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Like isnt there way to put the function value inside and then I would get the same value as the uniform probability function?

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closed as off-topic by Lee David Chung Lin, zz20s, Eevee Trainer, Thomas Shelby, Leucippus Mar 26 at 5:02

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The mgf is given by $$ M_X(t) = \mathbb{E}\left[e^{tX}\right] = \int_\mathbb{R} e^{tx} f(x)\ dx, $$ so $M_X(0) = 1$ and $$M_X'(0) = \mathbb{E}[X]$$ which is a quick shortcut you can use to check things. In your case, $\mathbb{E}[X] = 1/2$ by symmetry.

But if you want a formal calculation, your way is correct, namely that $$ M_X(t) = \int_\mathbb{R} e^{tx} f(x)\ dx = \int_0^1e^{tx} dx = \frac{e^t-1}{t} $$ The above spot-check is easy to make since $$ M_X'(t) = e^t t^{-1} - \left(e^t-1\right)t^{-2} = \frac{(t-1)e^t+1}{t^2} $$ and it's easy to see by L'Hospital's Rule (twice) that $M_X'(t) \to 1/2$ as $t \to 0$...

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