Question

How do you use a Power Series to estimate the integral `int_0^0.01sin(x^2)dx` ?

Answer 1

Assuming that you know that the power series for `sinx` is:

`sinx=sum_(n=1)^infty ((-1)^(n-1)x^(2n-1))/(2n-1)=x-(x^3)/(3!)+(x^5)/(5!)+...`

then we can answer this fairly quickly. If not, perhaps that can be a separate question!

So, if:

`sinx=x-(x^3)/(3!)+(x^5)/(5!)+(x^7)/(7!)...`

Then:

`sinx^2=x^2-(x^2)^3/(3!)+((x^2)^5)/(5!)+((x^2)^7)/(7!)...`

Which can be re-written as:

`sinx^2=x^2-1/(3!)x^6+1/(5!)x^10+1/(7!)x^14...`

So then:

`int_0^0.01 sinx^2=int_0^0.01 x^2-1/(3!)int_0^0.01x^6+1/(5!)int_0^0.01x^10...`

`=((x^3)/3-1/(3!)(x^7)/7+1/(5!)(x^11)/11+...)|_0^0.01`

When we plug in zero for `x`, all the terms will disappear. So all you have to do is plug in `0.01` for x out to however many terms you want.

Hope this helps!