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?
 A: 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
A: Maclaurin series (or, in general, Taylor series) gives polynomials that approximates the function. If the approximation is good in an interval we can exploit this polynomial to estimate roughly an integral. In your case you can calculate easily the exact integral $2 (e-1)$, so the exercise looks to my eyes a bit artificial. Anyway if you plot the serie you calculate, you can check that the approximation of $\exp \left( \frac{x}{2} \right)$ is good between $0$ and $2$. Here we have the exponential in green and the polynomial in blue:

So the two areas are approximately the same (clearly we will find that approximate value is smaller). This means that if you develop to the 3rd degree and if the integral goes from $0$ to $2$, you can estimate the searched value integrating your polynomial instead of the function:
$$
\int_0^2 \left( 1 + \frac{x}{2} + \frac{x^2}{8} + \frac{x^3}{48} \right) dx
$$
...that is $\frac{41}{12}$. Notice that the true value is about $3.43656$ while the approximate one is $3.41\bar{6}$, not so bad. In addition, notice that the check that the two function are approximately the same is a fundamental step in this kind of evaluations: for example you can of course use your polynomial to estimate very well $\int_0^1 \exp \left( \frac{x}{2} \right) dx $ but you cannot use it to estimate $\int_0^{10} \exp \left( \frac{x}{2} \right) dx $: you wil find $\frac{515}{4}=128.75$, very far from the exact value $2 (e^5 -1) \approx 295$. This discrepancy is not surprising if we look to the plot of the 2 function in the considered region: near $x=10$ the polynomial and the function are very different:

To use this metod in this case you should calculate some more term in Maclaurin serie, to obtain a polynomial that give a good approximation of $\exp \left( \frac{x}{2} \right)$ so far from $x=0$).
A: Your professor's question isn't a "realistic" application of this method. It's a highly simplified problem to test you on whether you understand how the method works. Often teachers will ask you to solve problems using a method to demonstrate your ability even if that isn't the best way to solve the problem.
In practice we would immediately integrate this, we wouldn't involve either the the power series or the polynomial approximation, i.e. we would just say that
$\int^2_0 e^\frac{x}{2}dx=[2 e^\frac{x}{2}]^2_0=e^2-1$
converting $e^\frac{x}{2}$ into a Maclaurin just makes the process of finding the integral more complicated.
We use the method of using the Maclaurin series when it's hard to find the integral. For example $\int e^{-x^2}dx$ is not an integral we can find in terms of basic functions. So if we want to find $\int^1_0 e^{-x^2}dx$ then one way we might approach it is using the method your professor is testing you on.
The Maclaurin series of $e^{-x^2}$ is quite straightforward.
$e^{-x^2}=\sum^\limits\infty_{n=0}\frac{(-x^2)^n}{n!}=\sum^\limits\infty_{n=0}\frac{(-1)^n x^{2n}}{n!}$
So we could easily integrate the whole power series. So your question still remains, why would we stop at the $x^3$ term?
Some possible reasons:

*

*we are working with a function where we can't so easily find a formula for the coefficient of $x^n$ in the Maclaurin series,

*we only care about getting a certain level of accuracy, so summing to infinity isn't worthwhile,

*we don't want to spend too much computing power on this.

