How do you show that integration of $x^{m} e^{a x} d x = \frac{x^{m} e^{a x}}{a} - \frac{m}{a} \int x^{m - 1} e^{a x} d x$ $x^m e^(ax)dx = (x^m e^(ax) )/a - m/a int x^(m-1) e^(ax) dx$ ?

Question

How do you show that integration of $x^{m} e^{a x} d x = \frac{x^{m} e^{a x}}{a} - \frac{m}{a} \int x^{m - 1} e^{a x} d x$ $x^m e^(ax)dx = (x^m e^(ax) )/a - m/a int x^(m-1) e^(ax) dx$ ?

1 Answer

Impact of this question

4509 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-11T00:09:04

Set $u = x^{m}$ $u = x^m$ and $\frac{d v}{d x} = e^{a x}$ $(dv)/dx = e^(ax)$ and then apply the integration by parts formula.

Explanation:

The integration by parts formula states that for continuous functions $u (x)$ $u(x)$ and $v (x)$ $v(x)$

$\int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x$ $intu(dv)/(dx)dx = uv-intv(du)/(dx)dx$

(A short proof of the integration by parts formula can be found here)

Starting from
$\int x^{m} e^{a x} d x$ $intx^me^(ax)dx$

we set $u (x) = x^{m}$ $u(x) = x^m$ and $d \frac{v}{d x} = e^{a x}$ $dv/dx = e^(ax)$ .

Then, differentiating and integrating gives us
$\frac{d u}{d x} = m x^{m - 1}$ $(du)/dx = mx^(m-1)$ and $v (x) = \frac{e^{a x}}{a}$ $v(x) = (e^(ax))/a$

Substituting these into the integration by parts formula gives the desired result of
$\int x^{m} e^{a x} d x = x^{m} \cdot \frac{e^{a x}}{a} - \int \frac{e^{a x}}{a} \cdot m x^{m - 1} d x$ $intx^me^(ax)dx = x^m*e^(ax)/a - inte^(ax)/a*mx^(m-1)dx$

$= \frac{x^{m} e^{a x}}{a} - \frac{m}{a} \int x^{m - 1} e^{a x} d x$ $=(x^me^(ax))/a - m/a intx^(m-1)e^(ax)dx$

Science

Math

Humanities

... and beyond

How do you show that integration of $x^{m} e^{a x} d x = \frac{x^{m} e^{a x}}{a} - \frac{m}{a} \int x^{m - 1} e^{a x} d x$ $x^m e^(ax)dx = (x^m e^(ax) )/a - m/a int x^(m-1) e^(ax) dx$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you show that integration of xmeaxdx=xmeaxa−ma∫xm−1eaxdxx^m e^(ax)dx = (x^m e^(ax) )/a - m/a int x^(m-1) e^(ax) dx?

1 Answer

Explanation:

Related questions

Impact of this question

How do you show that integration of $x^{m} e^{a x} d x = \frac{x^{m} e^{a x}}{a} - \frac{m}{a} \int x^{m - 1} e^{a x} d x$ $x^m e^(ax)dx = (x^m e^(ax) )/a - m/a int x^(m-1) e^(ax) dx$ ?