Mixing
How do I integrate the function $sqrt{(6x + 2)}$?
$begingroup$
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 Jan 5 at 2:26
Eevee Trainer
5,7471936
asked Jan 5 at 2:06
bcloneybcloney
264
$endgroup$
|
show 6 more comments
$begingroup$
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 Jan 5 at 2:26
Eevee Trainer
5,7471936
asked Jan 5 at 2:06
bcloneybcloney
264
$endgroup$
1
$begingroup$
Welcome to Math.Stackexchange! What approaches have to tried so far?
$endgroup$
– user458276
Jan 5 at 2:07
3
$begingroup$
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
$endgroup$
– JMoravitz
Jan 5 at 2:08
$begingroup$
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
$endgroup$
– bcloney
Jan 5 at 2:12
1
$begingroup$
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 $sqrt x = x^{1/2}$.
$endgroup$
– Eevee Trainer
Jan 5 at 2:17
1
$begingroup$
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 $x^{1/2}$.
$endgroup$
– JMoravitz
Jan 5 at 2:25
|
show 6 more comments
$begingroup$
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 Jan 5 at 2:26
Eevee Trainer
5,7471936
asked Jan 5 at 2:06
bcloneybcloney
264
$endgroup$
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 Jan 5 at 2:26
Eevee Trainer
5,7471936
asked Jan 5 at 2:06
bcloneybcloney
264
edited Jan 5 at 2:26
Eevee Trainer
5,7471936
asked Jan 5 at 2:06
bcloneybcloney
264
edited Jan 5 at 2:26
Eevee Trainer
5,7471936
edited Jan 5 at 2:26
Eevee Trainer
5,7471936
edited Jan 5 at 2:26
Eevee Trainer
5,7471936
5,7471936
asked Jan 5 at 2:06
bcloneybcloney
264
asked Jan 5 at 2:06
bcloneybcloney
264
asked Jan 5 at 2:06
bcloneybcloney
264
264
1
$begingroup$
Welcome to Math.Stackexchange! What approaches have to tried so far?
$endgroup$
– user458276
Jan 5 at 2:07
3
$begingroup$
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
$endgroup$
– JMoravitz
Jan 5 at 2:08
$begingroup$
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
$endgroup$
– bcloney
Jan 5 at 2:12
1
$begingroup$
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 $sqrt x = x^{1/2}$.
$endgroup$
– Eevee Trainer
Jan 5 at 2:17
1
$begingroup$
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 $x^{1/2}$.
$endgroup$
– JMoravitz
Jan 5 at 2:25
|
show 6 more comments
1
$begingroup$
Welcome to Math.Stackexchange! What approaches have to tried so far?
$endgroup$
– user458276
Jan 5 at 2:07
3
$begingroup$
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
$endgroup$
– JMoravitz
Jan 5 at 2:08
$begingroup$
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
$endgroup$
– bcloney
Jan 5 at 2:12
1
$begingroup$
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 $sqrt x = x^{1/2}$.
$endgroup$
– Eevee Trainer
Jan 5 at 2:17
1
$begingroup$
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 $x^{1/2}$.
$endgroup$
– JMoravitz
Jan 5 at 2:25
1
1
$begingroup$
Welcome to Math.Stackexchange! What approaches have to tried so far?
$endgroup$
– user458276
Jan 5 at 2:07
$begingroup$
Welcome to Math.Stackexchange! What approaches have to tried so far?
$endgroup$
– user458276
Jan 5 at 2:07
3
3
$begingroup$
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
$endgroup$
– JMoravitz
Jan 5 at 2:08
$begingroup$
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
$endgroup$
– JMoravitz
Jan 5 at 2:08
$begingroup$
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
$endgroup$
– bcloney
Jan 5 at 2:12
$begingroup$
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
$endgroup$
– bcloney
Jan 5 at 2:12
1
1
$begingroup$
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 $sqrt x = x^{1/2}$.
$endgroup$
– Eevee Trainer
Jan 5 at 2:17
$begingroup$
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 $sqrt x = x^{1/2}$.
$endgroup$
– Eevee Trainer
Jan 5 at 2:17
1
1
$begingroup$
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 $x^{1/2}$.
$endgroup$
– JMoravitz
Jan 5 at 2:25
$begingroup$
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 $x^{1/2}$.
$endgroup$
– JMoravitz
Jan 5 at 2:25
|
show 6 more comments
3 Answers
3
active
oldest
votes
$begingroup$
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 Jan 5 at 2:07
Eevee TrainerEevee Trainer
5,7471936
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 2:24
Rhys HughesRhys Hughes
5,7861528
$endgroup$
add a comment |
$begingroup$
$$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.
answered Jan 5 at 3:20
Chris CusterChris Custer
11.5k3824
$endgroup$
add a comment |
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3 Answers
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3 Answers
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active
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votes
$begingroup$
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 Jan 5 at 2:07
Eevee TrainerEevee Trainer
5,7471936
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 2:07
Eevee TrainerEevee Trainer
5,7471936
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 2:07
Eevee TrainerEevee Trainer
5,7471936
$endgroup$
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 Jan 5 at 2:07
Eevee TrainerEevee Trainer
5,7471936
edited Jan 5 at 2:31
answered Jan 5 at 2:07
Eevee TrainerEevee Trainer
5,7471936
answered Jan 5 at 2:07
Eevee TrainerEevee Trainer
5,7471936
answered Jan 5 at 2:07
Eevee TrainerEevee Trainer
5,7471936
5,7471936
add a comment |
add a comment |
$begingroup$
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 Jan 5 at 2:24
Rhys HughesRhys Hughes
5,7861528
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 2:24
Rhys HughesRhys Hughes
5,7861528
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 2:24
Rhys HughesRhys Hughes
5,7861528
$endgroup$
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 Jan 5 at 2:24
Rhys HughesRhys Hughes
5,7861528
answered Jan 5 at 2:24
Rhys HughesRhys Hughes
5,7861528
answered Jan 5 at 2:24
Rhys HughesRhys Hughes
5,7861528
answered Jan 5 at 2:24
Rhys HughesRhys Hughes
5,7861528
5,7861528
add a comment |
add a comment |
$begingroup$
$$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.
answered Jan 5 at 3:20
Chris CusterChris Custer
11.5k3824
$endgroup$
add a comment |
$begingroup$
$$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.
answered Jan 5 at 3:20
Chris CusterChris Custer
11.5k3824
$endgroup$
add a comment |
$begingroup$
$$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.
answered Jan 5 at 3:20
Chris CusterChris Custer
11.5k3824
$endgroup$
$$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.
answered Jan 5 at 3:20
Chris CusterChris Custer
11.5k3824
answered Jan 5 at 3:20
Chris CusterChris Custer
11.5k3824
answered Jan 5 at 3:20
Chris CusterChris Custer
11.5k3824
answered Jan 5 at 3:20
Chris CusterChris Custer
11.5k3824
11.5k3824
add a comment |
add a comment |
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1
$begingroup$
Welcome to Math.Stackexchange! What approaches have to tried so far?
$endgroup$
– user458276
Jan 5 at 2:07
3
$begingroup$
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
$endgroup$
– JMoravitz
Jan 5 at 2:08
$begingroup$
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
$endgroup$
– bcloney
Jan 5 at 2:12
1
$begingroup$
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 $sqrt x = x^{1/2}$.
$endgroup$
– Eevee Trainer
Jan 5 at 2:17
1
$begingroup$
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 $x^{1/2}$.
$endgroup$
– JMoravitz
Jan 5 at 2:25