# Integration using Taylor approximations

I am stuck on two problems:

1) Prove that $$\int_0^1 \frac{1+x^{30}}{1+x^{60}}dx=1+\frac{c}{31}$$ where $0< c <1$.

2) Prove that $$0.493948<\int_0^{\frac{1}{2}}\frac{1}{1+x^4}dx<0.493958$$

Now for the first one I get Taylor expansion at $0$: $$\frac{1+x^{30}}{1+x^{60}}=1+E_0(x)$$ where $E_0(x)$ is the remainder. I suppose I need to use Lagrangian form of remainder, by differentianting the function and evaluating it at $c$ but I can't make any progress as the derivative is too clumsy.

For the second again I do not know how to incorporate the remainder in bounding the integral.

Hints are appreciated!

Edit

Considering the second problem, I just need help on bounding the remainder of Taylor expansion. I get:

$$\frac{1}{1+x^4}=1-x^4+x^8-x^{12}+x^{16}-Remainder$$

How do I specify bound for the remainder?

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The first part is a duplicate. An approach using power series expansions appears in an answer there. – David Mitra Jan 9 '13 at 18:27

For the first one, since $x \in (0,1)$ we have $1 < 1+x^{60} < 1 + x^{30}$. Hence, we get that $$1 < \dfrac{1+x^{30}}{1+x^{60}} < 1+x^{30}$$ Now you can get the answer.
For the second one, we have $$\dfrac1{1+x^4} = \sum_{k=0}^{\infty}(-x^4)^{k}$$ and hence $$\sum_{k=0}^{2n+1}(-x^4)^{k} < \dfrac1{1+x^4} < \sum_{k=0}^{2n}(-x^4)^{k}$$ Take $n=2$ to get what you want.
@Sarunas: If you mean Taylor series for 1), note that the Taylor series for $\frac{1+x}{1+x^2} = 1+x-x^2+x^3\cdots$. If $x \in [0,1)$, you have $\frac{1+x}{1+x^2} \leq 1+x$. Now replace $x$ by $x^{30}$ and you have Marvis' estimate. – copper.hat Jan 9 '13 at 18:28