# How do you integrate #(x^2e^(x/2))dx#?

Then:

So:

Hence:

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To integrate ( x^2e^{x/2} ) with respect to ( x ), you can use integration by parts.

Let ( u = x^2 ) and ( dv = e^{x/2} dx ).

Then, ( du = 2x dx ) and ( v = 2e^{x/2} ).

Now, apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values of ( u ) and ( v ):

[ \int x^2e^{x/2} , dx = x^2 \cdot 2e^{x/2} - \int 2e^{x/2} \cdot 2x , dx ]

Simplify:

[ = 2x^2e^{x/2} - 4\int xe^{x/2} , dx ]

Now, integrate ( \int xe^{x/2} , dx ) using integration by parts again:

Let ( u = x ) and ( dv = e^{x/2} dx ).

Then, ( du = dx ) and ( v = 2e^{x/2} ).

Apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values of ( u ) and ( v ):

[ \int xe^{x/2} , dx = x \cdot 2e^{x/2} - \int 2e^{x/2} , dx ]

Simplify:

[ = 2xe^{x/2} - 4\int e^{x/2} , dx ]

[ = 2xe^{x/2} - 8e^{x/2} + C ]

Now, substitute back into the original integration:

[ \int x^2e^{x/2} , dx = 2x^2e^{x/2} - 4(2xe^{x/2} - 8e^{x/2}) + C ]

[ = 2x^2e^{x/2} - 8xe^{x/2} + 32e^{x/2} + C ]

[ = e^{x/2}(2x^2 - 8x + 32) + C ]

So, ( \int x^2e^{x/2} , dx = e^{x/2}(2x^2 - 8x + 32) + C ), where ( C ) is the constant of integration.

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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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