Definite integral of $e^{x/2}$ using Maclaurin polynomial My professor asked us to find the 3rd degree Maclaurin polynomial of $e^{x/2}$ which I found to be 
$$1 +  \frac{x}{2}  +  \frac{x^2}{8}  +  \frac{x^3}{48}$$ 
I do know that that the series for $f(x)=e^{\frac{x}{2}}$ is
$$\sum_{n=0}^\infty \frac{x^n}{n! 2^n}$$
But then he said to approximate $\int_0^2 e^{\frac{x}{2}} dx$ using part a (the Maclaurin polynomial) 
How do I do that? Would I just integrate the polynomial and not the series?

There are times when you can transform $\int [\sum f(x)] dx$ into $\sum[\int f(x) dx]$, although explaining when this works is a complex topic. A key aspect here is that $n$ is always a non-negative integer. You are basically transforming $\int (c_0+c_1x+c_2x^2+...c_nx^n)\,dx$ into $\int c_0\,dx+\int c_1x \,dx+\int c_2x^2\,dx+...\int c_nx^n\,dx$ using the sum rule of integrals. Here, just know you can flip them to get
$$\int_0^2 \bigg[\sum_{n=0}^\infty \frac{x^n}{n! 2^n}\bigg]dx = \sum_{n=0}^\infty \bigg[\int_0^2\frac{x^n}{n! 2^n}\bigg]dx = 2\sum_{n=0}^{\infty} \frac{1}{n!(n+1)}$$ 
Note that this approaches $2(e-1)$, the same answer you get computing the definite integral using the Fundamental Theorem of Calculus directly
