Question

consider the integral $\int x^{4}e^{x}dx$: applying the integration by parts technique, let $u =$ and $dv = dx$ then $uv-\int vdu=$ $-\int dx$

Explanation:

Step1: Recall integration - by - parts formula

The integration - by - parts formula is \(\int u\;dv=uv-\int v\;du\). When choosing \(u\) and \(dv\) for \(\int x^{4}e^{x}dx\), we follow the LIATE (Logarithmic, Inverse - trigonometric, Algebraic, Trigonometric, Exponential) rule. Here, \(u = x^{4}\) (Algebraic function) and \(dv=e^{x}dx\) (Exponential function).

Step2: Find \(du\) and \(v\)

If \(u = x^{4}\), then by the power rule \(du = 4x^{3}dx\). If \(dv=e^{x}dx\), then integrating \(dv\) with respect to \(x\) gives \(v=\int e^{x}dx=e^{x}\).

Step3: Apply the integration - by - parts formula

Substitute \(u\), \(v\), \(du\) into the formula \(\int u\;dv = uv-\int v\;du\). We get \(uv=x^{4}e^{x}\) and \(v\;du = 4x^{3}e^{x}dx\). So \(uv-\int v\;du=x^{4}e^{x}-\int4x^{3}e^{x}dx\).

Answer:

\(u = x^{4}\), \(dv = e^{x}dx\), \(uv-\int v\;du=x^{4}e^{x}-\int4x^{3}e^{x}dx\)