Improper Integrals & the Mean Value of a Function
An unbounded region can still enclose a finite area — and averaging a function over an interval is just integration in disguise.
What makes an integral "improper"
The integral $\displaystyle\int_a^bf(x)\,dx$ is improper if either:
- one or both limits is infinite, or
- $f(x)$ is undefined at $x=a$, at $x=b$, or at some point strictly between $a$ and $b$.
Even though the region described is unbounded (either infinitely long, or approaching a vertical asymptote), the enclosed area can still be finite — the curve simply needs to decay fast enough. When the limiting process below produces a finite value, the integral is convergent; otherwise it's divergent.
Replace the problematic limit with a variable (commonly $t$), evaluate the resulting proper integral as a function of $t$ using the fundamental theorem of calculus as usual, then take the limit as $t$ tends to the value the problematic limit was approaching.
Replace $\infty$ with $t$: $\displaystyle\int_1^t x^{-3}\,dx=\left[-\frac12x^{-2}\right]_1^t=-\frac1{2t^2}-\left(-\frac12\right)=\frac12-\frac1{2t^2}$.
As $t\to\infty$, $\dfrac1{2t^2}\to0$, so the integral tends to $\dfrac12-0=\dfrac12$.
The limit exists and is finite, so the integral converges, with value $\dfrac12$.
$f(x)=\dfrac1{\sqrt x}$ is undefined at $x=0$, so replace the lower limit with $t\to0^+$: $\displaystyle\int_t^4x^{-1/2}\,dx=\Big[2x^{1/2}\Big]_t^4=4-2\sqrt t$.
As $t\to0^+$, $2\sqrt t\to0$, so the integral tends to $4-0=4$. It converges, with value $4$.
Not every improper integral converges. For $\displaystyle\int_1^\infty\dfrac1x\,dx$, the working gives $\Big[\ln x\Big]_1^t=\ln t-\ln1=\ln t$, and $\ln t\to\infty$ as $t\to\infty$ — the limit doesn't exist as a finite number, so the integral diverges. Always check the limit actually exists before declaring convergence.
Splitting integrals with an internal singularity
If $f(x)$ is undefined at a point $c$ strictly between $a$ and $b$, you must split the integral at $c$ and consider each piece as a separate improper integral: $\displaystyle\int_a^bf(x)\,dx=\int_a^cf(x)\,dx+\int_c^bf(x)\,dx$. Both pieces must converge individually for the whole integral to converge — if either piece diverges, so does the whole integral.
The mean value of a function
The mean value of $f(x)$ on the interval $[a,b]$ is
$$\bar f=\frac1{b-a}\int_a^bf(x)\,dx$$
Geometrically, this is the constant height of a rectangle with base $[a,b]$ whose area equals the area under $y=f(x)$ on that interval — a direct extension of the mean of a finite list of numbers, replacing "sum and divide by count" with "integrate and divide by interval length."
$\bar f=\dfrac1{4-0}\displaystyle\int_0^4\dfrac6{\sqrt{1+2x}}\,dx=\dfrac64\int_0^4(1+2x)^{-1/2}\,dx$.
Using the reverse chain rule, $\displaystyle\int(1+2x)^{-1/2}\,dx=(1+2x)^{1/2}+c$ (the coefficient of $x$ inside, $2$, cancels with the $\frac12$ power's chain-rule factor).
$=\dfrac64\Big[(1+2x)^{1/2}\Big]_0^4=\dfrac64\big(\sqrt9-\sqrt1\big)=\dfrac64(3-1)=\dfrac64\times2=3$.
If $f(x)$ has mean value $m$ on $[a,b]$, then $f(x)+c$ has mean value $m+c$ on the same interval — adding a constant to the function shifts every value, and hence the average, by that same constant. This follows directly from linearity of the integral: $\displaystyle\int_a^b(f(x)+c)\,dx=\int_a^bf(x)\,dx+c(b-a)$, so dividing by $b-a$ adds exactly $c$ to the mean value.
Exam-style questions
Question 1 6 marks
Show that $\displaystyle\int_0^{\infty}e^{-4x}\,dx$ converges and find its exact value.
$\displaystyle\int_0^te^{-4x}\,dx=\left[-\dfrac14e^{-4x}\right]_0^t=-\dfrac14e^{-4t}+\dfrac14$. (M1, A1)
As $t\to\infty$, $e^{-4t}\to0$, so the integral $\to\dfrac14-0=\dfrac14$. (M1, A1)
The limit exists and is finite, so the integral converges to $\dfrac14$. (A1, A1)
Question 2 7 marks
Find the mean value of $f(x)=\dfrac1{x\ln x}$ over the interval $[e,e^3]$.
$\bar f=\dfrac1{e^3-e}\displaystyle\int_e^{e^3}\dfrac1{x\ln x}\,dx$. Using the substitution $u=\ln x$, $du=\dfrac1x\,dx$: $\displaystyle\int\dfrac1{x\ln x}\,dx=\int\dfrac1u\,du=\ln|u|+c=\ln(\ln x)+c$. (M1, A1, A1)
Evaluating: $\Big[\ln(\ln x)\Big]_e^{e^3}=\ln(\ln e^3)-\ln(\ln e)=\ln3-\ln1=\ln3$. (M1, A1)
$\bar f=\dfrac{\ln3}{e^3-e}$. (A1)