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I must be doing something wrong here. - chrome - 2010-11-20
Helping my brother with his calculus homework and I gave him the right answer, but I used my calculator to integrate something that he doesn't understand. He says he tried to do said integration by hand and got something different; I then integrated it by hand and got what he got. The equation:
(a = 0, b = t)[SIZE="4"]∫[/SIZE][k/(t+1)^2 (dt)] = -k/t+1
Could someone map that out for me? I know it's relatively simple but I can't seem to do the integration by hand. When I do, I get:
(a = 0, b = t)[SIZE="4"]∫[/SIZE][k/(t+1)^2 (dt)]
= [-k/(t+1)] - [-k/(0+1)]
= -k/(t+1) + k
Now, the problem with this: The original equation, k/(t+1)^2, = f'(t), meaning by my calculations -k/(t+1) + k = f(t). The problem that my brother needs help with involves calculating f(0), which = -k. But:
f(t) = -k/(t+1) + k
f(0) = -k/(0+1) + k
= -k + k
= 0
The correct solution:
f(t) = -k/(t+1)
f(0) = -k/(0+1)
= -k
I must be forgetting an integration law or something.
I must be doing something wrong here. - Russt - 2010-11-20
Err...
Okay when you differentiate, you lose the constant term. So if you're only given f'(t), you need some kind of initial value in order to find f(t). Otherwise the best you can do is f(t) = -k/(t+1) + c, for some constant c.
Ask your brother if there's any other information in the problem.
I must be doing something wrong here. - KajitiSouls - 2010-11-20
Nvm...
I must be doing something wrong here. - Hazzy - 2010-11-20
If f(0) = -k and f(t) = -k/(t+1) + k + C, then C = -k. Assuming k is a constant.
I'm not entirely sure what the question is....
I must be doing something wrong here. - OB3LISK - 2010-11-20
Uhm no. I stared at this for such a long time..Gosh.
In your hand work, you actually already subbed in the b-a part, which is where you're getting that extra +k, but of course if you choose b = a, in this case a = 0 and b = 0, then of course it's coming out to 0.
In your "correct" solution you're just plugging in your 0 value to find what 0 equals, which is -k.
Do I make sense?
I must be doing something wrong here. - chrome - 2010-11-21
Well, (a = 0, b = t)[SIZE="4"]∫[/SIZE][k/(t+1)^2 (dt)] is the original problem my brother gave me; he specified that it was a definite integral, and I was a bit confused, since based on the information he gave me that's an indefinite integral. So I was like "...k" and attempted to solve for it. Both of our calculators insisted that it was -k/(t+1) and since I was going under the assumption that I was to solve for a definite integration problem, I got confused.
Misunderstanding, I guess. I'm aware that [SIZE="4"]∫[/SIZE][k/(t+1)^2 (dt)] = -k/(t+1) + C.
I must be doing something wrong here. - KajitiSouls - 2010-11-21
I seriously need to stop being so r3tarded...
What your brother specified as the original problem is actually a limit-dependent integral, which is most likely beyond the scope of what he's learning (calculus I would guess). You can't pass off that definite integral as an indefinite integral.
I must be doing something wrong here. - Corn - 2010-11-22
So it's related to calculus, so I'm posting it in here.
How in the world do you find the (first) derivative of y=x^2-3x+2 with respect to x^2-3x?
I must be doing something wrong here. - Hazzy - 2010-11-23
Corn Wrote:So it's related to calculus, so I'm posting it in here.
How in the world do you find the (first) derivative of y=x^2-3x+2 with respect to x^2-3x?
I would factor it out with a u sub:
y=u+2
So... y'=u = x^2-3x?
Wolfram was no help to me at all. ._.[url=http://www.wolframalpha.com/input/?i=derivative+of+%28x%5E2-3x%2B2%29+with+respect+to+%28x%5E2-3x%29][/url]
I must be doing something wrong here. - KajitiSouls - 2010-11-23
Hazzy Wrote:I would factor it out with a u sub:
y=u+2
So... y'=u = x^2-3x?
Wolfram was no help to me at all. ._.[url=http://www.wolframalpha.com/input/?i=derivative+of+%28x%5E2-3x%2B2%29+with+respect+to+%28x%5E2-3x%29][/url]
Um I think that wouldn't work.
Let's say that y = x^5 or something, and that u = x^5, therefore y = u.
y' = 1 ????/??/
(Actually it would be more like dy/du = 1. Don't ask me what that means.)
As for CoB's question, I have never encountered that, so I can't help there.
I must be doing something wrong here. - Hazzy - 2010-11-23
KajitiSouls Wrote:Um I think that wouldn't work.
Let's say that y = x^5 or something, and that u = x^5, therefore y = u.
y' = 1 ????/??/
(Actually it would be more like dy/du = 1. Don't ask me what that means.)
As for CoB's question, I have never encountered that, so I can't help there.
We don't want dy/dx. we want dy/du. When x^5 increases by a, how much does y increase? By a.
I must be doing something wrong here. - Riyuran - 2010-11-23
Hazzy Wrote:I would factor it out with a u sub:
y=u+2
So... y'=u = x^2-3x?
Wolfram was no help to me at all. ._.[url=http://www.wolframalpha.com/input/?i=derivative+of+%28x%5E2-3x%2B2%29+with+respect+to+%28x%5E2-3x%29][/url]
Uhh..failure.
If y=u+2, then dy/du=1. Not u.
Lol, why you need Wolfram for such an easy problem?
If you're differentiating wrt something other than x, then you treat that as the new variable.
d/d(x^2-3x)(x^2-3x+2) = 1.
I must be doing something wrong here. - Corn - 2010-11-23
No one figured out how to do it eh? D=
Edit: Ah he edited.
I must be doing something wrong here. - ThatHurts - 2010-11-23
Riyuran Wrote:Uhh..failure.
If y=u+2, then dy/du=1. Not u.
Lol, why you need Wolfram for such an easy problem?
If you're differentiating wrt something other than x, then you treat that as the new variable.
d/d(x^2-3x)(x^2-3x+2) = 1.
ITT this is correct. That's how I looked at it at least.
I must be doing something wrong here. - Corn - 2010-11-23
Another question: what is the integral of [x]? (It's an integer function if you forgot what those brackets meant).
I must be doing something wrong here. - Riyuran - 2010-11-23
Corn Wrote:Another question: what is the integral of [x]? (It's an integer function if you forgot what those brackets meant).
It's not possible to express the indefinite integral of that. However, a definite integral wouldn't be too hard to calculate (areas of rectangles).
Wolfram is your friend.
I must be doing something wrong here. - Corn - 2010-11-23
How would you do definite integrals on wolfram?
I must be doing something wrong here. - John11 - 2010-11-23
Corn Wrote:How would you do definite integrals on wolfram?
integrate round(x) x=0..1
Change 0 and 1 to the values you need.