Use a taylor series to approximate definite integral of sin(x^2)dx from x=0 to x=0.2

Question

anonymous · Answer

fo you know series for sine?

anonymous · Answer

yes the series is x^2 - x^6/6+x^10/120-x^14/5040+...

anonymous · Answer

replace $x$ in that series for $x^2$ an then take it out as far as you care to
series is 
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$$  replace $x$ by $x^2$ get 
$$x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-...$$

anonymous · Answer

just replace $x$ by $x^2$ not the denominators

anonymous · Answer

Yes so I got that part right. thank you. how do I write the sigma notation ?

anonymous · Answer

since you are close to zero, you can probably get a good estimation by taking 
$$\int_0^{.2}x^2dx$$

anonymous · Answer

the sixth power term will not contribute that much

anonymous · Answer

thanks