Approximate the function f(x) = integral from 0 to x

Approximate the function $f(x) = \int_0^x \frac{t}{1+t^4} \, dt$ with a second-degree Maclaurin polynomial and use it to calculate the approximate value of $f\left(\frac{1}{\sqrt{10}}\right)$. Estimate the error.

We will find the Maclaurin polynomial of degree 2 for $f(x)$, use it to approximate $f\left(\frac{1}{\sqrt{10}}\right)$, and then estimate the error using Taylor's theorem with remainder.

Step 1: Maclaurin polynomial of degree 2

The Maclaurin polynomial of degree 2 is:

$$P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2$$

We need to find $f(0)$, $f'(0)$, and $f''(0)$.

Step 2: Calculate $f(0)$

$$f(0) = \int_0^0 \frac{t}{1+t^4} \, dt = 0$$

Step 3: Calculate $f'(x)$ and $f'(0)$

By the Fundamental Theorem of Calculus:

$$f'(x) = \frac{x}{1+x^4}$$

Therefore:

$$f'(0) = \frac{0}{1+0^4} = 0$$

Step 4: Calculate $f''(x)$ and $f''(0)$

We need to differentiate $f'(x) = \frac{x}{1+x^4}$ using the quotient rule:

$$f''(x) = \frac{d}{dx}\left(\frac{x}{1+x^4}\right) = \frac{(1+x^4) \cdot 1 - x \cdot 4x^3}{(1+x^4)^2} = \frac{1+x^4-4x^4}{(1+x^4)^2} = \frac{1-3x^4}{(1+x^4)^2}$$

Therefore:

$$f''(0) = \frac{1-3(0)^4}{(1+0^4)^2} = \frac{1}{1} = 1$$

Step 5: Construct the Maclaurin polynomial

$$P_2(x) = 0 + 0 \cdot x + \frac{1}{2!}x^2 = \frac{x^2}{2}$$

Step 6: Approximate $f\left(\frac{1}{\sqrt{10}}\right)$

Using the Maclaurin polynomial:

$$f\left(\frac{1}{\sqrt{10}}\right) \approx P_2\left(\frac{1}{\sqrt{10}}\right) = \frac{1}{2}\left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{2} \cdot \frac{1}{10} = \frac{1}{20}$$

Step 7: Estimate the error using Taylor's theorem

By Taylor's theorem with Lagrange remainder, the error is:

$$R_2(x) = f(x) - P_2(x) = \frac{f'''(c)}{3!}x^3$$

for some $c$ between 0 and $x$.

We need to find $f'''(x)$.

Calculate $f'''(x)$:

Starting from $f''(x) = \frac{1-3x^4}{(1+x^4)^2}$, we use the quotient rule:

$$f'''(x) = \frac{d}{dx}\left(\frac{1-3x^4}{(1+x^4)^2}\right)$$

Let $u = 1-3x^4$ and $v = (1+x^4)^2$. Then:

$$u' = -12x^3$$

$$v' = 2(1+x^4) \cdot 4x^3 = 8x^3(1+x^4)$$

Using the quotient rule:

$$f'''(x) = \frac{u'v - uv'}{v^2} = \frac{-12x^3(1+x^4)^2 - (1-3x^4) \cdot 8x^3(1+x^4)}{(1+x^4)^4}$$

Factor out $(1+x^4)$ from the numerator:

$$f'''(x) = \frac{(1+x^4)[-12x^3(1+x^4) - 8x^3(1-3x^4)]}{(1+x^4)^4} = \frac{-12x^3(1+x^4) - 8x^3(1-3x^4)}{(1+x^4)^3}$$

Expand the numerator:

$$-12x^3 - 12x^7 - 8x^3 + 24x^7 = -20x^3 + 12x^7 = 4x^3(-5 + 3x^4)$$

Therefore:

$$f'''(x) = \frac{4x^3(3x^4-5)}{(1+x^4)^3}$$

Step 8: Estimate $|f'''(c)|$ for $c \in \left[0, \frac{1}{\sqrt{10}}\right]$

For $x \in \left[0, \frac{1}{\sqrt{10}}\right]$:

Since $\frac{1}{\sqrt{10}} \approx 0.316$, we have $x^4 \leq \left(\frac{1}{\sqrt{10}}\right)^4 = \frac{1}{100}$.

For such small values, $3x^4 \leq \frac{3}{100} = 0.03 < 5$, so $3x^4 - 5 < 0$.

Therefore:

$$f'''(x) = \frac{4x^3(3x^4-5)}{(1+x^4)^3} < 0$$

For the error bound, we need $\max_{c \in [0, x]} |f'''(c)|$.

At $x = 0$: $f'''(0) = 0$

At $x = \frac{1}{\sqrt{10}}$:

$$f'''\left(\frac{1}{\sqrt{10}}\right) = \frac{4 \cdot \frac{1}{10^{3/2}} \cdot \left(\frac{3}{100} - 5\right)}{\left(1 + \frac{1}{100}\right)^3}$$

Let's simplify:

$$= \frac{4 \cdot \frac{1}{10\sqrt{10}} \cdot \left(\frac{3-500}{100}\right)}{\left(\frac{101}{100}\right)^3} = \frac{\frac{4}{10\sqrt{10}} \cdot \frac{-497}{100}}{\frac{101^3}{100^3}}$$

$$= \frac{4 \cdot (-497) \cdot 100^2}{10\sqrt{10} \cdot 100 \cdot 101^3} = \frac{-4 \cdot 497 \cdot 100}{10\sqrt{10} \cdot 101^3}$$

This is getting complicated. Let's use a simpler bound.

Simpler approach for error bound:

For $x \in \left[0, \frac{1}{\sqrt{10}}\right]$, note that:

$$|f'''(x)| = \frac{4|x|^3|3x^4-5|}{(1+x^4)^3}$$

Since $x^4 \leq \frac{1}{100}$, we have $3x^4 \leq 0.03$, so:

$$|3x^4 - 5| \leq 5$$

Also, $(1+x^4)^3 \geq 1$.

Therefore:

$$|f'''(x)| \leq \frac{4x^3 \cdot 5}{1} = 20x^3$$

At $x = \frac{1}{\sqrt{10}}$:

$$|f'''(c)| \leq 20 \cdot \left(\frac{1}{\sqrt{10}}\right)^3 = \frac{20}{10^{3/2}} = \frac{20}{10\sqrt{10}} = \frac{2}{\sqrt{10}} = \frac{2\sqrt{10}}{10} = \frac{\sqrt{10}}{5}$$

Step 9: Calculate the error bound

$$|R_2(x)| = \left|\frac{f'''(c)}{3!}x^3\right| \leq \frac{1}{6} \cdot \frac{\sqrt{10}}{5} \cdot \left(\frac{1}{\sqrt{10}}\right)^3$$

$$= \frac{1}{6} \cdot \frac{\sqrt{10}}{5} \cdot \frac{1}{10^{3/2}} = \frac{1}{6} \cdot \frac{\sqrt{10}}{5 \cdot 10\sqrt{10}} = \frac{1}{6} \cdot \frac{1}{50} = \frac{1}{300}$$

Final Answer

Approximate value:

$$\boxed{f\left(\frac{1}{\sqrt{10}}\right) \approx \frac{1}{20} = 0.05}$$

Error bound:

$$\boxed{|R_2| \leq \frac{1}{300} \approx 0.00333}$$

This means:

$$f\left(\frac{1}{\sqrt{10}}\right) = 0.05 \pm 0.00333$$

or

$$0.04667 \leq f\left(\frac{1}{\sqrt{10}}\right) \leq 0.05333$$

Verification (optional)

We can verify our approximation by noting that for small $x$:

$$\frac{t}{1+t^4} \approx t(1-t^4) = t - t^5$$

Therefore:

$$f(x) = \int_0^x \frac{t}{1+t^4} \, dt \approx \int_0^x (t - t^5) \, dt = \frac{x^2}{2} - \frac{x^6}{6}$$

For $x = \frac{1}{\sqrt{10}}$:

$$f\left(\frac{1}{\sqrt{10}}\right) \approx \frac{1}{2} \cdot \frac{1}{10} - \frac{1}{6} \cdot \frac{1}{1000} = \frac{1}{20} - \frac{1}{6000} = 0.05 - 0.000167 \approx 0.04983$$

This confirms our approximation and error estimate!

Summary

1. The Maclaurin polynomial of degree 2 for $f(x) = \int_0^x \frac{t}{1+t^4} \, dt$ is simply $P_2(x) = \frac{x^2}{2}$.
2. The approximate value is $f\left(\frac{1}{\sqrt{10}}\right) \approx \frac{1}{20} = 0.05$.
3. The error is bounded by $\frac{1}{300} \approx 0.0033$, which is quite small.
4. The approximation is very good because $\frac{1}{\sqrt{10}} \approx 0.316$ is close to 0, where Maclaurin polynomials work best.

Key lesson: Maclaurin polynomials provide excellent approximations for functions near $x = 0$, and the error can be rigorously bounded using Taylor's theorem with remainder.