To estimate the value of the integral

$\int_{0}^{0.3} e^{-x^{2}} \, dx,$

using a series expansion, we can use the Maclaurin series expansion of $e^{-x^2}$. This series expansion is given by:

$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \frac{x^8}{4!} - \cdots.$

We will approximate the integral by integrating the truncated series term by term.

Step 1: Write the integral as a sum

The integral can be written as:

$\int_{0}^{0.3} e^{-x^{2}} \, dx \approx \int_{0}^{0.3} \left( 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots \right) dx.$

Now we integrate each term of the series separately:

$\int_{0}^{0.3} 1 \, dx = x \bigg|_0^{0.3} = 0.3,$

$\int_{0}^{0.3} x^2 \, dx = \frac{x^3}{3} \bigg|_0^{0.3} = \frac{0.3^3}{3} = \frac{0.027}{3} = 0.009,$

$\int_{0}^{0.3} \frac{x^4}{2!} \, dx = \frac{1}{2} \cdot \frac{x^5}{5} \bigg|_0^{0.3} = \frac{1}{2} \cdot \frac{0.3^5}{5} = \frac{1}{2} \cdot \frac{0.00243}{5} = 0.000243,$

$\int_{0}^{0.3} \frac{x^6}{3!} \, dx = \frac{1}{6} \cdot \frac{x^7}{7} \bigg|_0^{0.3} = \frac{1}{6} \cdot \frac{0.3^7}{7} = \frac{1}{6} \cdot \frac{0.0002187}{7} = 0.00000522,$

$\int_{0}^{0.3} \frac{x^8}{4!} \, dx = \frac{1}{24} \cdot \frac{x^9}{9} \bigg|_0^{0.3} = \frac{1}{24} \cdot \frac{0.3^9}{9} = \frac{1}{24} \cdot \frac{0.00001944}{9} = 0.000000090.$

Step 2: Add the results

Now sum up all the integrals:

$0.3 - 0.009 + 0.000243 - 0.00000522 + 0.000000090 = 0.29023787.$

Step 3: Check the error

To ensure the error is less than $10^{-8}$, consider the next term in the series. The next term involves $x^{10}$, which gives a very small value (approximately $0.00000000043$). This is sufficiently small, so the error due to truncating after the $x^8$-term will be less than $10^{-8}$.

Final answer:

Thus, the estimate for the integral is:

$\int_{0}^{0.3} e^{-x^{2}} \, dx \approx 0.29023787.$

This result is rounded to eight decimal places.

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Would you like more details on this method or a further explanation of any step? Here are some related questions:

1. How do you estimate the error when truncating a series?
2. Can we use this same method for other functions, like $e^{x^2}$?
3. What happens if we include more terms in the series for higher accuracy?
4. How would this series expansion change for an integral over a different interval?
5. What is the importance of series expansions in numerical integration?

Tip: When working with series approximations, the more terms you include, the more accurate your result will be, but you also increase the complexity of the calculations. Always ensure the error is below the required threshold.