Home » Blog » Evaluate the indefinite integral of x2e-2x

Evaluate the indefinite integral of x2e-2x

Compute the following indefinite integral,

    \[ \int x^2 e^{-2x} \, dx. \]


To compute this integral we will integrate by parts twice. First, let

    \begin{align*}  u &= x^2 & du &= 2x \, dx \\ dv &= e^{-2x} \, dx & v &= -\frac{1}{2} e^{-2x}. \end{align*}

Therefore we have

    \[  \int x^2 e^{-2x} \, dx = -\frac{x^2}{2} e^{-2x} + \int xe^{-2x} \, dx. \]

To evaluate this next integral we use integration by parts a second time with

    \begin{align*}  u &= x & du &= dx \\ dv &= e^{-2x}\, dx & v &= -\frac{1}{2}e^{-2x}. \end{align*}

Giving us

    \begin{align*}   \int xe^{-2x} \, dx &= -\frac{x}{2}e^{-2x} + \frac{1}{2} \int e^{-2x} \, dx \\  &= -\frac{x}{2}e^{-2x} - \frac{1}{4} e^{-2x} + C. \end{align*}

So, putting this back into our formula we have

    \begin{align*}  \int x^2 e^{-2x} \, dx &= -\frac{x^2}{2} e^{-2x} + \int xe^{-2x} \, dx \\  &= -\frac{x^2}{2} e^{-2x} - \frac{x}{2}e^{-2x} - \frac{1}{4} e^{-2x} + C \\  &= -\frac{1}{2} e^{-2x} \left( x^2 + x + \frac{1}{2} \right) + C. \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):