To estimate the integral

$I = \int_0^1 f(x) \, dx \quad \text{where} \quad f(x) = e^{-x^2 / 2}$

using the degree 2 Taylor polynomial centered at the origin, we follow these steps:

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1. Compute the Taylor series of $f(x)$ around $x = 0$

The Taylor series expansion of $f(x) = e^{-x^2 / 2}$ is obtained by calculating derivatives at $x = 0$:

1. $f(x) = e^{-x^2 / 2}$
2. $f'(x) = -x e^{-x^2 / 2}$
3. $f''(x) = (-1 + x^2)e^{-x^2 / 2}$

Evaluate these derivatives at $x = 0$:

• $f(0) = 1$,
• $f'(0) = 0$,
• $f''(0) = -1$.

Thus, the degree 2 Taylor polynomial for $f(x)$ is:

$T_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 = 1 - \frac{1}{2}x^2.$

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2. Approximate the integral using $T_2(x)$:

$I \approx \int_0^1 T_2(x) \, dx = \int_0^1 \left( 1 - \frac{1}{2}x^2 \right) dx.$

Split the integral:

$\int_0^1 T_2(x) \, dx = \int_0^1 1 \, dx - \frac{1}{2} \int_0^1 x^2 \, dx.$

• $\int_0^1 1 \, dx = [x]_0^1 = 1$,
• $\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$.

Thus:

$\int_0^1 T_2(x) \, dx = 1 - \frac{1}{2} \cdot \frac{1}{3} = 1 - \frac{1}{6} = \frac{5}{6}.$

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Final Answer:

The estimate for the integral $I$ using the degree 2 Taylor polynomial is:

$I \approx \frac{5}{6}.$

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Do you have questions or need further details?

Here are 5 related questions you might explore:

1. How accurate is the Taylor polynomial approximation compared to the actual value of the integral?
2. Can we improve the estimate by using a degree 4 Taylor polynomial?
3. How is the error of Taylor polynomial approximations calculated?
4. What does the full Taylor series expansion of $e^{-x^2/2}$ look like?
5. How would this approximation change if the integral limits were altered?

Tip: When using Taylor series approximations for integrals, always check if the higher-order terms significantly impact the error.