Question

log 2 ∫ xe⁻ˣ dx ,Evaluate it. 0

Answer 1

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Answer 2

Answer:

$-log2[xe^{-x}\ +\ e^{-x}]$

Step-by-step explanation:

In this question,

We have been asked to evaluate

$log2\ \int{xe^{-x}\ dx}$

Using Integration by parts,

$\int{uv\ dx}\ =\ u\int{vdx}\ -\ \int{[\frac{du}{dx}\int{vdx}]dx}$

Here u = x

         v = $e^{-x}$

Putting the values we get,

So, $\int{xe^{-x}dx}\ =\ x\int{e^{-x}dx}\ -\ \int{[\frac{d}{dx}(x)\int{e^{-x}dx}]dx}$

$\int{xe^{-x}dx}\ =\ x(-e^{-x})\ -\ \int{1(-e^{-x})dx}$    {$\int{e^{-x}dx}\ =\ -e^{-x}$}

$\int{xe^{-x}dx}\ =\ -xe^{-x}\ +\ \int{e^{-x}dx}$

$\int{xe^{-x}dx}\ =\ -xe^{-x}\ -\ e^{-x}$          {$\int{e^{-x}dx}\ =\ -e^{-x}$}

Therefore, $log2\int{xe^{-x}dx}\ =\ -log2[xe^{-x}\ +\ e^{-x}]$