Integration by Parts — Question 4

Question 4

Define $$I_n=\int x^ne^x\,dx, \qquad n\geq 1.$$ Use integration by parts to derive a recurrence relation connecting $I_n$ and $I_{n-1}$. Then use the recurrence to evaluate $I_4$.

Question 4 – Solution

Start with $$I_n=\int x^ne^x\,dx.$$ To reduce the power of $x$, choose $$u=x^n,\quad dv=e^x\,dx, \qquad du=nx^{n-1}\,dx,\quad v=e^x.$$ Using $\int u\,dv=uv-\int v\,du$, we obtain $$\begin{align*} I_n &=x^ne^x-\int e^x\left(nx^{n-1}\right)\,dx \\ &=x^ne^x-n\int x^{n-1}e^x\,dx. \end{align*}$$ Since $I_{n-1}=\int x^{n-1}e^x\,dx$, the recurrence relation is $$\boxed{I_n=x^ne^x-nI_{n-1}}.$$ Here, as usual for indefinite integrals, the constant of integration is added after the recurrence has been applied. The base case is $$I_0=\int e^x\,dx=e^x.$$

Now apply the recurrence repeatedly: $$\begin{align*} I_1 &=xe^x-I_0 =xe^x-e^x =e^x(x-1), \\[3pt] I_2 &=x^2e^x-2I_1 \\ &=x^2e^x-2e^x(x-1) =e^x(x^2-2x+2), \\[3pt] I_3 &=x^3e^x-3I_2 \\ &=x^3e^x-3e^x(x^2-2x+2) \\ &=e^x(x^3-3x^2+6x-6), \\[3pt] I_4 &=x^4e^x-4I_3 \\ &=x^4e^x-4e^x(x^3-3x^2+6x-6) \\ &=e^x(x^4-4x^3+12x^2-24x+24). \end{align*}$$ Therefore, $$\boxed{\displaystyle I_4=\int x^4e^x\,dx =e^x(x^4-4x^3+12x^2-24x+24)+C}$$

Indeed, differentiating the result gives $$\begin{align*} \frac{d}{dx}\left[e^x(x^4-4x^3+12x^2-24x+24)\right] &=e^x(x^4-4x^3+12x^2-24x+24)\\ &\quad+e^x(4x^3-12x^2+24x-24)\\ &=x^4e^x, \end{align*}$$ which confirms the answer.