Question

use integration by parts to prove the reduction formula:\(\int {{x^n}{e^x}} dx = {x^n}{e^x} - n\int {{x^{n - 1}}{e^x}} dx\)

Short answer

We can prove the following reduction formula by using integration by parts:

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

Step-by-step solution

Use the integration by parts

\(\begin{array}{l}uv = \int {vdu} + \int {udv} \\\int {udv = uv - \int {vdu\,\,\,\,\,\,\,\,\,\,........\left( 1 \right)} } \end{array}\)

Let \(\begin{array}{l}u = {x^n}\\du = n{x^{n - 1}}dx\\dv = {e^x}dx\\v = {e^x}\end{array}\)

Substitute \(u,v,du,dv\) in the above equation

\(\begin{array}{l}\int {udv = uv - \int {vdu} } \\\int {{x^n}{e^x}dx = {x^n}{e^x} - } \int {n{x^{n - 1}}{e^x}dx} \\\int {{x^n}{e^x}dx = {x^n}{e^x} - n} \int {{x^{n - 1}}{e^x}dx} \end{array}\)

Hence, we can prove that for the reduction formula by using integration by parts:

\(\int {{x^n}{e^x}dx = {x^n}{e^x} - n} \int {{x^{n - 1}}{e^x}dx} \).