Hello!
I've been working on a question:
find int(dx/((1+2x)^(3/2))
Here is how I set up the problem:
u = 1+2x
du = 2dx
1/2*int(2((u)^-3/2)
1/2*int(-4/(u^-1/2)
int (u^-3/2) = -2(u^-1/2); 2 * -2 = -4
-2/(1+2x)^(1/2)
I plug in the two values, 4 (low end) 7.5 (high end)
answer:
(-1/2) - (3/2) = -2
This answer is incorrect and I'd like to know what I did wrong.
here's what i did...
let u =1+2x
du=2dx
but since in the original integral there is only a dx we multiply both sides of the previous equation by 1/2
1/2du=dx
now substitute back in
1/2 int (u^(-3/2)du)
the limits are now 16 for upper and 9 for lower
these are found by using u=1+2x and using the original limits
so after integration you get...
.5(u^-.5)/-.5 which = -1/u^.5
plugging in the limits you get
-1/4+-1/3 = 1/12
let u =1+2x
du=2dx
but since in the original integral there is only a dx we multiply both sides of the previous equation by 1/2
1/2du=dx
now substitute back in
1/2 int (u^(-3/2)du)
the limits are now 16 for upper and 9 for lower
these are found by using u=1+2x and using the original limits
so after integration you get...
.5(u^-.5)/-.5 which = -1/u^.5
plugging in the limits you get
-1/4+-1/3 = 1/12
QUOTE (curlybit89+Nov 1 2009, 11:35 PM)
find int(dx/((1+2x)^(3/2))
Here is how I set up the problem:
u = 1+2x
du = 2dx
1/2*int(2((u)^-3/2)
1/2*int(-4/(u^-1/2)
Here is how I set up the problem:
u = 1+2x
du = 2dx
1/2*int(2((u)^-3/2)
1/2*int(-4/(u^-1/2)
First of all, you began to mess up when you got sloppy by finding almost du, and not all of du. Second you didn't do the integration right.
Start from basics. Real basics. 1 times anything leaves that anything unchanged. So let's color code this and follow each any every step in the substitutions, OK?
∫ (1 + 2x)^(-3/2) dx
You say u = 1 + 2x and du=2 dx. But while u is there, du is not (yet).
So we use the fact that 1 = ˝×2 And we have:
∫ (1 + 2x)^(-3/2) dx
= ∫ (1 + 2x)^(-3/2)× 1 dx
= ∫ (1 + 2x)^(-3/2)× ˝×2 dx
= ∫ (1 + 2x)^(-3/2)× ˝× 2 dx
And only now may we make the substitution:
= ∫ (u)^(-3/2)× ˝ du
= ˝× ∫ (u)^(-3/2) du
= ˝× (-2) u^(-1/2) + C
= -(u^(-1/2)) + C
= -((1 + 2x)^(-1/2)) + C
= -1/√(1 + 2x) + C
Got it?
The definite integral from 4 to 7.5 is just this expression evaluated:
(-1/√(1 + 2×hi)) -(-1/√(1 + 2×lo))
= (-1/√(1 + 2×7.5)) -(-1/√(1 + 2×4))
= (-1/√(1 + 15)) -(-1/√(1 + 8))
= (-1/√(16)) -(-1/√(9))
= (-1/4) -(-1/3)
= (-3/12) - (-4/12)
= ((-3) - (-4))/12
= (4 - 3)/12
= 1/12
Start from basics. Real basics. 1 times anything leaves that anything unchanged. So let's color code this and follow each any every step in the substitutions, OK?
∫ (1 + 2x)^(-3/2) dx
You say u = 1 + 2x and du=2 dx. But while u is there, du is not (yet).
So we use the fact that 1 = ˝×2 And we have:
∫ (1 + 2x)^(-3/2) dx
= ∫ (1 + 2x)^(-3/2)× 1 dx
= ∫ (1 + 2x)^(-3/2)× ˝×2 dx
= ∫ (1 + 2x)^(-3/2)× ˝× 2 dx
And only now may we make the substitution:
= ∫ (u)^(-3/2)× ˝ du
= ˝× ∫ (u)^(-3/2) du
= ˝× (-2) u^(-1/2) + C
= -(u^(-1/2)) + C
= -((1 + 2x)^(-1/2)) + C
= -1/√(1 + 2x) + C
Got it?
The definite integral from 4 to 7.5 is just this expression evaluated:
(-1/√(1 + 2×hi)) -(-1/√(1 + 2×lo))
= (-1/√(1 + 2×7.5)) -(-1/√(1 + 2×4))
= (-1/√(1 + 15)) -(-1/√(1 + 8))
= (-1/√(16)) -(-1/√(9))
= (-1/4) -(-1/3)
= (-3/12) - (-4/12)
= ((-3) - (-4))/12
= (4 - 3)/12
= 1/12
Thank you rpenner,
I appreciate the color coding. It makes things easier when you're trying to keep your ducks in a row. So the discrepancy appears to be a conceptual misunderstanding. I did not think to break down the implied 1 of dx - my method was to just tack a 2 * 1/2 on the end to compensate for the inconsistencies between dx and du.
I tried to give you positive feedback, but n00bs aren't allowed to do that.
I appreciate the color coding. It makes things easier when you're trying to keep your ducks in a row. So the discrepancy appears to be a conceptual misunderstanding. I did not think to break down the implied 1 of dx - my method was to just tack a 2 * 1/2 on the end to compensate for the inconsistencies between dx and du.
I tried to give you positive feedback, but n00bs aren't allowed to do that.
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