Let's discuss the problem statement.If Y ~ Uniform(0,1), we have to find E(Y^k) using My(s).
So, let's start with the solution,Using the definition of moment generating function (MGF), we can find E(Y^k) using My(s) as below:$$M_y(s) = E(e^{sy}) = \int_{-\infty}^\infty e^{sy} f_Y(y)dy$$Here, $f_Y(y)$ is the PDF of Y, which is $f_Y(y)=1$ for $0\le y\le1$, otherwise $0$.
Thus, substituting the values, we have,$$M_y(s) = \int_{0}^1 e^{sy} dy = \left[\frac{e^{sy}}{s}\right]_0^1 = \frac{e^s-1}{s}$$Now, using the Taylor series expansion of $\frac{e^s-1}{s}$ about $s=0$ we have,$$\frac{e^s-1}{s} = 1 + \frac{s}{2!} + \frac{s^2}{3!} + \frac{s^3}{4!} + ...$$Comparing this expansion with the definition of MGF, we can see that the $k^{th}$ moment of Y is given by,$$E(Y^k) = M_y^{(k)}(0) = \frac{d^k}{ds^k} \left[\frac{e^s-1}{s}\right]_{s=0}$$Differentiating $\frac{e^s-1}{s}$, we have,$$\frac{d}{ds}\left[\frac{e^s-1}{s}\right] = \frac{se^s - e^s + 1}{s^2}$$$$\frac{d^2}{ds^2}\left[\frac{e^s-1}{s}\right] = \frac{s^2e^s - 3se^s + 2e^s}{s^3}$$$$\frac{d^3}{ds^3}\left[\frac{e^s-1}{s}\right] = \frac{s^3e^s - 6s^2e^s + 11se^s - 6e^s}{s^4}$$Putting $s=0$, we get the following values for different values of k:$$E(Y^1) = M_y^{(1)}(0) = \left[\frac{d}{ds}\left[\frac{e^s-1}{s}\right]\right]_{s=0} = 1$$$$E(Y^2) = M_y^{(2)}(0) = \left[\frac{d^2}{ds^2}\left[\frac{e^s-1}{s}\right]\right]_{s=0} = \frac{1}{3}$$$$E(Y^3) = M_y^{(3)}(0) = \left[\frac{d^3}{ds^3}\left[\frac{e^s-1}{s}\right]\right]_{s=0} = \frac{1}{2}$$$$E(Y^4) = M_y^{(4)}(0) = \left[\frac{d^4}{ds^4}\left[\frac{e^s-1}{s}\right]\right]_{s=0} = \frac{1}{5}$$Therefore, the values of $E(Y^k)$ using My(s) are,$$E(Y^1) = 1$$$$E(Y^2) = \frac{1}{3}$$$$E(Y^3) = \frac{1}{2}$$$$E(Y^4) = \frac{1}{5}$$Hence, this is the final solution.