- #1

bhartish

- 26

- 0

integrate with respect to ζ in the limits 0 and t

Help will be greatly appreciated

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- Thread starter bhartish
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- #1

bhartish

- 26

- 0

integrate with respect to ζ in the limits 0 and t

Help will be greatly appreciated

- #2

HallsofIvy

Science Advisor

Homework Helper

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I think you are letting complicated lookig **constants** confuse you.

If I read the parentheses correctly that is

[tex]\int (A+ 0.5\sqrt{Be^{-2(t- ζ)/ζ_0}})dζ[/tex]

with A and B representing those rather complicated**constants** in your integral.

I presume you know that [itex]\int Adt= At+ C[/itex]. For the second integral, let [itex]u= -2(t- ζ)/ζ_0[/itex], so that [itex]dζ= (ζ_0/2)du[/itex], so the second integral becomes

[tex]B\frac{\zeta_0}{2}\int e^u du[/tex]

If I read the parentheses correctly that is

[tex]\int (A+ 0.5\sqrt{Be^{-2(t- ζ)/ζ_0}})dζ[/tex]

with A and B representing those rather complicated

I presume you know that [itex]\int Adt= At+ C[/itex]. For the second integral, let [itex]u= -2(t- ζ)/ζ_0[/itex], so that [itex]dζ= (ζ_0/2)du[/itex], so the second integral becomes

[tex]B\frac{\zeta_0}{2}\int e^u du[/tex]

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- #3

Mark44

Mentor

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- 8,293

If I read the parentheses correctly that is

[tex]\int (A+ 0.5\sqrt{Be^{-2(t- ζ)/ζ_0}})dζ[/tex]

with A and B representing those rather complicatedconstantsin your integral.

I presume you know that [itex]\int Adt= At+ C[/itex]. For the second integral, let [itex]u= -2(t- ζ)/ζ_0[/itex], so that [itex]dζ= (ζ_0/2)du[/itex], so the second integral becomes

[tex]B\frac{\zeta_0}{2}\int e^u du[/tex]

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