How do you use integration by parts to find #intxe^-x dx#?
Explanation
Using Integration by Parts,
Similarly following for the problem,
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To integrate ( x e^{-x} ) using integration by parts, you would typically follow these steps:
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Choose which part of the integrand to designate as ( u ) and which part to designate as ( dv ) in the integration by parts formula: ( \int u , dv = uv - \int v , du ).
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Differentiate ( u ) to find ( du ), and integrate ( dv ) to find ( v ).
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Substitute the expressions for ( u ), ( du ), ( v ), and ( dv ) into the integration by parts formula.
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Simplify and evaluate the resulting integral.
For ( \int x e^{-x} , dx ), a common choice is to let ( u = x ) and ( dv = e^{-x} , dx ).
Then, ( du = dx ) and ( v = -e^{-x} ).
Substituting into the integration by parts formula:
( \int x e^{-x} , dx = -xe^{-x} - \int (-e^{-x}) , dx )
( = -xe^{-x} + \int e^{-x} , dx )
( = -xe^{-x} - e^{-x} + 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.
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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