Evaluate the integral using the substitution rule
The integral is $$int_0^4 frac{2t}{sqrt{1+2t}} ,dt$$
I set $u = 1+2t$ and got that $du = 2 dt$, which makes $dt = du/2$. After this, I am confused as to what to do because $2t$ is still in the equation.
(Also, I'm not sure how to format the equation on here so I apologize if there is any confusion)
calculus definite-integrals
edited Nov 28 '18 at 11:58
N. F. Taussig
43.6k93355
asked Nov 27 '18 at 19:59
CodeGuy7153
92
add a comment |
The integral is $$int_0^4 frac{2t}{sqrt{1+2t}} ,dt$$
I set $u = 1+2t$ and got that $du = 2 dt$, which makes $dt = du/2$. After this, I am confused as to what to do because $2t$ is still in the equation.
(Also, I'm not sure how to format the equation on here so I apologize if there is any confusion)
calculus definite-integrals
edited Nov 28 '18 at 11:58
N. F. Taussig
43.6k93355
asked Nov 27 '18 at 19:59
CodeGuy7153
92
You started correctly, and you are aware of importance to have only $u$ after the substitution. Good! Now replace in your integral $2t$ by $u-1$ and you can continue
– user376343
Nov 27 '18 at 20:05
If $u=2t+1$ then $2t=u-1$ right? Replace it and it's done.
– Ramiro Scorolli
Nov 27 '18 at 20:06
add a comment |
The integral is $$int_0^4 frac{2t}{sqrt{1+2t}} ,dt$$
I set $u = 1+2t$ and got that $du = 2 dt$, which makes $dt = du/2$. After this, I am confused as to what to do because $2t$ is still in the equation.
(Also, I'm not sure how to format the equation on here so I apologize if there is any confusion)
calculus definite-integrals
edited Nov 28 '18 at 11:58
N. F. Taussig
43.6k93355
asked Nov 27 '18 at 19:59
CodeGuy7153
92
The integral is $$int_0^4 frac{2t}{sqrt{1+2t}} ,dt$$
I set $u = 1+2t$ and got that $du = 2 dt$, which makes $dt = du/2$. After this, I am confused as to what to do because $2t$ is still in the equation.
(Also, I'm not sure how to format the equation on here so I apologize if there is any confusion)
calculus definite-integrals
calculus definite-integrals
edited Nov 28 '18 at 11:58
N. F. Taussig
43.6k93355
asked Nov 27 '18 at 19:59
CodeGuy7153
92
edited Nov 28 '18 at 11:58
N. F. Taussig
43.6k93355
asked Nov 27 '18 at 19:59
CodeGuy7153
92
edited Nov 28 '18 at 11:58
N. F. Taussig
43.6k93355
edited Nov 28 '18 at 11:58
N. F. Taussig
43.6k93355
edited Nov 28 '18 at 11:58
N. F. Taussig
43.6k93355
43.6k93355
asked Nov 27 '18 at 19:59
CodeGuy7153
92
asked Nov 27 '18 at 19:59
CodeGuy7153
92
asked Nov 27 '18 at 19:59
CodeGuy7153
92
92
You started correctly, and you are aware of importance to have only $u$ after the substitution. Good! Now replace in your integral $2t$ by $u-1$ and you can continue
– user376343
Nov 27 '18 at 20:05
If $u=2t+1$ then $2t=u-1$ right? Replace it and it's done.
– Ramiro Scorolli
Nov 27 '18 at 20:06
add a comment |
You started correctly, and you are aware of importance to have only $u$ after the substitution. Good! Now replace in your integral $2t$ by $u-1$ and you can continue
– user376343
Nov 27 '18 at 20:05
If $u=2t+1$ then $2t=u-1$ right? Replace it and it's done.
– Ramiro Scorolli
Nov 27 '18 at 20:06
You started correctly, and you are aware of importance to have only $u$ after the substitution. Good! Now replace in your integral $2t$ by $u-1$ and you can continue
– user376343
Nov 27 '18 at 20:05
You started correctly, and you are aware of importance to have only $u$ after the substitution. Good! Now replace in your integral $2t$ by $u-1$ and you can continue
– user376343
Nov 27 '18 at 20:05
If $u=2t+1$ then $2t=u-1$ right? Replace it and it's done.
– Ramiro Scorolli
Nov 27 '18 at 20:06
If $u=2t+1$ then $2t=u-1$ right? Replace it and it's done.
– Ramiro Scorolli
Nov 27 '18 at 20:06
add a comment |
3 Answers
3
active
oldest
votes
HINT
Note that $2t = u-1$ and you get
$$
int frac{u-1}{sqrt{u}}frac{du}{2}
$$
which cleanly splits into 2 integrals
answered Nov 27 '18 at 20:03
gt6989b
33k22452
I can't believe I missed that, thank you!
– CodeGuy7153
Nov 27 '18 at 20:10
add a comment |
Keep going.
Well, students forget that the sub they make can be manipulated. Let $u = 1 + 2t$, so then $2t = u -1.$ Now the integral is
$$ int_1 ^9 frac{u-1}{2sqrt{u}} du $$
which is easy to do once you properly separate the fraction.
answered Nov 27 '18 at 20:04
Sean Roberson
6,40531327
add a comment |
An other aproach
$$I=int_0^4frac{1+2t-1}{sqrt{1+2t}}dt=$$
$$int_0^4sqrt{1+2t}dt-int_0^4frac{dt}{sqrt{1+2t}}$$
answered Nov 27 '18 at 20:09
hamam_Abdallah
38k21634
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
HINT
Note that $2t = u-1$ and you get
$$
int frac{u-1}{sqrt{u}}frac{du}{2}
$$
which cleanly splits into 2 integrals
answered Nov 27 '18 at 20:03
gt6989b
33k22452
I can't believe I missed that, thank you!
– CodeGuy7153
Nov 27 '18 at 20:10
add a comment |
HINT
Note that $2t = u-1$ and you get
$$
int frac{u-1}{sqrt{u}}frac{du}{2}
$$
which cleanly splits into 2 integrals
answered Nov 27 '18 at 20:03
gt6989b
33k22452
I can't believe I missed that, thank you!
– CodeGuy7153
Nov 27 '18 at 20:10
add a comment |
HINT
Note that $2t = u-1$ and you get
$$
int frac{u-1}{sqrt{u}}frac{du}{2}
$$
which cleanly splits into 2 integrals
answered Nov 27 '18 at 20:03
gt6989b
33k22452
HINT
Note that $2t = u-1$ and you get
$$
int frac{u-1}{sqrt{u}}frac{du}{2}
$$
which cleanly splits into 2 integrals
answered Nov 27 '18 at 20:03
gt6989b
33k22452
answered Nov 27 '18 at 20:03
gt6989b
33k22452
answered Nov 27 '18 at 20:03
gt6989b
33k22452
answered Nov 27 '18 at 20:03
gt6989b
33k22452
33k22452
I can't believe I missed that, thank you!
– CodeGuy7153
Nov 27 '18 at 20:10
add a comment |
I can't believe I missed that, thank you!
– CodeGuy7153
Nov 27 '18 at 20:10
I can't believe I missed that, thank you!
– CodeGuy7153
Nov 27 '18 at 20:10
I can't believe I missed that, thank you!
– CodeGuy7153
Nov 27 '18 at 20:10
add a comment |
Keep going.
Well, students forget that the sub they make can be manipulated. Let $u = 1 + 2t$, so then $2t = u -1.$ Now the integral is
$$ int_1 ^9 frac{u-1}{2sqrt{u}} du $$
which is easy to do once you properly separate the fraction.
answered Nov 27 '18 at 20:04
Sean Roberson
6,40531327
add a comment |
Keep going.
Well, students forget that the sub they make can be manipulated. Let $u = 1 + 2t$, so then $2t = u -1.$ Now the integral is
$$ int_1 ^9 frac{u-1}{2sqrt{u}} du $$
which is easy to do once you properly separate the fraction.
answered Nov 27 '18 at 20:04
Sean Roberson
6,40531327
add a comment |
Keep going.
Well, students forget that the sub they make can be manipulated. Let $u = 1 + 2t$, so then $2t = u -1.$ Now the integral is
$$ int_1 ^9 frac{u-1}{2sqrt{u}} du $$
which is easy to do once you properly separate the fraction.
answered Nov 27 '18 at 20:04
Sean Roberson
6,40531327
Keep going.
Well, students forget that the sub they make can be manipulated. Let $u = 1 + 2t$, so then $2t = u -1.$ Now the integral is
$$ int_1 ^9 frac{u-1}{2sqrt{u}} du $$
which is easy to do once you properly separate the fraction.
answered Nov 27 '18 at 20:04
Sean Roberson
6,40531327
answered Nov 27 '18 at 20:04
Sean Roberson
6,40531327
answered Nov 27 '18 at 20:04
Sean Roberson
6,40531327
answered Nov 27 '18 at 20:04
Sean Roberson
6,40531327
6,40531327
add a comment |
add a comment |
An other aproach
$$I=int_0^4frac{1+2t-1}{sqrt{1+2t}}dt=$$
$$int_0^4sqrt{1+2t}dt-int_0^4frac{dt}{sqrt{1+2t}}$$
answered Nov 27 '18 at 20:09
hamam_Abdallah
38k21634
add a comment |
An other aproach
$$I=int_0^4frac{1+2t-1}{sqrt{1+2t}}dt=$$
$$int_0^4sqrt{1+2t}dt-int_0^4frac{dt}{sqrt{1+2t}}$$
answered Nov 27 '18 at 20:09
hamam_Abdallah
38k21634
add a comment |
An other aproach
$$I=int_0^4frac{1+2t-1}{sqrt{1+2t}}dt=$$
$$int_0^4sqrt{1+2t}dt-int_0^4frac{dt}{sqrt{1+2t}}$$
answered Nov 27 '18 at 20:09
hamam_Abdallah
38k21634
An other aproach
$$I=int_0^4frac{1+2t-1}{sqrt{1+2t}}dt=$$
$$int_0^4sqrt{1+2t}dt-int_0^4frac{dt}{sqrt{1+2t}}$$
answered Nov 27 '18 at 20:09
hamam_Abdallah
38k21634
answered Nov 27 '18 at 20:09
hamam_Abdallah
38k21634
answered Nov 27 '18 at 20:09
hamam_Abdallah
38k21634
answered Nov 27 '18 at 20:09
hamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
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You started correctly, and you are aware of importance to have only $u$ after the substitution. Good! Now replace in your integral $2t$ by $u-1$ and you can continue
– user376343
Nov 27 '18 at 20:05
If $u=2t+1$ then $2t=u-1$ right? Replace it and it's done.
– Ramiro Scorolli
Nov 27 '18 at 20:06