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

Intuition How to use this page

Every problem below rebuilds one muscle of the Fundamental Theorem of Calculus. Try each before opening its collapsible solution. The two engines you keep reusing:

Part 1 — the derivative of an "area-so-far" function $F (x) = \int_{a}^{x} f (t) d t$ is just the integrand at the top limit: $F^{'} (x) = f (x)$ .
Part 2 — a definite integral is endpoints of any antiderivative: $\int_{a}^{b} f = G (b) - G (a)$ .

The hypothesis both engines demand. Everything on this page assumes $f$ is continuous on the interval $[a, b]$ . That is not decoration: Part 1's proof used the Extreme Value Theorem and the Squeeze Theorem, both of which need Continuity. So the precise statements are $if f is continuous on [a, b], \frac{d}{d x} \int_{a}^{x} f (t) d t = f (x) (Part 1),$ $if f is continuous and G^{'} = f, \int_{a}^{b} f (x) d x = G (b) - G (a) (Part 2) .$ Whenever you differentiate $\int f$ or evaluate one below, silently check "is $f$ continuous here?" — for every problem on this page it is.

A quick reminder of what every symbol means, so no notation is used before it is earned:

$\int_{a}^{b} f (x) d x$ ::: the signed area under the curve $y = f (x)$ from $x = a$ to $x = b$ (area above the axis counts $+$ , below counts $-$ ).
$G^{'} (x)$ ::: the derivative — the instantaneous rate of change / slope of $G$ at $x$ .
"antiderivative of $f$ " ::: any function $G$ whose derivative is $f$ , i.e. $G^{'} = f$ .

Level 1 — Recognition

Can you spot which engine (Part 1 or Part 2) applies, and turn the crank once?

L1.1 Compute $\int_{1}^{4} x^{2} d x$ .

L1.2 Find $\frac{d}{d x} \int_{0}^{x} 1 + t^{4} d t$ .

L1.3 Compute $\int_{c}^{c} cos t d t$ for any number $c$ .

Recall Solution L1.1

This is a definite integral with constant limits → Part 2 (and $f (x) = x^{2}$ is continuous, so we are allowed to use it). We need an antiderivative $G$ with $G^{'} (x) = x^{2}$ . Reversing the power rule ( $\frac{d}{d x} x^{n} = n x^{n - 1}$ , so to undo it we add one to the power and divide): $G (x) = \frac{x ^{3}}{3}, G^{'} (x) = \frac{3 x ^{2}}{3} = x^{2} . ✓$ Now plug in the endpoints (top minus bottom): $\int_{1}^{4} x^{2} d x = G (4) - G (1) = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21.$

Recall Solution L1.2

The upper limit is exactly $x$ (not $x^{2}$ , not $3 x$ ), and $1 + t^{4}$ is continuous, so Part 1 applies directly. Part 1 says: differentiating an area-so-far function hands you back the integrand, evaluated at the top limit. $\frac{d}{d x} \int_{0}^{x} 1 + t^{4} d t = 1 + x^{4} .$ Notice we did not find an antiderivative — $1 + t^{4}$ has no elementary one. That is Part 1's whole point: it answers "what is the rate of area accumulation?" without ever summing rectangles.

Recall Solution L1.3

The limits are equal, so we are asked for the area of a slab of zero width. $\int_{c}^{c} cos t d t = 0.$ Via Part 2 it is also obvious: $G (c) - G (c) = 0$ for any antiderivative $G$ . This tiny fact is the "base value" that makes Part 1's $F (a) = \int_{a}^{a} f = 0$ .

Level 2 — Application

Turn the crank in slightly disguised situations.

L2.1 Compute $\int_{0}^{π /2} cos x d x$ .

L2.2 Compute $\int_{1}^{2} (3 x^{2} - \frac{1}{x ^{2}}) d x$ .

L2.3 Compute $\int_{- 1}^{1} x^{3} d x$ and explain the answer with a picture.

Recall Solution L2.1

Part 2 (and $cos x$ is continuous). We need $G$ with $G^{'} = cos x$ . Since $\frac{d}{d x} sin x = cos x$ , take $G (x) = sin x$ . $\int_{0}^{π /2} cos x d x = sin \frac{π}{2} - sin 0 = 1 - 0 = 1.$

Recall Solution L2.2

Break the integral into pieces (integrals add across sums). Rewrite $\frac{1}{x ^{2}} = x^{- 2}$ . Note $3 x^{2} - x^{- 2}$ is continuous on $[1, 2]$ (no zero in the way), so Part 2 is legal.

Antiderivative of $3 x^{2}$ is $x^{3}$ .
Antiderivative of $- x^{- 2}$ : use the power rule "add one to the power, divide by the new power." The new power is $- 2 + 1 = - 1$ , so $\int - x^{- 2} d x = - \frac{x ^{- 1}}{- 1} = - (x^{- 1} \cdot (- 1)) = x^{- 1} = \frac{1}{x} .$ The two minus signs (one from the coefficient, one from dividing by $- 1$ ) cancel. Check: $\frac{d}{d x} \frac{1}{x} = \frac{d}{d x} x^{- 1} = - x^{- 2} . ✓$ So $G (x) = x^{3} + \frac{1}{x}$ . $\int_{1}^{2} = G (2) - G (1) = (8 + \frac{1}{2}) - (1 + 1) = 8.5 - 2 = \frac{13}{2} .$

Recall Solution L2.3

Antiderivative of $x^{3}$ is $\frac{x ^{4}}{4}$ . $\int_{- 1}^{1} x^{3} d x = \frac{1 ^{4}}{4} - \frac{( - 1 ) ^{4}}{4} = \frac{1}{4} - \frac{1}{4} = 0.$ Why zero? Look at the figure. $x^{3}$ is an odd function: the piece from $- 1$ to $0$ lies below the axis (negative signed area) and is a mirror image of the piece from $0$ to $1$ above the axis (positive). They cancel exactly.

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

Level 3 — Analysis

Now the upper limit is a function of $x$ , so Part 1 must team up with the Chain Rule.

L3.1 Find $\frac{d}{d x} \int_{0}^{x^{2}} sin t d t$ .

L3.2 Find $\frac{d}{d x} \int_{x}^{5} e^{- t^{2}} d t$ .

L3.3 Find $\frac{d}{d x} \int_{x^{2}}^{x^{3}} ln (1 + t^{2}) d t$ .

Recall Solution L3.1

The top limit is $x^{2}$ , not $x$ , so Part 1 alone is not enough — we must account for how fast the limit itself moves. Set $u = x^{2}$ and $H (u) = \int_{0}^{u} sin t d t$ . By Part 1 ( $sin t$ continuous), $H^{'} (u) = sin u$ . By the chain rule (rate through the inner variable): $\frac{d}{d x} H (x^{2}) = H^{'} (x^{2}) \cdot \frac{d}{d x} (x^{2}) = sin (x^{2}) \cdot 2 x .$ General rule earned here: $\frac{d}{d x} \int_{a}^{g (x)} f (t) d t = f (g (x)) g^{'} (x)$ .

Recall Solution L3.2

The variable is the lower limit. Flip the limits first (swapping limits flips the sign): $\int_{x}^{5} e^{- t^{2}} d t = - \int_{5}^{x} e^{- t^{2}} d t .$ Now the top limit is $x$ , and $e^{- t^{2}}$ is continuous, so Part 1 gives: $\frac{d}{d x} (- \int_{5}^{x} e^{- t^{2}} d t) = - e^{- x^{2}} .$ Takeaway: a moving lower limit contributes with a minus sign.

Recall Solution L3.3

Both limits move. Split at any constant $k$ (interval additivity), where $k$ is just some fixed number we pick as a "meeting point": $\int_{x^{2}}^{x^{3}} = \int_{k}^{x^{3}} - \int_{k}^{x^{2}} .$ Apply the general rule from L3.1 to each piece (with $f (t) = ln (1 + t^{2})$ , which is continuous): $\frac{d}{d x} = ln (1 + (x^{3})^{2}) \cdot 3 x^{2} - ln (1 + (x^{2})^{2}) \cdot 2 x$ $= 3 x^{2} ln (1 + x^{6}) - 2 x ln (1 + x^{4}) .$ Why the answer doesn't depend on $k$ : the split introduces the constant pieces $\int_{k}^{x^{3}}$ and $\int_{k}^{x^{2}}$ , but every $k$ -flavoured contribution is baked into the lower limit $k$ , which is a constant — and the derivative of a constant limit is $0$ . So changing $k$ shifts each of the two integrals by a fixed number that differentiates away. The moving limits $x^{3}$ and $x^{2}$ are all that survive. Top limit adds; bottom limit subtracts — each carrying its own chain-rule factor.

Level 4 — Synthesis

Combine FTC with limits, roots, and the definition-of-derivative machinery.

L4.1 Let $F (x) = \int_{2}^{x} (t^{2} - 4 t + 3) d t$ . Find every $x$ where $F$ has a local maximum or minimum, and classify each.

L4.2 Evaluate $x \to 0 lim \frac{1}{x} \int_{0}^{x} cos (t^{2}) d t$ .

L4.3 Find $\int_{0}^{2} ∣ x - 1 ∣ d x$ (a definite integral of a function with a corner).

Recall Solution L4.1

By Part 1 (the integrand $t^{2} - 4 t + 3$ is continuous), $F^{'} (x) = x^{2} - 4 x + 3 = (x - 1) (x - 3)$ . Critical points where $F^{'} = 0$ : $x = 1$ and $x = 3$ . Sign chart of $F^{'} (x) = (x - 1) (x - 3)$ :

$x < 1$ : both factors negative → $F^{'} > 0$ (rising).
$1 < x < 3$ : $(x - 1) > 0, (x - 3) < 0$ → $F^{'} < 0$ (falling).
$x > 3$ : both positive → $F^{'} > 0$ (rising).

So $F$ rises, then falls, then rises: local maximum at $x = 1$ , local minimum at $x = 3$ .

We never needed the messy formula for

F

itself — Part 1 handed us

F^{'}

directly.

Recall Solution L4.2

As $x \to 0$ the integral $\to 0$ and $x \to 0$ , giving the indeterminate form $\frac{0}{0}$ . Two clean routes:

Route A — recognise the derivative. Let $A (x) = \int_{0}^{x} cos (t^{2}) d t$ , so $A (0) = 0$ and by Part 1 ( $cos (t^{2})$ continuous) $A^{'} (x) = cos (x^{2})$ . Then $lim_{x \to 0} \frac{A ( x )}{x} = lim_{x \to 0} \frac{A ( x ) - A ( 0 )}{x - 0} = A^{'} (0) = cos (0) = 1.$ That limit is the definition of the derivative of $A$ at $0$ .

Route B — L'Hôpital. $\frac{0}{0}$ form, differentiate top and bottom: top $\to cos (x^{2})$ (Part 1), bottom $\to 1$ . Limit $= cos (0) = 1$ . $x \to 0 lim \frac{1}{x} \int_{0}^{x} cos (t^{2}) d t = 1.$

Recall Solution L4.3

The integrand has a corner at $x = 1$ where the inside of $∣ ∣$ changes sign. Split at the corner and drop the absolute value with the correct sign on each piece:

On $[0, 1]$ : $x - 1 \leq 0$ , so $∣ x - 1∣ = 1 - x$ .
On $[1, 2]$ : $x - 1 \geq 0$ , so $∣ x - 1∣ = x - 1$ .

$\int_{0}^{1} (1 - x) d x = [x - \frac{x ^{2}}{2}]_{0}^{1} = (1 - \frac{1}{2}) - 0 = \frac{1}{2} .$ $\int_{1}^{2} (x - 1) d x = [\frac{x ^{2}}{2} - x]_{1}^{2} = (2 - 2) - (\frac{1}{2} - 1) = 0 - (- \frac{1}{2}) = \frac{1}{2} .$ Total $= \frac{1}{2} + \frac{1}{2} = 1$ . (Geometrically: two right triangles, each area $\frac{1}{2}$ .)

Level 5 — Mastery

Prove-and-generalise. These test whether you own the theorem, not just its formulas.

L5.1 Suppose $f$ is continuous and satisfies $\int_{0}^{x} f (t) d t = x e^{2 x}$ for all $x$ . Find $f (x)$ , and find $f (0)$ .

L5.2 Let $F (x) = \int_{0}^{x} (\int_{0}^{u} \frac{1}{1 + s ^{2}} d s) d u$ . Compute $F^{''} (0)$ .

L5.3 (Conceptual.) Give a continuous $f$ on $[- 1, 1]$ for which $\int_{- 1}^{1} f = 0$ but $f$ is not identically zero — and explain in one sentence why this does not contradict Part 1.

Recall Solution L5.1

Differentiate both sides. The left side is an area-so-far function with top limit $x$ (and $f$ is given continuous), so Part 1 gives $\frac{d}{d x} \int_{0}^{x} f = f (x)$ . The right side needs the product rule: $\frac{d}{d x} (x e^{2 x}) = e^{2 x} + x \cdot 2 e^{2 x} = e^{2 x} (1 + 2 x) .$ Therefore $f (x) = e^{2 x} (1 + 2 x) .$ At $x = 0$ : $f (0) = e^{0} (1 + 0) = 1.$

Recall Solution L5.2

Peel one layer at a time. Let $g (u) = \int_{0}^{u} \frac{1}{1 + s ^{2}} d s$ ; the integrand $\frac{1}{1 + s ^{2}}$ is continuous. By Part 1, $F^{'} (x) = g (x) = \int_{0}^{x} \frac{1}{1 + s ^{2}} d s .$ Differentiate again, using Part 1 on $g$ : $F^{''} (x) = g^{'} (x) = \frac{1}{1 + x ^{2}} .$ Hence $F^{''} (0) = \frac{1}{1 + 0} = 1.$ (Two nested integrals ⇒ two applications of Part 1 to strip them.)

Recall Solution L5.3

Take $f (x) = x$ (or any odd continuous function). Then $\int_{- 1}^{1} x d x = [\frac{x ^{2}}{2}]_{- 1}^{1} = \frac{1}{2} - \frac{1}{2} = 0,$ yet $f (x) = x$ is not the zero function. Why no contradiction with Part 1: Part 1 talks about the derivative of the variable-limit function $F (x) = \int_{- 1}^{x} f$ , saying $F^{'} (x) = f (x)$ . It says nothing forcing a single definite integral over a fixed interval to be nonzero. Signed area can cancel to $0$ while the height $f$ is genuinely nonzero.

Active recall — the reusable rules

Recall The four crank-turns to remember (all assume

$f$ continuous) Constant-limit definite integral ::: $\int_{a}^{b} f = G (b) - G (a)$ (Part 2), any antiderivative $G$ . Top limit is $x$ ::: $\frac{d}{d x} \int_{a}^{x} f = f (x)$ (Part 1). Top limit is $g (x)$ ::: $\frac{d}{d x} \int_{a}^{g (x)} f = f (g (x)) g^{'} (x)$ (Part 1 + Chain Rule). Both limits move ::: $\frac{d}{d x} \int_{p (x)}^{q (x)} f = f (q) q^{'} - f (p) p^{'}$ (split at a constant).

Prerequisite threads if any step felt shaky: Riemann Sums and the Definite Integral, Antiderivatives and Indefinite Integrals, Continuity, Mean Value Theorem, Extreme Value Theorem, Squeeze Theorem, Improper Integrals.