# How do you find the Improper integral #int (x^2)e^[(-x^2)/2] dx # from x=-∞ to x=∞?

These two methods require external/prior knowledge, which is highlighted below

METHOD 2

that's the same answer with two slightly different approaches.

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To find the improper integral ∫(x^2)e^(-x^2/2) dx from x=-∞ to x=∞, we can use the properties of even functions and integrate over the entire real line. Since the integrand is even, we have:

∫(-∞ to ∞) (x^2)e^(-x^2/2) dx = 2 * ∫(0 to ∞) (x^2)e^(-x^2/2) dx

We can then use integration by parts to evaluate this integral. Let u = x^2 and dv = e^(-x^2/2) dx. Then, du = 2x dx and v = -e^(-x^2/2).

Applying the integration by parts formula:

∫ u dv = uv - ∫ v du

We get:

2 * ∫(0 to ∞) (x^2)e^(-x^2/2) dx = -2x^2 e^(-x^2/2) ∣(0 to ∞) + 4 * ∫(0 to ∞) x e^(-x^2/2) dx

Since e^(-x^2/2) approaches 0 as x approaches ∞, the first term in the integral evaluates to 0. Now, we can use a substitution, letting z = -x^2/2, so -dz = x dx:

4 * ∫(0 to ∞) x e^(-x^2/2) dx = -4 ∫(0 to ∞) e^z dz = -4e^z ∣(0 to ∞) = -4(0 - e^0) = -4(-1) = 4

Therefore, the improper integral ∫(-∞ to ∞) (x^2)e^(-x^2/2) dx converges and its value is 4.

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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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