Can n be treated as a constant in this integral?

[kingwinner]

Homework Statement

n
∫2^n-tdt
0

Homework Equations

N/A

The Attempt at a Solution

I've been wondering about the correct way to deal with this type of integral for quite a long time. To me, the above integral looks like something of the form:
n
∫f(n,t)dt
0
n appears in the integrand AND in the limits of integration, how can I integrate in this case?

I am just wondering whether n can be treated as a "constant" in the above integral, i.e. can I treat 2^n as a constant and pull the 2^n OUT of the integral and evaluate
n
∫2^-tdt ?
0

Thank you for explaining!

 

[morphism]

Yes, the n is a constant in this case.

 

[HallsofIvy]

First, 2^n-t= 2ⁿ 2^-t.

Second, the derivative of 2^t is (ln 2)2^t so the anti-derivative is 2^t/ln(2).

 

[kingwinner]

Yes, the n is a constant in this case.

So even though "n" appears in the integrand and also appears in the limits of integration, we can still treat the "n" in the integrand as a constant and use the property ∫cf(t)dt=c∫f(t)dt ?

 

[Defennder]

If you integrating with respect to t, you don't have to worry about anything else unless n is a function of t, which it is not stated to be.