Power Series and Estimation of Integrals
Key Questions

Assuming that you know that the power series for
#sinx# is:#sinx=sum_(n=1)^infty ((1)^(n1)x^(2n1))/(2n1)=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 rewritten as:
#sinx^2=x^21/(3!)x^6+1/(5!)x^10+1/(7!)x^14...# So then:
#int_0^0.01 sinx^2=int_0^0.01 x^21/(3!)int_0^0.01x^6+1/(5!)int_0^0.01x^10...# #=((x^3)/31/(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# #=[x+x^3/3+x^5/{10}+cdots]_0^{0.01}# #=0.01+(0.01)^3/3+(0.01)^5/10+cdots# #approx 0.01# I hope that this was helpful.

Answer:
Integrate term by term
Explanation:
Replace f by its Taylor expansion. f is now a sum
#f = sum a^n x^n#
Integrate term by term
#int f dx = sum n a^n x^(n1)#
Questions
Power Series

Introduction to Power Series

Differentiating and Integrating Power Series

Constructing a Taylor Series

Constructing a Maclaurin Series

Lagrange Form of the Remainder Term in a Taylor Series

Determining the Radius and Interval of Convergence for a Power Series

Applications of Power Series

Power Series Representations of Functions

Power Series and Exact Values of Numerical Series

Power Series and Estimation of Integrals

Power Series and Limits

Product of Power Series

Binomial Series

Power Series Solutions of Differential Equations