The Mean Value Theorem for Integrals

The unweighted theorem is the direct stepping stone to the The First Fundamental Theorem of Calculus: to show that $F(x) = \int_a^x f(t)\,dt$ satisfies $F'(x) = f(x)$, one writes the difference quotient $\frac{F(x+h)-F(x)}{h} = \frac{1}{h}\int_x^{x+h} f$, recognizes the right side as an average value of $f$ over a shrinking interval, and uses exactly this theorem to replace that average by $f(c_h)$ for some $c_h$ between $x$ and $x+h$; continuity of $f$ then forces $f(c_h) \to f(x)$. The weighted version underlies error estimates in numerical integration: the remainder term in the trapezoidal rule and Simpson's rule is expressed using a weighted Mean Value Theorem for Integrals, where the weight $g$ is a fixed-sign kernel arising from Taylor's theorem with integral remainder. In probability and statistics, the theorem justifies that the expected value of a continuous random variable's transformation, when computed as a weighted integral against a density, equals the transformation evaluated at some representative point — a heuristic made rigorous by exactly this tool. In physics and engineering, it certifies that quantities like average temperature, average force, or average electric field over a region are values genuinely attained somewhere in that region, not artifacts of the averaging process, provided the underlying field is continuous.

:::quiz[If $f(x)=x^2$ on $[1,4]$, what is the average value $\bar f$?]

• 5
• 7
• 21/3 ✓
• 9

:::quiz[In the weighted Mean Value Theorem for Integrals, what happens if $g$ changes sign on $[a,b]$?]

• Nothing changes; the theorem still holds as stated
• The hypothesis is violated and the conclusion is not guaranteed by this proof ✓
• $f$ must then be constant
• The integral $\int_a^b g$ automatically becomes zero

:::quiz[For the step function $k(x)=0$ on $[0,1/2]$ and $k(x)=2$ on $(1/2,1]$, what is $\bar k$, and is it attained?]

• $\bar k = 2$, attained at $x=1$
• $\bar k = 1$, not attained anywhere on $[0,1]$ ✓
• $\bar k = 1$, attained at $x=1/2$
• $\bar k = 0$, attained everywhere on $[0,1/2]$

Ask the tutor: How exactly does this Mean Value Theorem get used inside the proof that $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$?

Picture the graph of a continuous function $f$ over $[a,b]$, sitting above and below the horizontal axis in some irregular way. The quantity $\int_a^b f$ is the signed area trapped between the graph and the axis. Dividing by the width $b-a$ converts that area into a height: the height of the rectangle with base $[a,b]$ whose area matches the signed area under $f$. The theorem says this rectangle's height is not some abstract number invented by the arithmetic of integration — it is a height the function itself actually reaches at some point $c$. Geometrically, the horizontal line $y=\bar f$ must cross the graph of $f$ somewhere in $[a,b]$, because the graph oscillates above and below that line in a way that exactly balances out (that is what "average" means), and a continuous curve cannot jump over a height without passing through it. The weighted version generalizes this by allowing the horizontal strip to be measured with a nonuniform ruler $g(x)$: instead of asking $f$ to match the plain average, it asks $f$ to match the $g$-weighted average, which emphasizes regions where $g$ is large. The proof idea is identical — sandwich, then apply the Intermediate Value Theorem — but now to the quantity $\int_a^b fg \,/\int_a^b g$.

:::svg[The average value $\bar f$ is a height the continuous curve must actually reach at some $c\in[a,b]$]

Example 1 (unweighted, asymmetric interval and function). Let $f(x) = x^3 - 2x + 1$ on $[0,3]$. This $f$ is a polynomial, hence continuous on $[0,3]$, so the Mean Value Theorem for Integrals applies. First compute the average value:

So

We must solve $f(c) = 4.75$ for some $c \in [0,3]$, i.e.

Check endpoints of a bracketing interval: $f(2) = 8-4+1=5$ and $f(1.9) = 6.859 - 3.8 + 1 = 4.059$. Since $f$ is continuous and $f(1.9) = 4.059 < 4.75 < 5 = f(2)$, the Intermediate Value Theorem (applied directly to $f$ on $[1.9,2]$, consistent with the proof) guarantees a root $c\in(1.9,2)$. A quick refinement: $f(1.97) = 1.97^3 - 2(1.97)+1 = 7.645... - 3.94 + 1 = 4.705$, and $f(1.99)=7.881-3.98+1=4.901$, so $c$ lies between $1.97$ and $1.99$, consistent with existence but — as guaranteed by the theorem — no closed form is needed; existence, not explicit computation, is the content of the theorem. This illustrates that the theorem guarantees $c$ exists in $[0,3]$ without requiring us to solve the cubic exactly.

Example 2 (weighted version, genuinely non-constant weight). Let $f(x) = \cos x$ and $g(x) = x^2$ on $[0,\pi]$. Here $f$ is continuous everywhere, and $g$ is continuous (hence integrable) and satisfies $g(x)=x^2\ge 0$ on $[0,\pi]$, so both hypotheses of the weighted theorem hold. Compute $\int_0^\pi g(x)\,dx = \int_0^\pi x^2\,dx = \frac{\pi^3}{3}$, which is positive, so we are in the main (non-degenerate) case. Compute $\int_0^\pi f(x)g(x)\,dx = \int_0^\pi x^2\cos x\,dx$ using integration by parts twice. With $u=x^2, dv=\cos x\,dx$:

Evaluate from $0$ to $\pi$: at $x=\pi$, $\sin\pi = 0, \cos\pi=-1$, giving $\pi^2\cdot 0 + 2\pi(-1) - 2\cdot 0 = -2\pi$. At $x=0$, everything is $0$. So

The theorem predicts a $c \in [0,\pi]$ with

Since $\cos$ is continuous on $[0,\pi]$ and ranges continuously from $\cos 0 = 1$ down to $\cos\pi=-1$, and $-0.6079$ lies in $[-1,1]$, the Intermediate Value Theorem guarantees a (unique, since $\cos$ is strictly decreasing on $[0,\pi]$) solution $c = \arccos(-6/\pi^2) \approx 2.226$ radians, which indeed lies in $[0,\pi] \approx [0, 3.1416]$. This confirms the theorem's prediction with an explicit, checkable value of $c$.