Single-Variable-Calculus · Unit 15 · Video 3 · Interactive Practice

Six Rules for Definite Integrals and Bounds on eᵇ

IKey Formulas

Rule	Formula	Condition
1–2 · Sums and constants	$\int_a^b \big(f(x) + g(x)\big)\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx$ $\int_a^b c\,f(x)\,dx = c\int_a^b f(x)\,dx$	$c$ is a constant, not a function of $x$
3 · Additivity	$\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$	$a$ , $b$ , $c$ in any order, once rule 5 is in place
4–5 · Empty and reversed intervals	$\int_a^a f(x)\,dx = 0$ $\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$	rule 5 is a definition, chosen so the FTC holds in both directions
6 · Estimation	$f(x) \le g(x) \implies \int_a^b f(x)\,dx \le \int_a^b g(x)\,dx$	only if $a < b$

Key Insight: Integrating $e^x \ge 1$ from $0$ to $b$ gives $e^b \ge 1 + b$ , and integrating that bound again gives $e^b \ge 1 + b + \frac{b^2}{2}$ . Each round adds one term, and each result is proved only for $b \ge 0$ .

IIVisualization 1 — Additivity in Any Order

With signed integrals, $\int_a^b f\,dx + \int_b^c f\,dx = \int_a^c f\,dx$ holds wherever the three points sit.

IIIVisualization 2 — Integrating an Inequality, Round by Round

Integrating $e^x \ge$ the previous bound from $0$ to $b$ produces the next bound, one term longer.

Start from b 1.20

💡 Challenge: starting from $e^x \ge 1 + x$ , find a $b < 0$ where the new bound is false. The argument never claimed it there, because estimation needs the lower limit $0$ below the upper limit $b$ .

IVVisualization 3 — Estimation Needs Forward Limits

From $f \le g$ , estimation concludes $\int_a^b f\,dx \le \int_a^b g\,dx$ only when $a < b$ .

Functions a 1 b 0

VQuiz Questions

Problem 1 · Sums and Constants

Given: $\int_0^3 f(x)\,dx = 5$ and $\int_0^3 g(x)\,dx = -2$ . Find $\int_0^3 \big(2f(x) - 3g(x)\big)\,dx$ .

✅ Correct! Rules 1 and 2 split the integral into

2 \cdot 5 - 3 \cdot (-2) = 10 + 6 = 16

❌ Close, but… the integral of

g

-2

, so

-3 \cdot (-2) = +6

, not

-6

❌ Not quite.

5 - (-2) = 7

drops the constants

2

and

3

; rule 2 factors them out, it does not erase them.

❌ Not quite. Factor out each constant, then substitute:

2\int_0^3 f\,dx - 3\int_0^3 g\,dx

Show solution

By rule 1 (sums) the integral splits, and by rule 2 (constants) each constant comes out:

\int_0^3 \big(2f(x) - 3g(x)\big)\,dx = 2\int_0^3 f(x)\,dx - 3\int_0^3 g(x)\,dx = 2(5) - 3(-2) = 10 + 6 = 16.

Problem 2 · Reversed Limits

Given: $\int_0^5 f(x)\,dx = 8$ and $\int_3^5 f(x)\,dx = 3$ . Find $\int_3^0 f(x)\,dx$ .

✅ Correct! Additivity gives

\int_0^3 f\,dx = 8 - 3 = 5

, and reversing the limits flips the sign.

❌ Close, but…

5

\int_0^3 f\,dx

. The limits here run from

3

down to

0

, and rule 5 flips the sign.

❌ Not quite. The pieces

[0, 3]

and

[3, 5]

add up to

[0, 5]

, so

\int_0^3 f\,dx

is the difference

8 - 3

, not the sum.

❌ Not quite. Find

\int_0^3 f\,dx

by additivity first, then apply rule 5.

Show solution

Additivity (rule 3) with $0 < 3 < 5$ :

\int_0^3 f(x)\,dx + \int_3^5 f(x)\,dx = \int_0^5 f(x)\,dx \quad\Longrightarrow\quad \int_0^3 f(x)\,dx = 8 - 3 = 5.

Reversed limits (rule 5):

\int_3^0 f(x)\,dx = -\int_0^3 f(x)\,dx = -5.

Problem 3 · One More Round

Given: $e^x \ge 1 + x + \frac{x^2}{2}$ for $x \ge 0$ . Integrate both sides from $0$ to $b$ (with $b \ge 0$ ) and set $b = 1$ . Find the lower bound this gives for $e$ .

✅ Correct! The new bound is

e^b \ge 1 + b + \frac{b^2}{2} + \frac{b^3}{6}

, and at

b = 1

it gives

e \ge \frac{8}{3} \approx 2.667

, closer to

e \approx 2.718

than

\frac{5}{2}

❌ Not quite.

\frac{5}{2}

comes from the bound you started with. Integrating once more adds the term

\frac{b^3}{6}

❌ Not quite. The antiderivative of

\frac{x^2}{2}

\frac{x^3}{6}

, not

\frac{x^3}{3}

. Note that

\frac{17}{6} \approx 2.83

exceeds

e

, so it could not be a lower bound.

❌ Close, but… the left side is

\int_0^b e^x\,dx = e^b - 1

. You still need to add the

1

back to isolate

e^b

Show solution

Both sides are integrated from $0$ to $b$ with $0 \le b$ , so estimation (rule 6) keeps the inequality:

\int_0^b e^x\,dx \ge \int_0^b \left(1 + x + \frac{x^2}{2}\right)dx.

Left side: $e^x\Big|_0^b = e^b - 1$ . Right side: $\left(x + \frac{x^2}{2} + \frac{x^3}{6}\right)\Big|_0^b = b + \frac{b^2}{2} + \frac{b^3}{6}$ . So

e^b \ge 1 + b + \frac{b^2}{2} + \frac{b^3}{6} \quad (b \ge 0).

At $b = 1$ : $e \ge 1 + 1 + \frac{1}{2} + \frac{1}{6} = \frac{8}{3} \approx 2.667$ .

Problem 4 · Which Estimate Is Valid?

Given: $x^2 \le x$ for $0 \le x \le 1$ . Which inequality does estimation (rule 6) prove?

✅ Correct! On

[0, 1]

with

0 < 1

both conditions hold, and indeed

\frac{1}{3} \le \frac{1}{2}

❌ Not quite. The limits run backwards: the two sides are

-\frac{1}{3}

and

-\frac{1}{2}

, and

-\frac{1}{3} \le -\frac{1}{2}

is false.

❌ Not quite. Estimation needs

f \le g

on the whole interval, and

x^2 \le x

fails there. The inequality is in fact false: the sides are

\frac{8}{3}

and

2

[0, 2]

, and

\frac{1}{3}

and

-\frac{1}{2}

[-1, 0]

❌ Not quite. Check both conditions of rule 6:

f \le g

on the interval, and lower limit below upper limit.

Show solution

Rule 6 needs two things: $f(x) \le g(x)$ on the interval, and $a < b$ .

(a) On $[0, 1]$ , $x^2 \le x$ and $0 < 1$ , so $\int_0^1 x^2\,dx \le \int_0^1 x\,dx$ , i.e. $\frac{1}{3} \le \frac{1}{2}$ . ✓

(b) The limits run from $1$ down to $0$ , so both sides change sign: $-\frac{1}{3} \le -\frac{1}{2}$ is false.

(d) On $[-1, 0)$ , $x^2 > x$ , so the hypothesis fails; indeed $\frac{1}{3} > -\frac{1}{2}$ .

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