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I have the following definite integral:

$\displaystyle \int^{g(a)}_{a} f(x) \, dx$

and I am asked to perform a shift of the variable x, so that it transforms in $x + T$ (T is just some constant). How do the limit of integration change under this transformation? I would expect that they shift too, so that the transformed integral reads

$\displaystyle \int^{g(a) +T}_{a+T} f(x+T) \, dx$

But later on I started having some doubts on how to treat the upper limit and at some point I started to think that it should become

$\displaystyle \int^{g(a+T)}_{a+T}$

Can someone give me a hint on this? It is a simple problem, but the fact that the upper limit is a function of the lower one is confusing for me (this is part of a homework problem by the way).

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1 Answer 1

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Actually, the bounds should be $\displaystyle a-T$ and $\displaystyle g(a) - T$. Note the minus signs.

The original integral is $\displaystyle \int_a^{g(a)} f(x)dx$. Substitute $\displaystyle x = u+T$. Now $\displaystyle dx = du$.

The integral (dropping the bounds for now) becomes $\displaystyle \int f(u+T)du$.

The original bounds correspond to $\displaystyle x=a$ and $\displaystyle x=g(a)$. The values of $\displaystyle u \ (u = x-T)$ at these bounds are $\displaystyle a-T$ and $\displaystyle g(a)-T$, respectively.

Hence the definite integral is $\displaystyle \int_{a-T}^{g(a)-T} f(u+T)du = \int_{a-T}^{g(a)-T} f(x+T)dx $, where the last step involves a straight swap of $x$ for $u$ because in a definite integral, the variable of integration is just a dummy variable.

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