# How do you find the definite integral for: #xe^(x^2 + 2) # for the intervals #[0, 2]#?

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To find the definite integral of (xe^{x^2 + 2}) over the interval ([0, 2]), you can use the technique of substitution. Let (u = x^2 + 2), then (du = 2x , dx). Rearranging, we get ( \frac{1}{2} du = x , dx). Now, we substitute (u) and (du) into the integral:

[ \int_{0}^{2} xe^{x^2 + 2} , dx = \int_{0}^{2} e^u \cdot \frac{1}{2} du ]

Now, we integrate with respect to (u) from the transformed limits:

[ \int_{0}^{2} e^u \cdot \frac{1}{2} du = \frac{1}{2} \int_{2}^{6} e^u , du ]

Now, integrating (e^u) with respect to (u), we get:

[ \frac{1}{2} \left[ e^u \right]_{2}^{6} = \frac{1}{2} \left( e^6 - e^2 \right) ]

So, the definite integral of (xe^{x^2 + 2}) over the interval ([0, 2]) is (\frac{1}{2} \left( e^6 - e^2 \right)).

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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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