# How do you find the integral #int_0^1x*e^(-x^2)dx# ?

We begin by making a substitution.

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[ \int_{0}^{1} x \cdot e^{-x^2} , dx ] can be solved using integration by substitution.

Let ( u = -x^2 ) and ( du = -2x , dx ).

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

[ = -\frac{1}{2} \left[ e^u \right]_{0}^{1} ]

[ = -\frac{1}{2} (e^1 - e^0) ]

[ = -\frac{1}{2} (e - 1) ]

[ = \frac{1}{2} (1 - e) ]

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To find the integral ( \int_{0}^{1} x \cdot e^{-x^2} , dx ), you can use integration by substitution. Let ( u = -x^2 ), then ( du = -2x , dx ). Solving for ( dx ), we have ( dx = -\frac{1}{2x} , du ).

Substituting these values into the integral, we get: [ \int_{0}^{1} x \cdot e^{-x^2} , dx = \int_{u(0)}^{u(1)} -\frac{1}{2} e^{u} , du ]

Now, evaluate the integral with respect to ( u ): [ \int_{u(0)}^{u(1)} -\frac{1}{2} e^{u} , du = -\frac{1}{2} \int_{0}^{-1} e^{u} , du ]

The integral of ( e^{u} ) with respect to ( u ) is ( e^{u} ). So, we have: [ -\frac{1}{2} \int_{0}^{-1} e^{u} , du = -\frac{1}{2} \left[ e^{u} \right]_{0}^{-1} ]

Substitute the limits of integration and simplify: [ -\frac{1}{2} \left[ e^{-1} - e^0 \right] = -\frac{1}{2} \left( e^{-1} - 1 \right) ]

Therefore, the value of the integral ( \int_{0}^{1} x \cdot e^{-x^2} , dx ) is ( -\frac{1}{2} \left( e^{-1} - 1 \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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