# Power Series and Estimation of Integrals

## Key Questions

• Assuming that you know that the power series for $\sin x$ 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!

• Since

e^x=1+x+x^2/{2!}+cdots,

e^{x^2}=1+x^2+x^4/{2!}+cdots

int_0^{0.01}e^{x^2}dx=int_0^{0.01} (1+x^2+x^4/{2!}+cdots)dx

$= {\left[x + {x}^{3} / 3 + {x}^{5} / \left\{10\right\} + \cdots\right]}_{0}^{0.01}$

$= 0.01 + {\left(0.01\right)}^{3} / 3 + {\left(0.01\right)}^{5} / 10 + \cdots$

$\approx 0.01$

I hope that this was helpful.

$f = \sum {a}^{n} {x}^{n}$
$\int f \mathrm{dx} = \sum n {a}^{n} {x}^{n - 1}$