Fundamental theorem and mean value theorems for integrals

Every definite integral you have ever evaluated by "finding the antiderivative and plugging in the limits" rests on a single theorem — the Fundamental Theorem of Calculus (FTC) — that most students use for years without ever seeing why it works. In MA10002 you meet it properly: FTC Part 1 says that a definite integral with a variable upper limit is itself a differentiable function whose derivative is just the original integrand back again; FTC Part 2 is the shortcut you already use, turning $\int_a^b f\,dx$ into $G(b)-G(a)$ for any antiderivative $G$ . Sitting beside FTC are the mean value theorems for integrals, which guarantee that the average height of a continuous function over an interval is a value the function genuinely reaches somewhere inside that interval, not just an abstract formula. Together these results are the theoretical spine connecting differentiation and integration, and this node builds them from statement to proof sketch to fully worked numbers.

The core insight

FTC Part 1 differentiates an integral back to its integrand, FTC Part 2 evaluates an integral via any antiderivative, and the mean value theorems guarantee a continuous function's average value on $[a,b]$ is actually attained at some interior point $c$ .

The core idea

FTC Part 1 — differentiating an integral. If $f$ is continuous on $[a,b]$ and $F(x) = \displaystyle\int_a^x f(t)\,dt$ for $x \in [a,b]$ , then $F$ is differentiable on $[a,b]$ and

\frac{d}{dx}\int_a^x f(t)\,dt = f(x).

where: $F(x)$ is the "area-so-far" function built by integrating $f$ from a fixed point $a$ up to a moving point $x$ ; the theorem says its derivative is simply the integrand evaluated at $x$ . Continuity of $f$ is what makes this legal — it is not a symbolic trick, it is a genuine limit statement about how the accumulated area changes as the upper limit inches forward.

FTC Part 2 — evaluating an integral. If $f$ is continuous on $[a,b]$ and $G$ is any antiderivative of $f$ (so $G'=f$ ), then

\int_a^b f(x)\,dx = G(b) - G(a).

where: $G$ can be any one antiderivative — the arbitrary constant of integration cancels automatically in the subtraction. This is exactly the "find the antiderivative, plug in the limits, subtract" routine from school, now justified rather than assumed: it works because a definite integral over $[a,b]$ equals $F(b)-F(a)$ for the specific antiderivative $F$ of Part 1, and any two antiderivatives of the same $f$ differ only by a constant that cancels.

First Mean Value Theorem (MVT) for integrals. If $f$ is continuous on $[a,b]$ , there exists at least one $c \in (a,b)$ such that

\int_a^b f(x)\,dx = f(c)(b-a).

where: $f(c) = \dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx$ is the average value of $f$ on $[a,b]$ ; the theorem says a continuous function's average height is not merely a computed number but a height the graph actually reaches at some interior $c$ . Geometrically the curved area under $y=f(x)$ equals the area of a flat rectangle of width $(b-a)$ and height $f(c)$ , for some point $c$ in between.

Second (generalized) MVT for integrals. If $f$ and $g$ are continuous on $[a,b]$ and $g$ does not change sign on $[a,b]$ , there exists $c \in (a,b)$ such that

\int_a^b f(x)g(x)\,dx = f(c)\int_a^b g(x)\,dx.

where: $g$ acts as a fixed "weight" that must stay one sign throughout $[a,b]$ — this hypothesis is essential, not decorative. Setting $g(x)=1$ collapses this statement exactly to the first MVT, so the second theorem is a genuine generalization, not a different idea.

Figure

A curve over an interval with the area under it shaded and a dashed rectangle of the same area whose height equals the average value at an interior point c

Derivation of the first mean value theorem from FTC

Derivation ADeriving the first mean value theorem for integrals from the Fundamental Theorem

The goal is to show, using nothing beyond FTC Part 1 and the ordinary Mean Value Theorem for derivatives (Lagrange's MVT) that a student already knows from differential calculus, that a continuous $f$ on $[a,b]$ must satisfy $\int_a^b f(x)\,dx = f(c)(b-a)$ for some $c$ strictly between $a$ and $b$ — turning a statement about integrals into a direct consequence of a statement about derivatives.

Step 1. Build the accumulation function. Since $f$ is continuous on $[a,b]$ , define $F(x) = \displaystyle\int_a^x f(t)\,dt$ . By FTC Part 1, $F$ is differentiable on $[a,b]$ with $F'(x) = f(x)$ for every $x$ in the interval.

Step 2. Apply Lagrange's Mean Value Theorem to $F$ . Since $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , Lagrange's theorem guarantees some $c \in (a,b)$ with

F'(c) = \frac{F(b) - F(a)}{b-a}.

Step 3. Translate both sides back into integrals. On the left, $F'(c) = f(c)$ by Step 1. On the right, $F(b) = \int_a^b f(t)\,dt$ and $F(a) = \int_a^a f(t)\,dt = 0$ , so $F(b)-F(a) = \int_a^b f(t)\,dt$ . Substituting gives

f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx,

which is exactly the first mean value theorem after multiplying both sides by $(b-a)$ .

Step 4. Read off why this matters. The mean value theorem for integrals is not an independent fact — it is Lagrange's Mean Value Theorem for derivatives applied to the specific function $F$ that FTC builds. This is the deepest reason FTC sits at the centre of the syllabus: it converts a differentiation fact into an integration fact almost for free.

Worked example

Problem. Verify the first mean value theorem for $f(x) = x^2$ on $[1,4]$ : find the guaranteed value of $c$ , and interpret $f(c)$ as an average.

Step 1. Compute the definite integral using FTC Part 2. An antiderivative of $x^2$ is $x^3/3$ , so

\int_1^4 x^2\,dx = \left[\frac{x^3}{3}\right]_1^4 = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21.

Step 2. Set up the MVT equation and solve for $c$ . By the theorem, $21 = f(c)(4-1) = 3c^2$ , so $c^2 = 7$ , giving $c = \sqrt7 \approx 2.6458$ .

Step 3. Confirm $c$ lies inside the open interval. Since $1 < \sqrt7 < 4$ , the theorem is confirmed: this value of $c$ genuinely sits inside $(1,4)$ , not at an endpoint.

Step 4. Interpret $f(c)$ as an average. The average value of $f(x)=x^2$ on $[1,4]$ is $f(c) = c^2 = 7$ , and directly, $\dfrac{1}{4-1}\displaystyle\int_1^4 x^2\,dx = \dfrac{21}{3} = 7$ — both computations agree, as they must. The guaranteed point is $c = \sqrt7 \approx 2.6458$ , and the average value of $x^2$ on $[1,4]$ is $7$ .

Competitive stretch

For the second MVT with $f(x)=x$ , $g(x)=1-x$ on $[0,1/2]$ : since $g(x)=1-x>0$ on this interval it does not change sign, so the theorem applies. The left side is $\int_0^{1/2}(x-x^2)\,dx = \tfrac18-\tfrac1{24}=\tfrac1{12}$ , and $\int_0^{1/2}(1-x)\,dx=\tfrac38$ , so $\tfrac1{12}=c\cdot\tfrac38$ gives $c=\tfrac{2}{9}$ , which lies inside $(0,1/2)$ as required.

Common traps

Confusing FTC Part 1 (differentiating an integral returns the integrand) with FTC Part 2 (evaluating an integral via an antiderivative) — they answer different questions and are often paired in one exam question.
Forgetting the chain-rule factor when a limit of integration is itself a function of $x$ : $\dfrac{d}{dx}\displaystyle\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x)$ , not just $f(g(x))$ .
When both limits are variable, dropping a term: $\dfrac{d}{dx}\displaystyle\int_{a(x)}^{b(x)} f(t)\,dt = f(b(x))b'(x) - f(a(x))a'(x)$ .
Treating the MVT's $c$ as unique or findable by inspection — it only guarantees existence of at least one $c$ in the open interval; you must solve for it algebraically.
Applying the second MVT without checking $g$ keeps one sign on $[a,b]$ — without that hypothesis the conclusion can fail.

Ask the tutor: if the jump from Lagrange's Mean Value Theorem for derivatives to the mean value theorem for integrals feels rushed, ask the AI tutor to walk through the derivation step by step with a second example.

Why does FTC Part 2 let me ignore the constant of integration when evaluating a definite integral?

Because you subtract $G(b)-G(a)$ . If you pick any antiderivative $G+C$ instead of $G$ , the constant appears in both $G(b)+C$ and $G(a)+C$ and cancels exactly when you subtract: $(G(b)+C)-(G(a)+C) = G(b)-G(a)$ . That is why for definite integrals you can drop the " $+C$ " that is mandatory for indefinite integrals — any single antiderivative gives the same final number.

Does the mean value theorem tell me how to find c, or just that it exists?

Only that it exists — existence, not a formula or a shortcut for locating it. To actually find $c$ you set up the MVT equation $\int_a^b f\,dx = f(c)(b-a)$ (or the analogous second-MVT equation) and solve it algebraically for your specific $f$ , exactly as in the worked example above. For some functions that equation may have more than one valid root inside $(a,b)$ ; the theorem only promises at least one.

Why does the second MVT need g to keep a constant sign, when the first MVT has no such condition?

The first MVT is really the special case $g(x)=1$ , which trivially never changes sign, so the condition is silently satisfied and easy to forget it was ever needed. For a general weight $g$ , if it changes sign the weighted integral $\int_a^b f(x)g(x)\,dx$ can behave like a difference of two areas with opposite orientation, and the clean "average height times total weight" reading that the theorem relies on breaks down — the proof genuinely needs $g$ to be one-signed throughout.