Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs: Worked examples

Before anything, two words we will use constantly:

Integrand — the function inside the integral sign (the thing being added up). In $\int_{a}^{b} f (x) d x$ the integrand is $f (x)$ .
Limits of integration — the two numbers (or expressions) $a$ (bottom) and $b$ (top) that say where the adding starts and stops.

The scenario matrix

Every FTC problem falls into one of these cells. The examples below are labelled with the cell they hit.

Cell	What makes it tricky	Example
A	Plain Part 2, positive area	Ex 1
B	Integrand goes negative → signed area	Ex 2
C	Degenerate: equal limits ( $a = b$ ) → zero	Ex 3
D	Reversed limits ( $a > b$ ) → sign flip	Ex 3
E	Part 1 with variable upper limit = $x$	Ex 4
F	Variable limit is a function of $x$ → chain rule	Ex 5
G	Both limits move	Ex 6
H	Integrand has a jump discontinuity → FTC breaks	Ex 7
I	Real-world word problem (accumulation)	Ex 8
J	Improper / limiting behaviour (infinite limit)	Ex 9
K	Exam twist: solve for an unknown limit	Ex 10

Prerequisites we lean on: Antiderivatives and Indefinite Integrals, Riemann Sums and the Definite Integral, Chain Rule, Continuity, Improper Integrals.

Cell B — signed (negative) area

The definite integral is a signed area: pieces of the curve below the $x$ -axis count as negative.

Figure — Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs

Worked example Ex 2 — Cell B

Compute $\int_{0}^{π} cos x d x$ and also the total geometric area between $cos x$ and the axis.

Forecast: on $[0, π]$ the cosine curve is above the axis on $[0, \frac{π}{2}]$ and below on $[\frac{π}{2}, π]$ . The two halves are mirror images, so the signed integral should be zero, while the geometric area is positive.

In Figure s01 the blue curve is $cos x$ . The green shaded lobe on the left ( $0$ to $\frac{π}{2}$ ) sits above the axis and counts $+ 1$ ; the red shaded lobe on the right ( $\frac{π}{2}$ to $π$ ) sits below and counts $- 1$ . The dashed vertical line marks the sign change at $x = \frac{π}{2}$ — exactly where we must split for geometric area. Notice the green and red regions are congruent reflections, which is why they cancel.

Step 1. Antiderivative of $cos x$ is $sin x$ (since $\frac{d}{d x} sin x = cos x$ ). Why this step? Part 2 needs an antiderivative.

Step 2. Signed integral: $\int_{0}^{π} cos x d x = [sin x]_{0}^{π} = sin π - sin 0 = 0 - 0 = 0.$ Why this step? The positive hump ( $+ 1$ area, green) and negative dip ( $- 1$ area, red) cancel — the integral measures signed area.

Step 3. Geometric area — split at the sign change $x = \frac{π}{2}$ (the dashed line) and take absolute values: $= 1 \int_{0}^{π /2} cos x d x + = - 1 \int_{π /2}^{π} cos x d x = 1 + 1 = 2.$ Why this step? "Area" (unsigned) needs each below-axis piece flipped positive; you must split at every root of the integrand.

Verify: $[sin x]_{0}^{π /2} = 1 - 0 = 1$ and $[sin x]_{π /2}^{π} = 0 - 1 = - 1$ . Sum of magnitudes $= 2$ , sum with signs $= 0$ . ✓

Cells C & D — degenerate and reversed limits

These are the "edge inputs" people forget on exams.

Worked example Ex 3 — Cells C and D

(a) Compute $\int_{5}^{5} e^{t^{2}} d t$ . (b) Compute $\int_{3}^{1} 2 x d x$ .

Forecast: (a) a strip of zero width has zero area → $0$ . (b) integrating "backwards" should just flip the sign of the forward integral.

Step 1 (a). Equal limits: $a = b = 5$ . $\int_{5}^{5} e^{t^{2}} d t = 0.$ Why this step? By definition $\int_{a}^{a} f = G (a) - G (a) = 0$ — no accumulation happens over zero length. Note we never needed an antiderivative of $e^{t^{2}}$ (it has no elementary one).

Step 2 (b). Reversed limits: top $= 1$ is below bottom $= 3$ . Use the rule $\int_{b}^{a} f = - \int_{a}^{b} f$ . $\int_{3}^{1} 2 x d x = [x^{2}]_{3}^{1} = 1 - 9 = - 8.$ Why this step? Part 2 blindly computes $G (top) - G (bottom) = G (1) - G (3)$ ; the ordering handles the sign automatically.

Verify: Forward, $\int_{1}^{3} 2 x d x = [x^{2}]_{1}^{3} = 9 - 1 = 8$ ; reversing gives $- 8$ . ✓ And $\int_{5}^{5}$ of anything is $0$ . ✓

Cell E — Part 1, variable upper limit equal to $x$

Worked example Ex 4 — Cell E

Find $\frac{d}{d x} \int_{2}^{x} 1 + t^{4} d t$ .

Forecast: Part 1 says "the rate area accumulates = the curve's height at the top edge." So the answer should just be the integrand with $t$ replaced by $x$ . No antiderivative required.

Step 1. Recognise the exact form of Part 1: upper limit is literally $x$ , lower limit is a constant. Why this step? Part 1 applies only when the top limit is $x$ and the integrand is continuous — $1 + t^{4}$ is continuous everywhere.

Step 2. Read off the answer: $\frac{d}{d x} \int_{2}^{x} 1 + t^{4} d t = 1 + x^{4} .$ Why this step? This is FTC Part 1: $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$ . The lower limit $2$ is a constant and disappears under differentiation.

Verify (sanity by numbers): approximating $F (x) = \int_{2}^{x} 1 + t^{4} d t$ , the difference quotient $\frac{F ( 2.001 ) - F ( 2 )}{0.001}$ should be $\approx 1 + 2^{4} = 17 \approx 4.123$ . The thin sliver has width $0.001$ and height $\approx 17$ , giving that ratio. ✓

Cell F — variable limit that is a function of $x$

Worked example Ex 5 — Cell F

Find $\frac{d}{d x} \int_{0}^{s i n x} ln (1 + u^{2}) d u$ .

Forecast: the top limit is $sin x$ , not $x$ . Part 1 gives the rate with respect to the limit; but the limit itself moves as $x$ changes. Two rates chained together → expect an extra factor $cos x$ .

Figure s02 shows the two stacked rates. The red curve is the inner limit $w = sin x$ ; its steepness at a chosen $x$ (red arrow) is $cos x$ — how fast the top edge slides. The blue curve is the outer accumulated value $H (sin x)$ ; the blue arrow marks how fast that stored area grows per unit of $w$ , which Part 1 says is $ln (1 + w^{2})$ . The final rate is the product of the red slope and the blue slope — that is exactly what the chain rule multiplies together.

Step 1. Name the inner limit. Let $w = sin x$ and define $H (w) = \int_{0}^{w} ln (1 + u^{2}) d u$ . Why this step? Splitting into $H$ (an integral with limit $w$ ) and $w = sin x$ (a plain function) separates the two dependencies so we can apply each rule cleanly.

Step 2. By Part 1, $H^{'} (w) = ln (1 + w^{2})$ . Why this step? Now the upper limit is the variable $w$ — exact Part 1 form.

Step 3. By the Chain Rule, $\frac{d}{d x} H (sin x) = H^{'} (sin x) \cdot \frac{d}{d x} (sin x)$ . $= ln (1 + sin^{2} x) \cdot cos x .$ Why this step? The chain rule multiplies the "outer rate" $H^{'}$ (blue arrow) by the "inner rate" $\frac{d w}{d x} = cos x$ (red arrow) — the figure shows why both stack.

Verify: At $x = 0$ : $sin 0 = 0$ , so the answer is $ln (1) \cdot cos 0 = 0 \cdot 1 = 0$ . Independently, near $x = 0$ the top limit barely moves off $0$ where the integrand $ln (1 + u^{2}) \approx 0$ , so the rate is $\approx 0$ . ✓

Cell G — both limits moving

Worked example Ex 6 — Cell G

Find $\frac{d}{d x} \int_{x^{2}}^{x^{3}} \frac{1}{1 + t ^{2}} d t$ .

Forecast: both ends slide. Break the interval at a fixed point, differentiate each piece with the chain rule, and combine — top piece positive, bottom piece negative.

Step 1. Split at a constant $a$ (say $a = 0$ ): $\int_{x^{2}}^{x^{3}} = \int_{0}^{x^{3}} - \int_{0}^{x^{2}} .$ Why this step? Part 1 and the chain rule only handle a variable upper limit against a constant lower one; splitting turns a moving lower limit into a subtracted moving upper limit.

Step 2. Differentiate each with Cell F's method. With $g (t) = \frac{1}{1 + t ^{2}}$ : $\frac{d}{d x} \int_{0}^{x^{3}} g = g (x^{3}) \cdot 3 x^{2}, \frac{d}{d x} \int_{0}^{x^{2}} g = g (x^{2}) \cdot 2 x .$ Why this step? Each is Part 1 + chain rule (Cell F).

Step 3. Subtract: $\frac{d}{d x} \int_{x^{2}}^{x^{3}} \frac{d t}{1 + t ^{2}} = \frac{3 x ^{2}}{1 + x ^{6}} - \frac{2 x}{1 + x ^{4}} .$ Why this step? The upper limit contributes $+$ , the lower limit contributes $-$ (it was subtracted).

Verify: At $x = 1$ both limits equal $1$ so the interval collapses (Cell C) — the derivative need not be zero, but the two terms give $\frac{3}{2} - \frac{2}{2} = \frac{1}{2}$ , a finite number, as expected for a smooth accumulation. ✓

Cell H — jump discontinuity: FTC's hypothesis fails

Worked example Ex 7 — Cell H

Let $f (t) = {1, 3, t < 1 t \geq 1$ and $F (x) = \int_{0}^{x} f (t) d t$ . Is $F^{'} (1) = f (1)$ ?

Forecast: Part 1 needs a continuous integrand. Here $f$ jumps at $t = 1$ . So the "rate area accumulates" should change abruptly at $x = 1$ , and $F^{'} (1)$ may not exist. Guess: no clean $F^{'} (1)$ .

Figure s03 has two panels. On the left, the orange integrand $f (t)$ leaps from height $1$ to height $3$ at $t = 1$ (open circle = value not taken, filled circle = value taken). On the right, the blue accumulated area $F (x)$ rises with slope $1$ before $x = 1$ and slope $3$ after — the red dot marks the corner where the two straight pieces meet. Because the left slope ( $1$ ) and right slope ( $3$ ) disagree, the graph has a sharp kink, so no single tangent exists there.

Step 1. Compute $F$ on each side. For $x \leq 1$ : $F (x) = \int_{0}^{x} 1 d t = x$ . For $x \geq 1$ : $F (x) = \int_{0}^{1} 1 1 + \int_{1}^{x} 3 d t = 1 + 3 (x - 1) = 3 x - 2$ . Why this step? Piecewise integrands are integrated piece by piece (interval additivity).

Step 2. Compare one-sided slopes at $x = 1$ (the corner in the figure). $F^{'} (1^{-}) = slope of x = 1, F^{'} (1^{+}) = slope of 3 x - 2 = 3.$ Why this step? A derivative exists only if the left and right slopes agree.

Step 3. They differ ( $1 \neq = 3$ ), so $F^{'} (1)$ does not exist — Part 1's conclusion fails exactly where continuity fails. Why this step? This is why the theorem demands continuity: EVT and the Squeeze in the proof both broke here.

Verify: $F$ is continuous everywhere ( $F (1) = 1$ from both formulas: $x \to 1$ gives $1$ , $3 x - 2 \to 1$ ) but has a corner — continuous yet non-differentiable, matching the Continuity vs differentiability gap. ✓

Cell I — real-world accumulation

Worked example Ex 8 — Cell I

Water flows into a tank at rate $r (t) = 6 t$ litres per minute, $t$ in minutes. How many litres enter between $t = 1$ and $t = 4$ minutes?

Forecast: total water = accumulated flow = area under the rate curve. Since the rate grows linearly, expect the answer to be the area of a trapezoid.

Step 1. Total added $= \int_{1}^{4} r (t) d t = \int_{1}^{4} 6 t d t$ . Why this step? By FTC, integrating a rate over time gives the total change of the quantity — this is Part 2 read as "endpoints of the amount function."

Step 2. Antiderivative of $6 t$ is $3 t^{2}$ ; apply Part 2: $\int_{1}^{4} 6 t d t = [3 t^{2}]_{1}^{4} = 48 - 3 = 45 litres .$ Why this step? $\frac{d}{d t} (3 t^{2}) = 6 t$ , so $3 t^{2}$ is the amount-so-far function; subtract endpoint values.

Verify (units + geometry): rate is L/min, time in min → integral in litres ✓. Trapezoid area: heights $r (1) = 6$ , $r (4) = 24$ , width $3$ : $\frac{( 6 + 24 )}{2} \cdot 3 = 45$ . ✓

Cell J — improper / limiting behaviour

Worked example Ex 9 — Cell J

Evaluate the improper integral $\int_{1}^{\infty} \frac{1}{x ^{2}} d x$ .

Forecast: the top limit is infinite. We can't "plug in $\infty$ ." Instead accumulate to a finite $b$ and watch the limit as $b \to \infty$ . The tail $1/ x^{2}$ shrinks fast, so the total area should be finite.

Step 1. Replace $\infty$ by a variable $b$ and take a limit: $\int_{1}^{\infty} \frac{d x}{x ^{2}} = lim_{b \to \infty} \int_{1}^{b} \frac{d x}{x ^{2}} .$ Why this step? FTC (Part 2) only applies over a closed finite interval; the improper integral is defined as this limit.

Step 2. Antiderivative of $x^{- 2}$ is $- x^{- 1}$ ; apply Part 2: $\int_{1}^{b} \frac{d x}{x ^{2}} = [- \frac{1}{x}]_{1}^{b} = - \frac{1}{b} + 1 = 1 - \frac{1}{b} .$ Why this step? $\frac{d}{d x} (- x^{- 1}) = x^{- 2}$ .

Step 3. Take the limit: $lim_{b \to \infty} (1 - \frac{1}{b}) = 1.$ Why this step? As $b \to \infty$ , $\frac{1}{b} \to 0$ ; the leftover area in the far tail vanishes.

Verify: each finite piece $1 - \frac{1}{b} < 1$ and increases toward $1$ — bounded and monotone, so the value $1$ is consistent. ✓

Cell K — exam twist: solve for an unknown limit

Active recall

Recall Which cell? Match the trigger to the fix.

Integrand dips below the axis ::: Cell B — split at each zero, take absolute values for geometric area. Top and bottom limits are equal ::: Cell C — the integral is $0$ regardless of the integrand. Bottom limit bigger than top ::: Cell D — flip sign, $\int_{b}^{a} = - \int_{a}^{b}$ . Upper limit is $g (x)$ , not $x$ ::: Cell F — Part 1 then multiply by $g^{'} (x)$ (chain rule). Both limits are functions of $x$ ::: Cell G — split at a constant, subtract two chain-rule terms. Integrand has a jump ::: Cell H — Part 1 fails; $F$ may have a corner (no derivative there). A limit is $\infty$ ::: Cell J — rewrite as $lim_{b \to \infty}$ , then apply FTC to the finite piece.

Worked examples — Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs