How to Calculate the Definite Integral of 1 - e^{-x^2}

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The definite integral of a complex function such as 1 - e^{-x^2} over the interval [0, 1] cannot be expressed in terms of elementary functions. However, we can still evaluate it using series expansion and numerical approximation techniques. Let's break down the process step-by-step.

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The Definite Integral and Initial Split

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To evaluate the integral ( I int_{0}^{1} (1 - e^{-x^2}) , dx ), we start by splitting the integral into two separate integrals:

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[ I int_{0}^{1} 1 , dx - int_{0}^{1} e^{-x^2} , dx ]

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The first integral is straightforward:

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[ int_{0}^{1} 1 , dx 1 ]

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Thus, we can rewrite the integral ( I ) as:

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[ I 1 - int_{0}^{1} e^{-x^2} , dx ]

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We denote the second integral by ( J ):

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[ J int_{0}^{1} e^{-x^2} , dx ]

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Series Expansion of the Exponential Function

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The exponential function ( e^x ) can be expanded as an infinite series:

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[ e^x sum_{n0}^{infty} frac{x^n}{n!} ]

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By substituting (-x^2) for ( x ) in the series expansion, we get:

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[ e^{-x^2} sum_{n0}^{infty} frac{(-x^2)^n}{n!} sum_{n0}^{infty} frac{(-1)^n x^{2n}}{n!} ]

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Substituting into the Integral

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Substituting this series expansion into the integral ( J ), we get:

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[ J int_{0}^{1} e^{-x^2} , dx int_{0}^{1} sum_{n0}^{infty} frac{(-1)^n x^{2n}}{n!} , dx ]

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Since the series converges uniformly, we can interchange the summation and integration:

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[ J sum_{n0}^{infty} frac{(-1)^n}{n!} int_{0}^{1} x^{2n} , dx ]

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Evaluating the Series Integral

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Each term of the series can be integrated as follows:

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[ int_{0}^{1} x^{2n} , dx left[ frac{x^{2n 1}}{2n 1} right]_{0}^{1} frac{1}{2n 1} ]

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Therefore, the series becomes:

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[ J sum_{n0}^{infty} frac{(-1)^n}{(2n 1)n!} ]

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Numerical Approximation

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Since the series converges slowly, we can approximate ( J ) by summing the first few terms of the series until the terms become small enough to be negligible. Let's calculate the first few terms:

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[ J approx 1 - frac{1}{3 cdot 1!} - frac{1}{10 cdot 2!} - frac{1}{42 cdot 3!} ]

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Evaluating each term:

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[ 1 1 ]

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[ -frac{1}{3 cdot 1!} -frac{1}{3} approx -0.333 ]

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[ -frac{1}{10 cdot 2!} -frac{1}{20} -0.05 ]

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[ -frac{1}{42 cdot 3!} -frac{1}{252} approx -0.004 ]

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Adding these terms together:

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[ J approx 1 - 0.333 - 0.05 - 0.004 0.613 ]

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Final Calculation of the Integral

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Now, we substitute the approximation of ( J ) into the expression for ( I ):

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[ I 1 - J approx 1 - 0.613 0.387 ]

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Therefore, the value of the integral ( I ) is approximately:

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[ boxed{I approx 0.26} ]