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5. \( y=e^{-x^{2}}, \quad y=0, \quad x=0, \quad x=1 \)

Ask by Murray Rogers. in Yemen
Feb 25,2025

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Answer

The area of the region bounded by \( y = e^{-x^{2}} \), \( y = 0 \), \( x = 0 \), and \( x = 1 \) is \( \frac{\sqrt{\pi}}{2} \) times the error function evaluated at 1, which is approximately 0.7468 square units.

Solution

We want to find the area of the region bounded by the curves   y = e^(–x²), y = 0, x = 0, and x = 1. This region lies above the x-axis and under the curve y = e^(–x²) from x = 0 to x = 1. Step 1. Write the integral for the area. The area A can be found by integrating the function with respect to x from 0 to 1:   A = ∫[0 to 1] e^(–x²) dx Step 2. Notice that the antiderivative of e^(–x²) cannot be written in terms of elementary functions. However, we can express the answer in terms of the error function, erf(x). Recall that the error function is defined as   erf(x) = (2/√π) ∫[0 to x] e^(–t²) dt Thus,   ∫[0 to 1] e^(–x²) dx = (√π/2) erf(1) Step 3. Write the final answer. The exact area is   A = (√π/2) erf(1) If you need a numerical approximation, note that erf(1) ≈ 0.8427007929. Therefore,   A ≈ (√π/2) × 0.8427007929 ≈ (1.77245385/2) × 0.8427007929 ≈ 0.7468 So, the area of the region is (√π/2) erf(1), which is approximately 0.7468 square units.

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Extra Insights

The function \( y = e^{-x^2} \) is known as the Gaussian function and plays a crucial role in statistics, particularly in normal distribution. The area under the curve from \( x = 0 \) to \( x = 1 \) can be computed using definite integrals. This area represents the probability of a random variable following a normal distribution being between these two points, and it's approximately 0.3413 when calculated. When evaluating the definite integral to find the area under this curve, many make the mistake of trying to find a simple antiderivative. In reality, the integral cannot be expressed in elementary terms, requiring numerical methods or special functions like the error function (\( \text{erf} \)). So, don't worry if you can't find the antiderivative directly—numerical approximation techniques can be your best buddies!

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