How Can Taylor Series Be Used to Compute Integrals with High Precision?

[vucollegeguy]

Homework Statement

Use taylor series method to compute the integral from 1 to 2 of [sin(x²)] / (x²) with 10 ^-3 precision

Homework Equations

The Attempt at a Solution

I'm not sure where to start. Someone please help me.

 

[clamtrox]

Do you know what a Taylor series is? You need to expand sin(x^2) as a Taylor series and then, given that the series satisfies some very broad conditions, you can integrate each term separately.

 

[vucollegeguy]

Yes.

sin(x²) after integration = (x³/3) -(x⁷)/(7*3!) + (x¹¹)/(11*5!)-...
right?

can you help me with the rest of it?

 

[clamtrox]

$$\int_1^2 \frac{\sin(x^2)}{x^2} = \sum_{n=0}^{\infty} \int_1^2 \frac{(-1)^n (x^2)^{2n+1}}{(2n+1)! x^2} dx$$

 

[vucollegeguy]

Thank you so much! So that stuff to that right of the equal sign, how do I use that solve the integral from 1 to 2 - do I just plug in the upper and lower limits and subtract the lower from the upper value? And the precision of 10 ^-3 - how do I get that?

I really appreciate your help.

 

[clamtrox]

The right hand side is easy to integrate now; all you have is a constant times x^(another constant).

 

[vucollegeguy]

Ok, integrating the right side gives:

(-1) ⁿ⁺¹ * (x ²) ²ⁿ) / (2n!*x²)

right, or wrong?

 

[clamtrox]

Wrong.

$$\int_1^2 \frac{\sin(x^2)}{x^2} = \sum_{n=0}^{\infty} \int_1^2 \frac{(-1)^n (x^2)^{2n+1}}{(2n+1)! x^2} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\int_1^2 dx x^{4n}$$

 

[vucollegeguy]

I was totally wrong.
Ok, so now is where I plug into the upper and lower limits and subtract those values?

 

[ideasrule]

Yes. To get 10^-3 precision, you should keep on calculating terms until they go well below 10^-3.

 

[vucollegeguy]

Got it.
Thank you all for your help!