# How do you use the integral test to determine whether #int e^(-x^2)# converges or diverges from #[0,oo)#?

is equivalent to the convergence of the sum:

based on the integral test theorem.

so the integral must also be convergent.

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To determine whether the integral of ( e^{-x^2} ) from ( 0 ) to ( \infty ) converges or diverges, we use the integral test for convergence. This involves comparing the function's integral to a known convergent or divergent integral. However, the integral of ( e^{-x^2} ) does not have a elementary antiderivative, so we cannot integrate it directly. Consequently, we employ a different approach, considering its behavior within the integral.

The function ( e^{-x^2} ) is always positive, continuous, and decreasing on the interval ([0, \infty)). Thus, we can't use the integral test directly. However, we can compare it to the Gaussian integral ( \int_{0}^{\infty} e^{-x^2} , dx ), which is a well-known convergent integral. Since ( e^{-x^2} \leq e^{-x} ) for all ( x \geq 1 ), and ( e^{-x} ) is integrable on ([1, \infty)), we can conclude that ( e^{-x^2} ) is also integrable on ([1, \infty)), and therefore on ([0, \infty)). This implies that the integral ( \int_{0}^{\infty} e^{-x^2} , dx ) converges.

Thus, by comparison with the convergent integral ( \int_{0}^{\infty} e^{-x^2} , dx ), we can conclude that ( \int_{0}^{\infty} e^{-x^2} , dx ) also converges.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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