Integration by Parts — Question 2

Question 2

Find the unique constant $a$ such that $$\int_0^1 xe^{ax}\,dx=1.$$ Give the defining equation and a numerical value.

Question 2 – Solution

First consider the special case $a=0$. In that case, $$\int_0^1 xe^{0x}\,dx =\int_0^1 x\,dx =\left.\frac{x^2}{2}\right|_0^1 =\frac12\ne1.$$ Therefore, $a=0$ is not a solution, so we may assume that $a\ne0$.

To evaluate the integral, use integration by parts with $$u=x, \qquad dv=e^{ax}\,dx.$$ Then $$du=dx, \qquad v=\frac{e^{ax}}{a}.$$ Using $\int u\,dv=uv-\int v\,du$, we obtain $$\begin{align*} \int_0^1 xe^{ax}\,dx &=\left.\frac{xe^{ax}}{a}\right|_0^1 -\frac1a\int_0^1 e^{ax}\,dx \\[4pt] &=\left.\frac{xe^{ax}}{a}\right|_0^1 -\left.\frac{e^{ax}}{a^2}\right|_0^1 \\[4pt] &=\frac{e^a}{a}-\frac{e^a-1}{a^2} \\[4pt] &=\frac{ae^a-e^a+1}{a^2} \\[4pt] &=\frac{e^a(a-1)+1}{a^2}. \end{align*}$$

We want this integral to equal $1$, so $$\begin{align*} \frac{e^a(a-1)+1}{a^2}&=1, \\[4pt] e^a(a-1)+1&=a^2. \end{align*}$$ Thus, the defining equation is $$\boxed{e^a(a-1)+1=a^2}.$$

Substituting $a=1$ verifies that it satisfies the equation: $$e^1(1-1)+1=1=1^2.$$ Hence, $a=1$ is a solution.

To see that it is the only solution, define $$F(a)=\int_0^1 xe^{ax}\,dx.$$ Differentiating with respect to $a$ gives $$F'(a)=\int_0^1 x^2e^{ax}\,dx>0$$ for every real $a$. Therefore, $F$ is strictly increasing and can equal $1$ for at most one value of $a$. Since $F(1)=1$, the unique constant is $$\boxed{a=1.0000}.$$