Day 18: The Definite Integral & the Fundamental Theorem

Properties of the Definite Integral: Linearity, Additivity, and Order

Just as derivatives have a sum rule and a constant-multiple rule, integrals have algebraic properties that let you break a hard integral into easy pieces or combine known results — especially when the integrand is only given by a graph or a few known integral values.

Core Theorem

For integrable functions

f, g

and constants

c, a, b

:
1. Sum rule:

\int_a^b [f(x) + g(x)] \, dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx

2. Constant multiple:

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

3. Additivity over intervals:

\int_a^c f(x)\,dx + \int_c^b f(x)\,dx = \int_a^b f(x)\,dx

4. Reversing the limits:

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

5. Zero width:

\int_a^a f(x)\,dx = 0

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Given

\int_0^2 f(x)\,dx = 5

and

\int_0^2 g(x)\,dx = 3

, find

\int_0^2 [2f(x) - g(x)]\,dx

.

How to think about it

Key idea: Use the constant-multiple rule and the sum rule.

Full solution

1. Split using linearity

\int_0^2 [2f(x) - g(x)]\,dx = 2\int_0^2 f(x)\,dx - \int_0^2 g(x)\,dx

.

2. Substitute the given values

= 2(5) - 3 = 7

.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Given

\int_0^3 h(x)\,dx = 8

and

\int_5^3 h(x)\,dx = 2

, find

\int_0^5 h(x)\,dx

.

Need a hint?

\int_5^3 h(x)\,dx

has its limits reversed relative to

\int_3^5 h(x)\,dx

.

Show solution

1. Fix the reversed limits

\int_5^3 h(x)\,dx = -\int_3^5 h(x)\,dx = 2 \implies \int_3^5 h(x)\,dx = -2

.

2. Combine with additivity

\int_0^5 h(x)\,dx = \int_0^3 h(x)\,dx + \int_3^5 h(x)\,dx = 8 + (-2) = 6

.

Practice 02Medium

The graph of

f

on

[0, 6]

encloses a triangle of area 4 above the

x

-axis on

[0, 3]

and a region of area 5 below the

x

-axis on

[3, 6]

. Find

\int_0^6 f(x)\,dx

and

\int_0^6 |f(x)|\,dx

.

Need a hint?

The plain integral uses signed area; the absolute-value integral uses total area.

Show solution

1. Signed area

\int_0^6 f(x)\,dx = (+4) + (-5) = -1

.

2. Total area

\int_0^6 |f(x)|\,dx = 4 + 5 = 9

.

Common Pitfalls

⚠
Assuming $\int (fg) = (\int f)(\int g)$There is no product rule for integrals. Linearity applies only to sums and constant multiples.

The Fundamental Theorem of Calculus, Part 1

Evaluating a definite integral with an antiderivative. If $F' = f$ , then $\int_a^b f(x)\,dx = F(b) - F(a)$ — this turns limit-of-Riemann-sum calculations into simple algebra.

Core Theorem

If

f

is continuous on

[a, b]

and

F

is any antiderivative of

f

(that is,

F'(x) = f(x)

), then

\int_a^b f(x) \, dx = F(b) - F(a) = \Big[F(x)\Big]_a^b

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Evaluate

\int_1^3 x^2 \, dx

.

How to think about it

Key idea: An antiderivative of $x^2$ is $\frac{x^3}{3}$ .

Full solution

1. Find an antiderivative

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

, since

F'(x) = x^2

.

2. Evaluate at the endpoints

F(3) = \frac{27}{3} = 9

and

F(1) = \frac{1}{3}

.

3. Subtract

\int_1^3 x^2 \, dx = 9 - \frac{1}{3} = \frac{26}{3}

.

Worked Example 02Medium

Evaluate

\int_0^{\pi} \sin(x) \, dx

.

How to think about it

Key idea: The antiderivative of $\sin(x)$ is $-\cos(x)$ .

Full solution

1. Antiderivative

F(x) = -\cos(x)

.

2. Evaluate

F(\pi) = -\cos(\pi) = 1

and

F(0) = -\cos(0) = -1

.

3. Subtract

1 - (-1) = 2

.

Step-by-Step SOP

1
Find F(x)
Determine an antiderivative of $f(x)$ (rewrite the integrand first if needed).
2
Evaluate at the bounds
Compute $F(b)$ and $F(a)$ .
3
Subtract
The value of the integral is $F(b) - F(a)$ .

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Evaluate

\int_0^1 (e^x + 2x) \, dx

.

Need a hint?

Integrate

e^x

and

2x

separately using the sum rule.

Show solution

1. Split the integral

\int_0^1 e^x \, dx + \int_0^1 2x \, dx

.

2. Antiderivatives

F(x) = e^x + x^2

.

3. Evaluate

(e^1 + 1^2) - (e^0 + 0^2) = (e + 1) - 1 = e

.

Practice 02Medium

Evaluate

\int_1^2 \dfrac{3x^2 - 1}{x}\,dx

.

Need a hint?

Divide through by

x

first:

3x - \frac{1}{x}

.

Show solution

1. Rewrite the integrand

\dfrac{3x^2 - 1}{x} = 3x - \dfrac{1}{x}

.

2. Antiderivative

F(x) = \dfrac{3x^2}{2} - \ln|x|

.

3. Evaluate

\left(6 - \ln 2\right) - \left(\dfrac{3}{2} - 0\right) = \dfrac{9}{2} - \ln 2

.

Common Pitfalls

⚠
Trying to Evaluate Without an AntiderivativeFTC Part 1 needs $F$ with $F' = f$ . If you cannot find one by hand, the AP context expects a calculator (Part A) or a geometry/accumulation argument.

The Fundamental Theorem of Calculus, Part 2

Differentiating a function defined by an integral. If $g(x) = \int_a^x f(t)\,dt$ , then $g'(x) = f(x)$ — differentiation undoes integration. With a function in the upper limit, multiply by its derivative (chain rule).

Core Theorem

If

f

is continuous and

g(x) = \int_a^x f(t) \, dt

, then

g'(x) = f(x)

. More generally, with a function

u(x)

in the upper limit:

\frac{d}{dx} \int_a^{u(x)} f(t) \, dt = f\big(u(x)\big) \cdot u'(x)

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Find

\frac{d}{dx} \int_0^x \sin(t) \, dt

.

How to think about it

Key idea: Apply FTC Part 2 directly — the upper limit is just $x$ .

Full solution

1. Recognize the form

This is

\int_a^x f(t)\,dt

with

f(t) = \sin(t)

.

2. Apply the theorem

\frac{d}{dx} \int_0^x \sin(t) \, dt = \sin(x)

.

Worked Example 02Medium

Find

\frac{d}{dx} \int_1^{x^2} e^t \, dt

.

How to think about it

Key idea: The upper limit is $u(x) = x^2$ , so use the chain-rule form.

Full solution

1. Identify f and u

f(t) = e^t

,

u(x) = x^2

, so

u'(x) = 2x

.

2. Apply the rule

f\big(u(x)\big)\cdot u'(x) = e^{x^2}\cdot 2x = 2x\,e^{x^2}

.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

If

g(x) = \int_2^{x} (3t^2 + 1)\,dt

, find

g'(x)

and

g(2)

.

Need a hint?

FTC Part 2 gives

g'

instantly. For

g(2)

, look at the interval of integration.

Show solution

1. Differentiate

g'(x) = 3x^2 + 1

.

2. Evaluate g(2)

g(2) = \int_2^2 (3t^2 + 1)\,dt = 0

(zero-width interval).

Practice 02Hard

Find

\frac{d}{dx} \int_x^{4} \cos(t^2)\,dt

.

Need a hint?

The variable is in the LOWER limit. Reverse the limits (introducing a minus sign) so

x

is on top.

Show solution

1. Reverse the limits

\int_x^4 \cos(t^2)\,dt = -\int_4^x \cos(t^2)\,dt

.

2. Differentiate

\frac{d}{dx}\left[-\int_4^x \cos(t^2)\,dt\right] = -\cos(x^2)

.

Common Pitfalls

⚠
Forgetting the chain-rule factorWhen the upper limit is $x^2$ (not just $x$ ), you must multiply by $2x$ . Leaving it out is the most common mistake.
⚠
Variable in the lower limitIf $x$ is the lower limit, $\frac{d}{dx}\int_x^b f(t)\,dt = -f(x)$ — flip the sign first using the reversal property.

The Definite Integral And Ftc

Properties of the Definite Integral: Linearity, Additivity, and Order

Worked Examples

1. Split using linearity

2. Substitute the given values

Practice — Your Turn

1. Fix the reversed limits

2. Combine with additivity

1. Signed area

2. Total area

The Fundamental Theorem of Calculus, Part 1

Worked Examples

1. Find an antiderivative

2. Evaluate at the endpoints

3. Subtract

1. Antiderivative

2. Evaluate

3. Subtract

Find F(x)

Evaluate at the bounds

Subtract

Practice — Your Turn

1. Split the integral

2. Antiderivatives

3. Evaluate

1. Rewrite the integrand

2. Antiderivative

3. Evaluate

The Fundamental Theorem of Calculus, Part 2

Worked Examples

1. Recognize the form

2. Apply the theorem

1. Identify f and u

2. Apply the rule

Practice — Your Turn

1. Differentiate

2. Evaluate g(2)

1. Reverse the limits

2. Differentiate

Gary Chang