Calculate the area of region between curve f and x axis using series

Question

Consider $f(x)=\sqrt{1+x^4}$

I need to approximately calculate the area of a region between a curve $f$ and the x-axis on [0,1].

However, I need to do this using the five first term non-null of the serie development in $f$ power.

I have no idea how to do this as I'm not sure what the question mean...

To the question is added an hint :

$\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \frac{5x^4}{128} + ...$

How am-I supposed to do that ?

Answer 1

Since this sounds like homework and you're just getting started, here are some steps to solve the problem:

1. Express the area of this region as an integral involving $f(x) = \sqrt{1 + x^4}$ in the usual way.
2. Find a power series expansion for $f(x)$. If you're allowed to use the hint without proof, then you already know $\sqrt{1+u} = 1 + \frac u 2 - \frac{u^2}{8} + \ldots$, but you don't want a power series for $\sqrt{1+u}$. Substitute an appropriate function of $x$ for $u$ in this formula to get a power series expansion for $f(x)$.
3. Once you have your power series for $f(x)$, get rid of all but the first $5$ (nonzero) terms in the series and replace $f(x)$ in the integral from part 1. with these 5 terms.
4. Evaluate the resulting integral.

Answer 2

Hint

In the expansion $$\sqrt{1+y} = 1 + \frac{y}{2} - \frac{y^2}{8} + \frac{y^3}{16} - \frac{5y^4}{128} + ...$$ just replace $y$ by $x^4$ and integrate with respect to $x$

Answer 3

Replace $x$ in $\sqrt{1+x}$ by $x^{4}$.

$f(x)=\sqrt{1+x^{4}}\approx \\\approx 1+\frac{x^{4}}{2}-\frac{x^{8}}{8}+\frac{x^{12}}{16}-\frac{5x^{16}}{128}$