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I must be doing something wrong here.
#1
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.
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Messages In This Thread
I must be doing something wrong here. - by chrome - 2010-11-20, 09:02 PM
I must be doing something wrong here. - by Russt - 2010-11-20, 10:56 PM
I must be doing something wrong here. - by Hazzy - 2010-11-20, 11:39 PM
I must be doing something wrong here. - by chrome - 2010-11-21, 01:16 AM
I must be doing something wrong here. - by Corn - 2010-11-22, 11:26 PM
I must be doing something wrong here. - by Hazzy - 2010-11-23, 12:11 AM
I must be doing something wrong here. - by Hazzy - 2010-11-23, 01:45 AM
I must be doing something wrong here. - by Corn - 2010-11-23, 04:22 PM
I must be doing something wrong here. - by Corn - 2010-11-23, 05:44 PM
I must be doing something wrong here. - by Corn - 2010-11-23, 06:04 PM
I must be doing something wrong here. - by John11 - 2010-11-23, 06:16 PM

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