3) Assume that equals its Maclaurin series for all x.
Use the Maclaurin series for to evaluate the integral . Your answer will be an infinite series. Use the first two terms to estimate its value.
I am not sure how to do this!
If you believe that $\displaystyle \sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$ then surely you have that $\displaystyle \sin(5x^2)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}$.
Plugging this into your integral you get:
$\displaystyle \int_0^{0.64}\sin(5x^2)\, dx=\int_0^{0.64}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx$
Next thing to do is to be bold enough to exchange $\displaystyle \int$ and $\displaystyle \sum$, like this:
$\displaystyle =\sum_{n=0}^\infty \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx\approx \sum_{n=0}^1 \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx=\ldots$
Apparently you must have made some mistake(s) when evaluating the remaning two integrals. To continue my own calculation:
$\displaystyle \ldots =5 \int_0^{0.64}x^2\, dx-\frac{125}{6}\int_0^{0.64}x^6\,dx= 5\cdot\Big[\frac{1}{3}x^3\Big]_{x=0}^{0.64}-\frac{125}{6}\cdot\Big[\frac{1}{7}x^7\Big]_{x=0}^{0.64}$
$\displaystyle =5\cdot\frac{1}{3}\cdot 0.64^3-\frac{125}{6}\cdot\frac{1}{7}\cdot 0.64^7\approx 0.306$
I'm sorry about that, but I cannot find a serious mistake in my calculation. If I ask a CAS to integrate $\displaystyle \int_0^{0.64}\sin(5*x^2)\, dx$ numerically I get the answer $\displaystyle \approx 0.32226$.
Now this does not fit exactly what I have found, namely $\displaystyle \approx 0.306$, but that's not particularly surprising because the next term in the series gives $\displaystyle \approx 0.0174$.
If that's the case, then I take it that your teacher must be mistaken, or, another possibility, not telling you the truth. Why could that be? - Well, I remember that during my studies I once was told that my solution of a somewhat more difficult exercise was wrong. I immediately startet arguing with the teacher. After a few exchanges as to why I thought his objections were not valid, he began to smile and told me that he now really did believe that I was right.
So at that time, my teacher was simply testing whether I had found the solution to this exercise on my own (or at least understood the solution so well that I was able to argue for it convincingly). If a student gets the solution from someone else however, he or she typically will not immediately try to argue for its correctness, but instead rather quickly withdraw...