Question

Approximating definite integrals Use a Taylor series to approximate the following definite integrals. Retain as many terms as needed to ensure the error is less than 10⁻⁴.∫₀⁰ᐧ²⁵ e⁻ˣ² dx

Accepted Answer

Recognize that the integral involves the function $$f(x) = e$$^{-x^2}$$, which does not have an elementary antiderivative, making it a good candidate for approximation using a Taylor series expansion. Write the Taylor series expansion of e$$^{-x^2}$$ centered at x=0$. Recall that the exponential function e^u$ can be expanded as $$\sum_{n=0}^{\infty} \frac{u^n}{n!}$$. Here, u$$ = $$-x^2$$, so the series becomes $$e^{-x^2}$$ = \sum_{n=0}^{\infty} \frac{(-1)^n $$x^{2n}$$}{n!}$$. Substitute the Taylor series into the integral to approximate it: $$\$$int_0$$^{0.25} $$e^{-x^2}$$ \, dx = \$$int_0$$^{0.25} \sum_{n=0}^{\infty} \frac{(-1)^n $$x^{2n}$$}{n!} \, dx$$. Interchange the integral and the summation (justified by uniform convergence on the interval) to get $$\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \$$int_0$$^{0.25} $$x^{2n}$$ \, dx$$. Evaluate the integral inside the summation: $$\$$int_0$$^{0.25} $$x^{2n}$$ \, dx = \frac{$$(0.25)^{2n+1}$$}{2n+1}$$. So the approximation becomes $$\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \cdot \frac{$$(0.25)^{2n+1}$$}{2n+1}$$. Determine how many terms to retain by estimating the remainder term to ensure the error is less than 10$$^{-4}$$. Use the alternating series remainder estimation or compare the magnitude of the next term to 10$$^{-4}$$. Stop adding terms once the next term's absolute value is smaller than this threshold.

Approximating definite integrals Use a Taylor series to approximate the following definite integrals. Retain as many terms as needed to ensure the error is less than 10⁻⁴.∫₀⁰ᐧ²⁵ e⁻ˣ² dx

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