,
$$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$
$endgroup$
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,
$$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$
$endgroup$
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,
$$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$
$endgroup$
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,
$$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$
answered 1 hour ago
Eevee TrainerEevee Trainer
6,55811237
6,55811237
add a comment |
add a comment |
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