Q.$\displaystyle\int_0^1 x^3e^{-2x}\,dx$

Step 1. List the successive derivatives of $u=x^3$. $u^{(1)}=3x^2,\ u^{(2)}=6x,\ u^{(3)}=6,\ u^{(4)}=0$ — the column terminates after three derivatives.

Step 2. List the successive anti-derivatives of $v=e^{-2x}$.

$v_{(1)}=\displaystyle\int e^{-2x}\,dx=-\dfrac12e^{-2x}$

$v_{(2)}=\displaystyle\int v_{(1)}\,dx=\dfrac14e^{-2x}$

$v_{(3)}=\displaystyle\int v_{(2)}\,dx=-\dfrac18e^{-2x}$

$v_{(4)}=\displaystyle\int v_{(3)}\,dx=\dfrac1{16}e^{-2x}$

Step 3. Apply Bernoulli's formula. $\int uv\,dx=uv_{(1)}-u^{(1)}v_{(2)}+u^{(2)}v_{(3)}-u^{(3)}v_{(4)}$ (the series stops here since $u^{(4)}=0$):

Step 4. Self-verify by differentiating $F(x)$. $F'(x)=-\dfrac18\Big[-2e^{-2x}(4x^3+6x^2+6x+3)+e^{-2x}(12x^2+12x+6)\Big]=-\dfrac18e^{-2x}\big[-8x^3-12x^2-12x-6+12x^2+12x+6\big]=-\dfrac18e^{-2x}(-8x^3)=x^3e^{-2x}$ ✓, matching the original integrand exactly.

Step 5. Evaluate $F(1)-F(0)$.

$F(1)=-\dfrac18e^{-2}(4+6+6+3)=-\dfrac{19}{8e^2}$

$F(0)=-\dfrac18(1)(0+0+0+3)=-\dfrac38$

Step 6. Combine.

Bernoulli's formula for $\int uv\,dx$ ($u=x^3$ polynomial, $v=e^{-2x}$)

• Sign error in the alternating sum $uv_{(1)}-u^{(1)}v_{(2)}+u^{(2)}v_{(3)}-\cdots$
• Stopping the series before $u^{(3)}$ contributes, since $u^{(3)}=6\ne0$ (only $u^{(4)}=0$ ends it)