We will use the degree 4 Taylor polynomial for e−x2 at x=0 to approxmate

∫0.170e−x2dx

First enter in the degree 4 Taylor polynomial (hint: you don't need to calculate it with derivatives, you can use the Taylor series for ex)

P4(x)=   

Now evaluate ∫0.170P4(x)dx to get the approximate value for the original integral, giving your answer accurate to 4 decimal places
 
 
Note that the answer will most likely give a different answer to a calculator for the original integral, but it won't be too far off (the calculator is also using an approximation like this, but a better one).

Let's try a bigger value. Use the same method to approximate ∫1.440e−x2dx
 
You should find now that it is is quite different to a calculator value. It turns out that this approximation starts to significantly diverge from the function above about x=0.6 and then goes badly wrong above x=1. You can see this for yourself if you plot them together on a graph. To get an approximation that works for larger x you need more terms from the Taylor Series. If you include enough terms you can get a good approximation of any x!
 

Let's break down the problem step by step.

### 1. Degree 4 Taylor Polynomial for \( e^{-x^2} \)

The Taylor series for \( e^x \) centered at \( x = 0 \) is:
\[
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots
\]

To obtain the Taylor series for \( e^{-x^2} \), we substitute \( -x^2 \) in place of \( x \):
\[
e^{-x^2} = 1 - x^2 + \frac{(-x^2)^2}{2!} - \frac{(-x^2)^3}{3!} + \frac{(-x^2)^4}{4!} + \dots
\]

Simplifying this gives the degree 4 polynomial \( P_4(x) \):
\[
P_4(x) = 1 - x^2 + \frac{x^4}{2}
\]

### 2. Approximate the Integral \( \int_{0}^{0.17} e^{-x^2} dx \)

We will approximate this integral using the degree 4 polynomial \( P_4(x) \) instead of \( e^{-x^2} \):
\[
\int_{0}^{0.17} P_4(x) dx = \int_{0}^{0.17} \left( 1 - x^2 + \frac{x^4}{2} \right) dx
\]

Let's compute this integral.

### 3. Compute the Integral \( \int_{0}^{1.44} e^{-x^2} dx \)

Similarly, we will approximate this integral using the same degree 4 polynomial:
\[
\int_{0}^{1.44} P_4(x) dx = \int_{0}^{1.44} \left( 1 - x^2 + \frac{x^4}{2} \right) dx
\]

Now, let's calculate both integrals.### Results:

1. **Approximate value for the integral** \( \int_{0}^{0.17} e^{-x^2} dx \):
   \[
   \int_{0}^{0.17} P_4(x) dx \approx 0.1684
   \]

2. **Approximate value for the integral** \( \int_{0}^{1.44} e^{-x^2} dx \):
   \[
   \int_{0}^{1.44} P_4(x) dx \approx 1.0638
   \]

These values provide approximations of the integrals using the degree 4 Taylor polynomial. The first approximation (from 0 to 0.17) should be close to the true value, while the second (from 0 to 1.44) will diverge more noticeably due to the limitations of using a low-degree polynomial for larger \( x \).

Would you like more details or have any other questions?

Here are 5 related questions to explore:
1. How does the error change if you increase the degree of the Taylor polynomial?
2. What is the exact value of the integrals computed by a more precise numerical method?
3. How does the Taylor series approximation behave for \( e^{-x^2} \) as \( x \) increases?
4. How would including the next term in the Taylor series improve the approximation for larger \( x \)?
5. Can you derive the general formula for the Taylor series of \( e^{-x^2} \) at \( x = 0 \)?

**Tip:** When using Taylor polynomials to approximate integrals or functions, the degree of the polynomial and the interval of approximation significantly affect accuracy.

Learn how to approximate integrals using Taylor polynomials. Explore the degree 4 Taylor polynomial for e^(-x^2) at x=0 and use it to approximate ∫0.170e^(-x^2) dx and ∫1.440e^(-x^2) dx. Understand the limitations of this method for larger x values. Suitable for advanced high school students studying calculus.

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