1. ## Maclaurin series

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!

2. Originally Posted by nhatie
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 $\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$ then surely you have that $\sin(5x^2)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}$.

Plugging this into your integral you get:

$\int_0^{0.64}\sin(5x^2)\, dx=\int_0^{0.64}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx$
Next thing to do is to be bold enough to exchange $\int$ and $\sum$, like this:

$=\sum_{n=0}^\infty \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx\approx \sum_{n=0}^1 \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx=\ldots$

3. Originally Posted by Failure
If you believe that $\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$ then surely you have that $\sin(5x^2)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}$.

Plugging this into your integral you get:

$\int_0^{0.64}\sin(5x^2)\, dx=\int_0^{0.64}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx$
Next thing to do is to be bold enough to exhange $\int$ and $\sum$, like this:

$=\sum_{n=0}^\infty \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx\approx \sum_{n=0}^1 \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx=\ldots$
I kind of understand how to do that step but I can't seem to know what will next take place Thank you

4. Originally Posted by nhatie
I kind of understand how to do that step but I can't seem to know what will next take place Thank you
So let me elaborate this a bit more:

$\ldots=\int_0^{0.64} 5x^2\, dx+\int_0^{0.64}\frac{-1}{3!}(5x^2)^3\, dx=5\int_0^{0.64}x^2\, dx-\frac{125}{6}\int_0^{0.64}x^6\,dx=\ldots$

5. so it would be
(5)(x) from 0 to .64 - (125/6)(1/5)(x^5) from .0 to .64?
my answer of it is : 3.2 - .44739 = 2.7526
but they said it was wrong? did i do something wrong??

6. Originally Posted by nhatie
so it would be
(5)(x) from 0 to .64 - (125/6)(1/5)(x^5) from .0 to .64?
my answer of it is : 3.2 - .44739 = 2.7526
but they said it was wrong? did i do something wrong??
Apparently you must have made some mistake(s) when evaluating the remaning two integrals. To continue my own calculation:

$\ldots =5 \int_0^{0.64}x^2\, dx-\frac{125}{6}\int_0^{0.64}x^6\,dx= 5\cdot\Big[\frac{1}{3}x^3\Big]_{x=0}^{0.64}-\frac{125}{6}\cdot\Big[\frac{1}{7}x^7\Big]_{x=0}^{0.64}$
$=5\cdot\frac{1}{3}\cdot 0.64^3-\frac{125}{6}\cdot\frac{1}{7}\cdot 0.64^7\approx 0.306$

7. i try it but they said it was wrong

8. Originally Posted by nhatie
i try it but they said it was wrong
I'm sorry about that, but I cannot find a serious mistake in my calculation. If I ask a CAS to integrate $\int_0^{0.64}\sin(5*x^2)\, dx$ numerically I get the answer $\approx 0.32226$.
Now this does not fit exactly what I have found, namely $\approx 0.306$, but that's not particularly surprising because the next term in the series gives $\approx 0.0174$.

9. it is so weird!

10. Originally Posted by nhatie
it is so weird!
Are you sure you have given us the correct problem statement?

11. Originally Posted by Failure
Are you sure you have given us the correct problem statement?
here
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.

12. Originally Posted by nhatie
here
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.
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...