So...It's not the gamma function?

Ok, my way is this,

using integration by parts

$\displaystyle \int { { x }^{ n }{ e }^{ -x }dx } \quad =\quad -{ x }^{ n }e^{ -x }+\int { n{ x }^{ n-1 }{ e }^{ -x } } \quad \\ \\ again\quad using\quad IBP\\ \\ -{ x }^{ n }e^{ -x }+\int { n{ x }^{ n-1 }{ e }^{ -x } } \quad =\quad { -x }^{ n }e^{ -x }-\quad n{ x }^{ n-1 }{ e }^{ -x }\quad +\quad \int { n{ (n-1)x }^{ n-2 }{ e }^{ -x } } \\$

in general

proceeding till power of x is 0, and putting limits

$\displaystyle {I(n,a)\quad =\quad -\sum _{ r=0 }^{ n }{ \frac { n! }{ (n-r)! } } { a }^{ n-r }{ e }^{ -a }\quad +\quad n!}$

now dividing by factorial of n and limiting,

we use the fact tha all terms having a defined numerator and an n term in denominator will tend to 0,

except the one that doesnt namely at r=n

and we have

$\displaystyle{\frac { I(n,a) }{ n! } \quad =\quad -\sum _{ r=0 }^{ n }{ \frac { 1 }{ (n-r)! } } { a }^{ n-r }{ e }^{ -a }\quad +\quad 1\\ \\ n-->\quad \infty \quad \frac { I(n,a) }{ n! } \quad =\quad -\sum _{ r=0 }^{ n }{ \frac { 1 }{ (n-r)! } } { a }^{ n-r }{ e }^{ -a }\quad +\quad 1\quad =\quad -{ e }^{ -a }\quad +\quad 1\\ \\ \\ (1+{ e }^{ -a }-1){ e }^{ a }=\quad { e }^{ a }\quad =\quad T(a)\\ \\ \int _{ 0 }^{ 2 }{ { e }^{ a }da } =\quad answer}$

hope i am right

I'll post a full solution if no else does so in the near future.

My claim is as follows: $(1-\frac{I(n,a)}{n!})e^a=1+\frac{a}{1!}+\frac{a^2}{2!}+...\frac{a^n}{n!}$ This holds for all $(n,a)$ and you can prove it by establishing a recursive relation between $I(n,a)$ and $I(n-1,a)$ .

If we let n tend to infinity we get: $T(a)=e^a$

What's interesting here is that, $I(n,a)$ has two parameters $n$ and $a$ .

Letting $a$ tend to infinity gives the gamma function and thus, an expression for n!

But if we let the other parameter tend to infinity, we get an expression for a seemingly unrelated quantity: $e^a$

The final answer becomes $\quad \int_{0}^{2}e^xdx=e^2-1$