Single-Variable-Calculus · Unit 16 · Video 3 · Interactive Practice

Proving FTC1 and FTC2 with the Mean Value Theorem

IKey Formulas

Step	Formula	Rests on
FTC 2: the strip	$\Delta G = G(x + \Delta x) - G(x) \approx f(x)\,\Delta x$	the strip is almost a rectangle
FTC 2: the limit	$G'(x) = \displaystyle\lim_{\Delta x \to 0}\frac{\Delta G}{\Delta x} = f(x)$	$f$ continuous
Same derivative	$F' = G' \;\Longrightarrow\; F = G + c$	the mean value theorem: $h' = 0 \Rightarrow h$ constant
FTC 1	$F(b) - F(a) = G(b) - G(a) = \displaystyle\int_a^b f(x)\,dx$	$c$ cancels; $G(a) = 0$

Key Insight: FTC 2 follows from a picture of a thin strip. FTC 1 follows from FTC 2 except for one step, that two functions with the same derivative differ by a constant, and that step is the mean value theorem.

IIVisualization 1 — The Strip Becomes a Rectangle

For $f(t) = 1 + \frac{t^2}{8}$ at $x = 2$ , the strip's area divided by its width tends to the height $f(2) = 1.5$ .

Width Δx 1

IIIVisualization 2 — The Constant Cancels

Every antiderivative $F(x) = -\frac{1}{x} + c$ of $\frac{1}{x^2}$ gives the same $F(2) - F(1) = \int_1^2 \frac{dt}{t^2} = \frac{1}{2}$ .

Constant c 0

IVQuiz Questions

Problem 1 · One Strip, Computed Exactly

Given: $f(t) = t^2$ and $G(x) = \int_0^x t^2\,dt$ . At $x = 1$ with $\Delta x = 0.1$ , find $\dfrac{\Delta G}{\Delta x} = \dfrac{G(1.1) - G(1)}{0.1}$ .

✅ Correct! The ratio lies between the heights

f(1) = 1

and

f(1.1) = 1.21

at the strip's edges, and it tends to

f(1) = 1

\Delta x \to 0

❌ $1 = f(1)$ is the limit as $\Delta x \to 0$ . At

\Delta x = 0.1

the strip is not yet a rectangle; compute

G(1.1) - G(1)

exactly.

❌ $1.21 = f(1.1)$ is the height at the right edge. The strip's average height lies strictly between

f(1)

and

f(1.1)

❌ That is $\Delta G$ itself. Divide by

\Delta x = 0.1

❌ Not quite. Use

G(x) = \frac{x^3}{3}

Show solution

Here $G(x) = \frac{x^3}{3}$ , so

\Delta G = \frac{1.1^3 - 1^3}{3} = \frac{0.331}{3}, \qquad \frac{\Delta G}{\Delta x} = \frac{0.331}{0.3} = \frac{331}{300} \approx 1.103.

As $\Delta x \to 0$ this tends to $f(1) = 1$ , which is FTC 2 at $x = 1$ .

Problem 2 · Where the Mean Value Theorem Enters

In the proof of FTC 1 from FTC 2, which step needs the mean value theorem?

✅ Correct! With

h = F - G

h' = 0

, and the mean value theorem gives

h(x) - h(a) = h'(s)(x - a) = 0

, so

h

is constant.

❌ That is FTC 2. It comes from the strip picture and the continuity of

f

, with no mean value theorem.

❌ That holds because the interval has length zero. No theorem is needed.

❌ That is arithmetic. The constant appears in both terms and cancels.

❌ Not quite.

Show solution

Let $h = F - G$ . Then $h' = F' - G' = f - f = 0$ . For any $x$ , the mean value theorem gives a point $s$ between $a$ and $x$ with

h(x) - h(a) = h'(s)\,(x - a) = 0,

so $h(x) = h(a)$ for every $x$ : $F = G + c$ . Every other step is FTC 2, the definition of $G$ , or arithmetic.

Problem 3 · Two Antiderivatives, One Integral

Given: $F(x) = -\frac{1}{x} + 7$ and $G(x) = \int_1^x \frac{dt}{t^2} = 1 - \frac{1}{x}$ , both antiderivatives of $\frac{1}{x^2}$ for $x > 0$ . Find the constant $c$ with $F = G + c$ , and $\int_1^4 \frac{dt}{t^2}$ .

What is $c$ ?

What is $\int_1^4 \frac{dt}{t^2}$ ?

✅ Correct!

F - G = 6

, and either antiderivative gives

\frac{3}{4}

: the

6

cancels in the difference.

❌ $G$ has its own constant, $1$ .

F - G = \left(-\frac{1}{x} + 7\right) - \left(1 - \frac{1}{x}\right)

❌ Check the order.

c = F - G

, and

F

is the larger of the two.

❌ That is $F(4)$ alone. FTC 1 needs the difference

F(4) - F(1)

❌ That is $F(1) - F(4)$ . Subtract the value at the lower limit from the value at the upper limit.

❌ Check $c$ . Subtract:

F(x) - G(x)

❌ Check the integral. Compute

F(4) - F(1)

G(4) - G(1)

Show solution

F(x) - G(x) = \left(-\tfrac{1}{x} + 7\right) - \left(1 - \tfrac{1}{x}\right) = 6.

With $F$ : $F(4) - F(1) = \left(-\tfrac{1}{4} + 7\right) - (-1 + 7) = \tfrac{3}{4}$ . With $G$ : $G(4) - G(1) = \left(1 - \tfrac{1}{4}\right) - 0 = \tfrac{3}{4}$ . The constant $6$ appears in both values of $F$ and cancels.

Problem 4 · Same Derivative, Different Start

Given: $F'(x) = G'(x)$ for every $x$ , with $F(0) = 2$ , $G(0) = -1$ and $G(3) = 4$ . Find $F(3)$ .

✅ Correct!

F = G + c

with

c = F(0) - G(0) = 3

, so

F(3) = 4 + 3 = 7

❌ Equal derivatives do not make the functions equal. They differ by a constant; find it from the values at

0

❌ Check the sign of the constant.

c = F(0) - G(0) = 2 - (-1) = 3

❌ Not quite. Write

F = G + c

and evaluate at

x = 0

Show solution

Since $F' = G'$ , the mean value theorem gives $F = G + c$ for a constant $c$ . At $x = 0$ :

c = F(0) - G(0) = 2 - (-1) = 3.

So $F(3) = G(3) + 3 = 7$ . Equivalently, $F(3) - F(0) = G(3) - G(0) = 5$ , which is the same cancellation that proves FTC 1.

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