How do I integrate the function $sqrt{(6x + 2)}$?
How do you integrate $sqrt{(6x + 2)}$?
I've tried to use the following substitutions: let $x = sin(u)$ and $dx = cos(u)$ (along the lines of the Yahoo Answers link). I tried looking for simple examples of integrals with square roots on Yahoo Answers and elsewhere by Googling, but couldn't find any simpler ones, and that substitution got me nowhere.
calculus integration indefinite-integrals
edited yesterday
Eevee Trainer
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asked yesterday
bcloney
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show 6 more comments
How do you integrate $sqrt{(6x + 2)}$?
I've tried to use the following substitutions: let $x = sin(u)$ and $dx = cos(u)$ (along the lines of the Yahoo Answers link). I tried looking for simple examples of integrals with square roots on Yahoo Answers and elsewhere by Googling, but couldn't find any simpler ones, and that substitution got me nowhere.
calculus integration indefinite-integrals
edited yesterday
Eevee Trainer
4,9971634
asked yesterday
bcloney
264
New contributor
bcloney is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Welcome to Math.Stackexchange! What approaches have to tried so far?
– user458276
yesterday
3
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
– JMoravitz
yesterday
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
– bcloney
yesterday
1
Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
– Eevee Trainer
yesterday
1
The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
– JMoravitz
yesterday
|
show 6 more comments
How do you integrate $sqrt{(6x + 2)}$?
I've tried to use the following substitutions: let $x = sin(u)$ and $dx = cos(u)$ (along the lines of the Yahoo Answers link). I tried looking for simple examples of integrals with square roots on Yahoo Answers and elsewhere by Googling, but couldn't find any simpler ones, and that substitution got me nowhere.
calculus integration indefinite-integrals
edited yesterday
Eevee Trainer
4,9971634
asked yesterday
bcloney
264
New contributor
bcloney is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How do you integrate $sqrt{(6x + 2)}$?
I've tried to use the following substitutions: let $x = sin(u)$ and $dx = cos(u)$ (along the lines of the Yahoo Answers link). I tried looking for simple examples of integrals with square roots on Yahoo Answers and elsewhere by Googling, but couldn't find any simpler ones, and that substitution got me nowhere.
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited yesterday
Eevee Trainer
4,9971634
asked yesterday
bcloney
264
New contributor
bcloney is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Eevee Trainer
4,9971634
asked yesterday
bcloney
264
New contributor
bcloney is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Eevee Trainer
4,9971634
edited yesterday
Eevee Trainer
4,9971634
edited yesterday
Eevee Trainer
4,9971634
4,9971634
asked yesterday
bcloney
264
New contributor
bcloney is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
bcloney
264
asked yesterday
bcloney
264
264
New contributor
bcloney is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
bcloney is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
bcloney is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Welcome to Math.Stackexchange! What approaches have to tried so far?
– user458276
yesterday
3
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
– JMoravitz
yesterday
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
– bcloney
yesterday
1
Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
– Eevee Trainer
yesterday
1
The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
– JMoravitz
yesterday
|
show 6 more comments
1
Welcome to Math.Stackexchange! What approaches have to tried so far?
– user458276
yesterday
3
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
– JMoravitz
yesterday
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
– bcloney
yesterday
1
Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
– Eevee Trainer
yesterday
1
The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
– JMoravitz
yesterday
1
1
Welcome to Math.Stackexchange! What approaches have to tried so far?
– user458276
yesterday
Welcome to Math.Stackexchange! What approaches have to tried so far?
– user458276
yesterday
3
3
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
– JMoravitz
yesterday
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
– JMoravitz
yesterday
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
– bcloney
yesterday
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
– bcloney
yesterday
1
1
Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
– Eevee Trainer
yesterday
Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
– Eevee Trainer
yesterday
1
1
The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
– JMoravitz
yesterday
The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
– JMoravitz
yesterday
|
show 6 more comments
3 Answers
3
active
oldest
votes
Hints:
- Make the $u$-substitution $u = 6x+2$.
- Don't forget that $sqrt u = u^{1/2}$ and that, for all $n neq -1$, we have
$$int x^n dx = frac{x^{n+1}}{n+1} + C$$
answered yesterday
Eevee Trainer
4,9971634
1
Roll Tide!!. I am a big fan of Alabama!!.
– Satish Ramanathan
yesterday
Thanks, I'll try that
– bcloney
yesterday
add a comment |
You are looking for:
$$int(6x+2)^frac12 dx$$
Notice that if we let $u=6x+2$, $frac{du}{dx}=6$ which leads to $dx=frac16 du$
In other words, the above integral is exactly the same as this:
$$int (u)^frac12 cdot frac 16 du$$
You can take the constant outside the integral to make this:
$$frac 16 int u^frac 12 du$$
And deal with that integral using the normal power rules.
At the end, don't forget to resubsitute $u=6x+2$ back in!
answered yesterday
Rhys Hughes
5,0441427
add a comment |
$$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.
answered yesterday
Chris Custer
10.9k3824
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hints:
- Make the $u$-substitution $u = 6x+2$.
- Don't forget that $sqrt u = u^{1/2}$ and that, for all $n neq -1$, we have
$$int x^n dx = frac{x^{n+1}}{n+1} + C$$
answered yesterday
Eevee Trainer
4,9971634
1
Roll Tide!!. I am a big fan of Alabama!!.
– Satish Ramanathan
yesterday
Thanks, I'll try that
– bcloney
yesterday
add a comment |
Hints:
- Make the $u$-substitution $u = 6x+2$.
- Don't forget that $sqrt u = u^{1/2}$ and that, for all $n neq -1$, we have
$$int x^n dx = frac{x^{n+1}}{n+1} + C$$
answered yesterday
Eevee Trainer
4,9971634
1
Roll Tide!!. I am a big fan of Alabama!!.
– Satish Ramanathan
yesterday
Thanks, I'll try that
– bcloney
yesterday
add a comment |
Hints:
- Make the $u$-substitution $u = 6x+2$.
- Don't forget that $sqrt u = u^{1/2}$ and that, for all $n neq -1$, we have
$$int x^n dx = frac{x^{n+1}}{n+1} + C$$
answered yesterday
Eevee Trainer
4,9971634
Hints:
- Make the $u$-substitution $u = 6x+2$.
- Don't forget that $sqrt u = u^{1/2}$ and that, for all $n neq -1$, we have
$$int x^n dx = frac{x^{n+1}}{n+1} + C$$
answered yesterday
Eevee Trainer
4,9971634
edited yesterday
answered yesterday
Eevee Trainer
4,9971634
answered yesterday
Eevee Trainer
4,9971634
answered yesterday
Eevee Trainer
4,9971634
4,9971634
1
Roll Tide!!. I am a big fan of Alabama!!.
– Satish Ramanathan
yesterday
Thanks, I'll try that
– bcloney
yesterday
add a comment |
1
Roll Tide!!. I am a big fan of Alabama!!.
– Satish Ramanathan
yesterday
Thanks, I'll try that
– bcloney
yesterday
1
1
Roll Tide!!. I am a big fan of Alabama!!.
– Satish Ramanathan
yesterday
Roll Tide!!. I am a big fan of Alabama!!.
– Satish Ramanathan
yesterday
Thanks, I'll try that
– bcloney
yesterday
Thanks, I'll try that
– bcloney
yesterday
add a comment |
You are looking for:
$$int(6x+2)^frac12 dx$$
Notice that if we let $u=6x+2$, $frac{du}{dx}=6$ which leads to $dx=frac16 du$
In other words, the above integral is exactly the same as this:
$$int (u)^frac12 cdot frac 16 du$$
You can take the constant outside the integral to make this:
$$frac 16 int u^frac 12 du$$
And deal with that integral using the normal power rules.
At the end, don't forget to resubsitute $u=6x+2$ back in!
answered yesterday
Rhys Hughes
5,0441427
add a comment |
You are looking for:
$$int(6x+2)^frac12 dx$$
Notice that if we let $u=6x+2$, $frac{du}{dx}=6$ which leads to $dx=frac16 du$
In other words, the above integral is exactly the same as this:
$$int (u)^frac12 cdot frac 16 du$$
You can take the constant outside the integral to make this:
$$frac 16 int u^frac 12 du$$
And deal with that integral using the normal power rules.
At the end, don't forget to resubsitute $u=6x+2$ back in!
answered yesterday
Rhys Hughes
5,0441427
add a comment |
You are looking for:
$$int(6x+2)^frac12 dx$$
Notice that if we let $u=6x+2$, $frac{du}{dx}=6$ which leads to $dx=frac16 du$
In other words, the above integral is exactly the same as this:
$$int (u)^frac12 cdot frac 16 du$$
You can take the constant outside the integral to make this:
$$frac 16 int u^frac 12 du$$
And deal with that integral using the normal power rules.
At the end, don't forget to resubsitute $u=6x+2$ back in!
answered yesterday
Rhys Hughes
5,0441427
You are looking for:
$$int(6x+2)^frac12 dx$$
Notice that if we let $u=6x+2$, $frac{du}{dx}=6$ which leads to $dx=frac16 du$
In other words, the above integral is exactly the same as this:
$$int (u)^frac12 cdot frac 16 du$$
You can take the constant outside the integral to make this:
$$frac 16 int u^frac 12 du$$
And deal with that integral using the normal power rules.
At the end, don't forget to resubsitute $u=6x+2$ back in!
answered yesterday
Rhys Hughes
5,0441427
answered yesterday
Rhys Hughes
5,0441427
answered yesterday
Rhys Hughes
5,0441427
answered yesterday
Rhys Hughes
5,0441427
5,0441427
add a comment |
add a comment |
$$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.
answered yesterday
Chris Custer
10.9k3824
add a comment |
$$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.
answered yesterday
Chris Custer
10.9k3824
add a comment |
$$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.
answered yesterday
Chris Custer
10.9k3824
$$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.
answered yesterday
Chris Custer
10.9k3824
answered yesterday
Chris Custer
10.9k3824
answered yesterday
Chris Custer
10.9k3824
answered yesterday
Chris Custer
10.9k3824
10.9k3824
add a comment |
add a comment |
bcloney is a new contributor. Be nice, and check out our Code of Conduct.
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1
Welcome to Math.Stackexchange! What approaches have to tried so far?
– user458276
yesterday
3
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
– JMoravitz
yesterday
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
– bcloney
yesterday
1
Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
– Eevee Trainer
yesterday
1
The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
– JMoravitz
yesterday