Having trouble computing $int_3^5fract1+0.1t dt $
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$begingroup$
$$int_3^5fract1+0.1t dt $$
For some reason this is equal to:
1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))
I have no idea how to reduce to that.
integration definite-integrals
edited Feb 20 at 2:20
Michael Rybkin
3,884420
asked Feb 18 at 13:51
ximxim
516
$endgroup$
add a comment |
$begingroup$
$$int_3^5fract1+0.1t dt $$
For some reason this is equal to:
1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))
I have no idea how to reduce to that.
integration definite-integrals
edited Feb 20 at 2:20
Michael Rybkin
3,884420
asked Feb 18 at 13:51
ximxim
516
$endgroup$
$begingroup$
If you let $x=1+0.1t$, then $t=10x-10$...
$endgroup$
– Eleven-Eleven
Feb 18 at 13:53
$begingroup$
Do you mean $$int_3^5fract1+frac110tdt$$?
$endgroup$
– Dr. Sonnhard Graubner
Feb 18 at 13:54
$begingroup$
This is not an improper integral, by the way. So, I removed that tag.
$endgroup$
– Michael Rybkin
Feb 18 at 14:29
add a comment |
$begingroup$
$$int_3^5fract1+0.1t dt $$
For some reason this is equal to:
1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))
I have no idea how to reduce to that.
integration definite-integrals
edited Feb 20 at 2:20
Michael Rybkin
3,884420
asked Feb 18 at 13:51
ximxim
516
$endgroup$
$$int_3^5fract1+0.1t dt $$
For some reason this is equal to:
1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))
I have no idea how to reduce to that.
integration definite-integrals
integration definite-integrals
edited Feb 20 at 2:20
Michael Rybkin
3,884420
asked Feb 18 at 13:51
ximxim
516
edited Feb 20 at 2:20
Michael Rybkin
3,884420
asked Feb 18 at 13:51
ximxim
516
edited Feb 20 at 2:20
Michael Rybkin
3,884420
edited Feb 20 at 2:20
Michael Rybkin
3,884420
edited Feb 20 at 2:20
Michael Rybkin
3,884420
3,884420
asked Feb 18 at 13:51
ximxim
516
asked Feb 18 at 13:51
ximxim
516
asked Feb 18 at 13:51
ximxim
516
516
$begingroup$
If you let $x=1+0.1t$, then $t=10x-10$...
$endgroup$
– Eleven-Eleven
Feb 18 at 13:53
$begingroup$
Do you mean $$int_3^5fract1+frac110tdt$$?
$endgroup$
– Dr. Sonnhard Graubner
Feb 18 at 13:54
$begingroup$
This is not an improper integral, by the way. So, I removed that tag.
$endgroup$
– Michael Rybkin
Feb 18 at 14:29
add a comment |
$begingroup$
If you let $x=1+0.1t$, then $t=10x-10$...
$endgroup$
– Eleven-Eleven
Feb 18 at 13:53
$begingroup$
Do you mean $$int_3^5fract1+frac110tdt$$?
$endgroup$
– Dr. Sonnhard Graubner
Feb 18 at 13:54
$begingroup$
This is not an improper integral, by the way. So, I removed that tag.
$endgroup$
– Michael Rybkin
Feb 18 at 14:29
$begingroup$
If you let $x=1+0.1t$, then $t=10x-10$...
$endgroup$
– Eleven-Eleven
Feb 18 at 13:53
$begingroup$
If you let $x=1+0.1t$, then $t=10x-10$...
$endgroup$
– Eleven-Eleven
Feb 18 at 13:53
$begingroup$
Do you mean $$int_3^5fract1+frac110tdt$$?
$endgroup$
– Dr. Sonnhard Graubner
Feb 18 at 13:54
$begingroup$
Do you mean $$int_3^5fract1+frac110tdt$$?
$endgroup$
– Dr. Sonnhard Graubner
Feb 18 at 13:54
$begingroup$
This is not an improper integral, by the way. So, I removed that tag.
$endgroup$
– Michael Rybkin
Feb 18 at 14:29
$begingroup$
This is not an improper integral, by the way. So, I removed that tag.
$endgroup$
– Michael Rybkin
Feb 18 at 14:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
$$fract1+0.1t = frac10cdot(1+0.1t) - 101+0.1t = 10 - frac101+0.1t$$
answered Feb 18 at 13:53
5xum5xum
91.4k394161
$endgroup$
add a comment |
$begingroup$
$$
fracx1+0.1x=fracx1+0.1xcdotfrac1010=
frac10x10+x=10left(fracx10+xright)=\
10left(frac-10+10+x10+xright)=
10left(frac-1010+x+frac10+x10+xright)=
10left(-frac1010+x+1right)=\
10left(1-frac1010+xright)=10-frac10010+x.
$$
$$
intleft(10-frac10010+xright),dx=
10int,dx-100intfrac110+xfracddx(10+x),dx=\
10x-100intfrac110+x,d(10+x)=
10x-100ln+C.
$$
$$
int_3^5fract1+0.1t,dt=
bigg[10t-100ln10+tbigg]_3^5=\
50-100ln15-(30-100ln13)=
20-100ln15+100ln13=\
20-100(ln15-ln13)=20-100lnfrac1513.
$$
The answer you gave is equivalent to what I got:
$$
frac10.1left(2-frac10.1left[ln1.5-ln1.3right]right)=
10left(2-10left[lnfrac1510-lnfrac1310right]right)=\
20-100lnleft(frac1510divfrac1310right)=
20-100lnfrac1513.
$$
answered Feb 18 at 13:59
Michael RybkinMichael Rybkin
3,884420
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$$fract1+0.1t = frac10cdot(1+0.1t) - 101+0.1t = 10 - frac101+0.1t$$
answered Feb 18 at 13:53
5xum5xum
91.4k394161
$endgroup$
add a comment |
$begingroup$
Hint:
$$fract1+0.1t = frac10cdot(1+0.1t) - 101+0.1t = 10 - frac101+0.1t$$
answered Feb 18 at 13:53
5xum5xum
91.4k394161
$endgroup$
add a comment |
$begingroup$
Hint:
$$fract1+0.1t = frac10cdot(1+0.1t) - 101+0.1t = 10 - frac101+0.1t$$
answered Feb 18 at 13:53
5xum5xum
91.4k394161
$endgroup$
Hint:
$$fract1+0.1t = frac10cdot(1+0.1t) - 101+0.1t = 10 - frac101+0.1t$$
answered Feb 18 at 13:53
5xum5xum
91.4k394161
answered Feb 18 at 13:53
5xum5xum
91.4k394161
answered Feb 18 at 13:53
5xum5xum
91.4k394161
answered Feb 18 at 13:53
5xum5xum
91.4k394161
91.4k394161
add a comment |
add a comment |
$begingroup$
$$
fracx1+0.1x=fracx1+0.1xcdotfrac1010=
frac10x10+x=10left(fracx10+xright)=\
10left(frac-10+10+x10+xright)=
10left(frac-1010+x+frac10+x10+xright)=
10left(-frac1010+x+1right)=\
10left(1-frac1010+xright)=10-frac10010+x.
$$
$$
intleft(10-frac10010+xright),dx=
10int,dx-100intfrac110+xfracddx(10+x),dx=\
10x-100intfrac110+x,d(10+x)=
10x-100ln+C.
$$
$$
int_3^5fract1+0.1t,dt=
bigg[10t-100ln10+tbigg]_3^5=\
50-100ln15-(30-100ln13)=
20-100ln15+100ln13=\
20-100(ln15-ln13)=20-100lnfrac1513.
$$
The answer you gave is equivalent to what I got:
$$
frac10.1left(2-frac10.1left[ln1.5-ln1.3right]right)=
10left(2-10left[lnfrac1510-lnfrac1310right]right)=\
20-100lnleft(frac1510divfrac1310right)=
20-100lnfrac1513.
$$
answered Feb 18 at 13:59
Michael RybkinMichael Rybkin
3,884420
$endgroup$
add a comment |
$begingroup$
$$
fracx1+0.1x=fracx1+0.1xcdotfrac1010=
frac10x10+x=10left(fracx10+xright)=\
10left(frac-10+10+x10+xright)=
10left(frac-1010+x+frac10+x10+xright)=
10left(-frac1010+x+1right)=\
10left(1-frac1010+xright)=10-frac10010+x.
$$
$$
intleft(10-frac10010+xright),dx=
10int,dx-100intfrac110+xfracddx(10+x),dx=\
10x-100intfrac110+x,d(10+x)=
10x-100ln+C.
$$
$$
int_3^5fract1+0.1t,dt=
bigg[10t-100ln10+tbigg]_3^5=\
50-100ln15-(30-100ln13)=
20-100ln15+100ln13=\
20-100(ln15-ln13)=20-100lnfrac1513.
$$
The answer you gave is equivalent to what I got:
$$
frac10.1left(2-frac10.1left[ln1.5-ln1.3right]right)=
10left(2-10left[lnfrac1510-lnfrac1310right]right)=\
20-100lnleft(frac1510divfrac1310right)=
20-100lnfrac1513.
$$
answered Feb 18 at 13:59
Michael RybkinMichael Rybkin
3,884420
$endgroup$
add a comment |
$begingroup$
$$
fracx1+0.1x=fracx1+0.1xcdotfrac1010=
frac10x10+x=10left(fracx10+xright)=\
10left(frac-10+10+x10+xright)=
10left(frac-1010+x+frac10+x10+xright)=
10left(-frac1010+x+1right)=\
10left(1-frac1010+xright)=10-frac10010+x.
$$
$$
intleft(10-frac10010+xright),dx=
10int,dx-100intfrac110+xfracddx(10+x),dx=\
10x-100intfrac110+x,d(10+x)=
10x-100ln+C.
$$
$$
int_3^5fract1+0.1t,dt=
bigg[10t-100ln10+tbigg]_3^5=\
50-100ln15-(30-100ln13)=
20-100ln15+100ln13=\
20-100(ln15-ln13)=20-100lnfrac1513.
$$
The answer you gave is equivalent to what I got:
$$
frac10.1left(2-frac10.1left[ln1.5-ln1.3right]right)=
10left(2-10left[lnfrac1510-lnfrac1310right]right)=\
20-100lnleft(frac1510divfrac1310right)=
20-100lnfrac1513.
$$
answered Feb 18 at 13:59
Michael RybkinMichael Rybkin
3,884420
$endgroup$
$$
fracx1+0.1x=fracx1+0.1xcdotfrac1010=
frac10x10+x=10left(fracx10+xright)=\
10left(frac-10+10+x10+xright)=
10left(frac-1010+x+frac10+x10+xright)=
10left(-frac1010+x+1right)=\
10left(1-frac1010+xright)=10-frac10010+x.
$$
$$
intleft(10-frac10010+xright),dx=
10int,dx-100intfrac110+xfracddx(10+x),dx=\
10x-100intfrac110+x,d(10+x)=
10x-100ln+C.
$$
$$
int_3^5fract1+0.1t,dt=
bigg[10t-100ln10+tbigg]_3^5=\
50-100ln15-(30-100ln13)=
20-100ln15+100ln13=\
20-100(ln15-ln13)=20-100lnfrac1513.
$$
The answer you gave is equivalent to what I got:
$$
frac10.1left(2-frac10.1left[ln1.5-ln1.3right]right)=
10left(2-10left[lnfrac1510-lnfrac1310right]right)=\
20-100lnleft(frac1510divfrac1310right)=
20-100lnfrac1513.
$$
answered Feb 18 at 13:59
Michael RybkinMichael Rybkin
3,884420
edited Feb 18 at 14:44
answered Feb 18 at 13:59
Michael RybkinMichael Rybkin
3,884420
answered Feb 18 at 13:59
Michael RybkinMichael Rybkin
3,884420
answered Feb 18 at 13:59
Michael RybkinMichael Rybkin
3,884420
3,884420
add a comment |
add a comment |
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$begingroup$
If you let $x=1+0.1t$, then $t=10x-10$...
$endgroup$
– Eleven-Eleven
Feb 18 at 13:53
$begingroup$
Do you mean $$int_3^5fract1+frac110tdt$$?
$endgroup$
– Dr. Sonnhard Graubner
Feb 18 at 13:54
$begingroup$
This is not an improper integral, by the way. So, I removed that tag.
$endgroup$
– Michael Rybkin
Feb 18 at 14:29