Simple Integration Watch

dholmes_7
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#1
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I'm having a strange day. I've forgotten how to integrate!

Can someone remind me how to integrate: y=12*((41/52)^x/8)
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moleo
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#2
Report 12 years ago
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lny = ln12((41/52)^x/8)
lny = ln12 + ln(41/52)^x/8
lny = ln12 + (x/8)*ln(41/52)
1/y . dy/dx = 1/8 * ln(41/52)
dy/ dx = y.1/8.ln(41/52)
dy / dx = 12*((41/52)^x/8).ln(41/52)
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dholmes_7
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(Original post by moleo)
lny = ln12((41/52)^x/8)
lny = ln12 + ln(41/52)^x/8
lny = ln12 + (x/8)*ln(41/52)
1/y . dy/dx = 1/8 * ln(41/52)
dy/ dx = y.1/8.ln(41/52)
dy / dx = 12*((41/52)^x/8).ln(41/52)
Thanks, but I needed it integrating, not differentiated!
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[email protected]
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(Original post by dholmes_7)
I'm having a strange day. I've forgotten how to integrate!

Can someone remind me how to integrate: y=12*((41/52)^x/8)
Do you mean:
y=12*((41/52)^(x/8))

or:
y=12*(((41/52)^x)/8)
???
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dholmes_7
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#5
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(Original post by [email protected])
Do you mean:
y=12*((41/52)^(x/8))

or:
y=12*(((41/52)^x)/8)
???
sorry, yes i should have mad the clearer. The first one: y=12*((41/52)^(x/8))
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[email protected]
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y=12*((41/52)^(x/8))

y/12 = ((41/52)^(x/8))

Lets integrate ((41/52)^(x/8)) then you multiply by 12 later.

Take logs then EXP.

e^ln [(41/52)^(x/8)] = e^[(x/8) * ln(41/52)]

Now by substitution take u = (x/8) * ln(41/52) then du = (1/8) * ln(41/52)dx
So dx = du/[(1/8) * ln(41/52)]

Now: Int[e^[(x/8) * ln(41/52)]] becomes: Int[(e^u)du/[(1/8) * ln(41/52)]
So we integrate: 1/[(1/8) * ln(41/52)] * Int e^u du
Which is:
(1/[(1/8) * ln(41/52)]) * e^u + C

Now sub in back what u is, then multiply by 12 as we divided it by 12 before.
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meef cheese
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#7
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INT ax dx = ax/lna

INT ax/8 dx = 8ax/8/lna , where a is a constant.

You should be able to it from there.
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