💡 Key Idea: The definite integral is the limit of Riemann sums as the number of rectangles approaches infinity!

Notation

$\int_a^b f(x)\,dx$

Components:

• $\int$ - integral symbol
• $a$ - lower limit of integration
• $b$ - upper limit of integration
• $f(x)$ - integrand
• $dx$ - with respect to $x$

Read as: "The integral from $a$ to $b$ of $f(x)$ with respect to $x$"

Definite vs Indefinite Integrals

Indefinite Integral

$\int f(x)\,dx = F(x) + C$

• No limits
• Result is a function (family of antiderivatives)
• Includes $+C$

Definite Integral

$\int_a^b f(x)\,dx = \text{number}$

• Has limits $a$ and $b$
• Result is a number (area)
• No $+C$ needed

The Fundamental Theorem of Calculus

This is one of the most important theorems in all of mathematics! It connects derivatives and integrals.

Part 1 (FTC1)

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then:

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

In words: To evaluate a definite integral, find an antiderivative, then subtract its values at the endpoints!

Notation for FTC1

We write $F(x)\Big|_a^b$ or $[F(x)]_a^b$ to mean $F(b) - F(a)$

Example: $\left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} - 0 = \frac{8}{3}$

Using FTC to Evaluate Integrals

Step-by-Step Process

Step 1: Find an antiderivative $F(x)$ of $f(x)$

(Don't include $+C$ for definite integrals!)

Step 2: Evaluate $F(b)$ and $F(a)$

Step 3: Subtract: $F(b) - F(a)$

Example 1: Basic Application

Evaluate $\int_0^2 x^2\,dx$

Step 1: Find antiderivative

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

Step 2: Apply FTC

$\int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2$

$= \frac{2^3}{3} - \frac{0^3}{3}$

$= \frac{8}{3} - 0 = \frac{8}{3}$

Answer: $\frac{8}{3}$ square units

Example 2: With Multiple Terms

Evaluate $\int_1^3 (2x + 1)\,dx$

Step 1: Find antiderivative

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

Step 2: Apply FTC

$\int_1^3 (2x+1)\,dx = [x^2 + x]_1^3$

$= (3^2 + 3) - (1^2 + 1)$

$= (9 + 3) - (1 + 1)$

$= 12 - 2 = 10$

Answer: $10$ square units

Example 3: Trigonometric

Evaluate $\int_0^{\pi} \sin x\,dx$

Step 1: Find antiderivative

$F(x) = -\cos x$

Step 2: Apply FTC

$\int_0^{\pi} \sin x\,dx = [-\cos x]_0^{\pi}$

$= -\cos(\pi) - (-\cos(0))$

$= -(-1) - (-1)$

$= 1 + 1 = 2$

Answer: $2$ square units

Properties of Definite Integrals

Property 1: Constant Multiple

$\int_a^b k \cdot f(x)\,dx = k\int_a^b f(x)\,dx$

Property 2: Sum/Difference

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

Property 3: Reversing Limits

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

Swapping limits changes the sign!

Property 4: Zero Width

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

No area when limits are the same!

Property 5: Additivity

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

You can split integrals at any point!

Example 4: Using Properties

Given $\int_0^2 f(x)\,dx = 5$ and $\int_2^4 f(x)\,dx = 3$, find $\int_0^4 f(x)\,dx$.

Solution:

Using additivity property:

$\int_0^4 f(x)\,dx = \int_0^2 f(x)\,dx + \int_2^4 f(x)\,dx$

$= 5 + 3 = 8$

Area and Signed Area

Signed Area

The definite integral gives signed area:

• Area above the x-axis: positive
• Area below the x-axis: negative

Example: Area Below Axis

$\int_0^{2\pi} \sin x\,dx = [-\cos x]_0^{2\pi} = -\cos(2\pi) + \cos(0) = -1 + 1 = 0$

Why zero? Because area above axis (0 to $\pi$) cancels area below axis ($\pi$ to $2\pi$)!

Finding Total Area

To find total area (all positive), split at x-intercepts:

$\text{Total Area} = \int_0^{\pi} \sin x\,dx + \left|\int_{\pi}^{2\pi} \sin x\,dx\right|$

$= 2 + |-2| = 2 + 2 = 4$

The Second Part of FTC

FTC Part 2

If $f$ is continuous on $[a, b]$, then:

$\frac{d}{dx}\left[\int_a^x f(t)\,dt\right] = f(x)$

In words: The derivative of an integral with respect to its upper limit is the integrand!

This shows that differentiation and integration are inverse operations!

Example 5: Using FTC Part 2

Find $\frac{d}{dx}\left[\int_0^x t^2\,dt\right]$

Solution:

By FTC Part 2: $\frac{d}{dx}\left[\int_0^x t^2\,dt\right] = x^2$

Verification: Do it the long way:

$\int_0^x t^2\,dt = \left[\frac{t^3}{3}\right]_0^x = \frac{x^3}{3} - 0 = \frac{x^3}{3}$

$\frac{d}{dx}\left[\frac{x^3}{3}\right] = \frac{3x^2}{3} = x^2$ ✓

Example 6: With Chain Rule

Find $\frac{d}{dx}\left[\int_0^{x^2} \sin t\,dt\right]$

Solution:

Upper limit is $x^2$, not just $x$. Need chain rule!

Let $u = x^2$, then by FTC Part 2 and chain rule:

$\frac{d}{dx}\left[\int_0^{x^2} \sin t\,dt\right] = \sin(x^2) \cdot \frac{d}{dx}[x^2]$

$= \sin(x^2) \cdot 2x = 2x\sin(x^2)$

Average Value of a Function

The average value of $f$ on $[a, b]$ is:

$f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$

Think: Total area divided by width = average height

Example 7: Average Value

Find the average value of $f(x) = x^2$ on $[0, 3]$.

Solution:

$f_{\text{avg}} = \frac{1}{3-0}\int_0^3 x^2\,dx$

$= \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3$

$= \frac{1}{3} \cdot \frac{27}{3} = \frac{1}{3} \cdot 9 = 3$

Answer: The average value is $3$.

Mean Value Theorem for Integrals

If $f$ is continuous on $[a, b]$, there exists $c$ in $[a, b]$ such that:

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

In words: There's a point where the function equals its average value!

$f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx = f_{\text{avg}}$

⚠️ Common Mistakes

Mistake 1: Including +C

WRONG: $\int_0^2 x^2\,dx = \frac{x^3}{3} + C\Big|_0^2$

RIGHT: $\int_0^2 x^2\,dx = \frac{x^3}{3}\Big|_0^2$

Definite integrals are numbers, no +C needed!

Mistake 2: Wrong Order of Subtraction

WRONG: $F(a) - F(b)$

RIGHT: $F(b) - F(a)$ (upper minus lower!)

Mistake 3: Forgetting Negative Areas

If the function dips below the x-axis, the integral can be negative!

For total area, use absolute values or split at zeros.

Mistake 4: Wrong Variable in FTC Part 2

$\frac{d}{dx}\left[\int_a^x f(t)\,dt\right] = f(x)$

Not $f(t)$! The answer is in terms of $x$ (the upper limit).

Why FTC is Profound

The Fundamental Theorem of Calculus reveals:

Differentiation and Integration are inverse operations!

• Integration "undoes" differentiation
• Differentiation "undoes" integration

This connects two seemingly different problems:

• Area problem (integration)
• Tangent problem (differentiation)

📝 Practice Strategy

1. For evaluation: Find antiderivative, evaluate at limits, subtract
2. Don't include +C for definite integrals
3. Remember: $F(b) - F(a)$ (upper minus lower)
4. Check limits: Is the answer positive/negative as expected?
5. For FTC Part 2: Derivative of integral = integrand (watch for chain rule)
6. For average value: $\frac{1}{b-a}\int_a^b f(x)\,dx$
7. For total area: Split at zeros if function crosses x-axis

Fundamental Theorem of Calculus:

$\int_1^3 (2x + 1) \, dx = [x^2 + x]_1^3$

$= (3^2 + 3) - (1^2 + 1)$

$= 12 - 2 = 10$

Part (b): Antiderivative of $\cos x$ is $\sin x$:

$\int_0^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2}$

$= \sin(\pi/2) - \sin(0)$

$= 1 - 0 = 1$

$= 7 + (-2) = 5$

Part 2: Find $\int_5^1 f(x)\,dx$

Use the property that reversing limits changes the sign:

$\int_5^1 f(x)\,dx = -\int_1^5 f(x)\,dx$

$= -5$

Answers:

• $\int_1^5 f(x)\,dx = 5$
• $\int_5^1 f(x)\,dx = -5$

Split the integral:

$\int_{-1}^2 |x| \, dx = \int_{-1}^0 (-x) \, dx + \int_0^2 x \, dx$

First integral:

$\int_{-1}^0 (-x) \, dx = \left[-\frac{x^2}{2}\right]_{-1}^0 = 0 - \left(-\frac{1}{2}\right) = \frac{1}{2}$

Second integral:

$\int_0^2 x \, dx = \left[\frac{x^2}{2}\right]_0^2 = \frac{4}{2} - 0 = 2$

Total:

$\int_{-1}^2 |x| \, dx = \frac{1}{2} + 2 = \frac{5}{2}$

Step 2: Simplify

$= \sqrt{1+x^{12}} \cdot 3x^2$

$= 3x^2\sqrt{1+x^{12}}$

Answer: $3x^2\sqrt{1+x^{12}}$

Key idea: When the upper limit is a function of $x$ (not just $x$), multiply by the derivative of that function (chain rule)!