Question

Use integration by parts to verify the formula. (For Exercises \(57-60\), assume that \(n\) is a positive integer.) $$ \int x^{n} e^{a x} d x=\frac{x^{n} e^{a x}}{a}-\frac{n}{a} \int x^{n-1} e^{a x} d x $$

Short answer

The given integral formula is verified using integration by parts. The integral \(\int x^{n} e^{a x} dx\) is indeed equal to \(\frac{x^{n} e^{a x}}{a}-\frac{n}{a} \int x^{n-1} e^{a x} dx\).

Step-by-step solution

Understand Integration by Parts

Integration by parts is a method for solving integrals, and is based on the rule of differentiation for the product of two functions. It is given by: \[\int u dv = uv - \int v du\].

Assign functions to \(u\) and \(dv\)

We must assign the functions to \(u\) and \(dv\) to proceed with the integration by parts. Here, assign \(u = x^{n}\) and \(dv = e^{ax} dx\)

Find \(du\) and \(v\)

After assigning \(u\) and \(dv\), calculate \(du\) and \(v\). The derivative of \(u\) is \(n x^{n - 1} dx\), so \(du = n x^{n - 1} dx\). Integration of \(dv = e^{ax} dx\) gives \(v = \frac{1}{a}e^{ax}\).

Apply the formula of Integration by Parts

Substitute \(u\), \(v\), \(du\), and \(dv\) into the integration by parts formula \[\int u dv = uv - \int v du\]\n to get \[\int x^{n} e^{ax} dx = x^{n} \cdot \frac{1}{a} e^{ax} - \int \frac{1}{a} e^{ax} n x^{n - 1} dx\]. This simplifies to \[\frac{x^{n} e^{ax}}{a} - \frac{n}{a} \int x^{n-1} e^{ax} dx\], which verifies the given formula.