Definite Integrals and the Fundamental Theorem

What You’ll Learn

In this lesson you’ll learn what a definite integral computes, how the Fundamental Theorem of Calculus connects differentiation and integration, and how to evaluate definite integrals.

The Concept

The Definite Integral

The definite integral from a to b of f(x) is written:

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

It represents the net signed area between the curve and the x-axis from x = a to x = b. Area above the x-axis counts as positive, area below counts as negative.

f(x) = x² area = definite integral from 1 to 3

The green shaded region is the definite integral of x² from 1 to 3. It’s the area trapped between the blue curve, the x-axis, and the vertical lines at x = 1 and x = 3.

$$\int_1^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3} \approx 8.67$$

The Fundamental Theorem of Calculus

This is the theorem that ties the entire course together.

Part 1 (Evaluation): if F is any antiderivative of f, then:

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

Find an antiderivative, plug in the upper bound, plug in the lower bound, subtract. That’s it.

How to Evaluate a Definite Integral (Step by Step)

Here’s the process, broken down:

Step 1: Find the antiderivative F(x) of the integrand f(x). This is the same as finding the indefinite integral, but you don’t need + C.

Step 2: Write the antiderivative with the bounds using the bracket notation:

$$\left[F(x)\right]_a^b$$

This means “F(x) evaluated from a to b.”

Step 3: Plug in the upper bound b to get F(b).

Step 4: Plug in the lower bound a to get F(a).

Step 5: Subtract: F(b) - F(a). That’s your answer.

No + C needed for definite integrals because it cancels out:

$$(F(b) + C) - (F(a) + C) = F(b) - F(a)$$

Part 2 (Derivative of an integral): if f is continuous, then:

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

Differentiation and integration undo each other. This is why they’re called inverse operations.

Worked Example

Example 1: Polynomial

Evaluate the integral of 3x² - 2x + 5 from 1 to 4.

$$\int_1^4 (3x^2 - 2x + 5) \, dx$$

Antiderivative: F(x) = x³ - x² + 5x.

$$F(4) - F(1) = (64 - 16 + 20) - (1 - 1 + 5) = 68 - 5 = 63$$

Example 2: Trig function

Evaluate the integral of sin(x) from 0 to pi/2.

$$\int_0^{\pi/2} \sin x \, dx$$

Antiderivative: F(x) = -cos(x).

$$F(\pi/2) - F(0) = -\cos(\pi/2) - (-\cos(0)) = 0 - (-1) = 1$$

The area under one arch of sin(x) from 0 to pi/2 is exactly 1.

Example 3: FTC Part 2

Find the derivative of the integral from 1 to x of the square root of (t³ + 1) dt.

By Part 2 of the FTC, the answer is just the integrand evaluated at x:

$$\frac{d}{dx}\left[\int_1^x \sqrt{t^3 + 1} \, dt\right] = \sqrt{x^3 + 1}$$

No antiderivative needed. The derivative of the integral is the original function. That’s the whole point of the theorem.

Real-World Application

Definite integrals compute total accumulation:

• Physics: total distance traveled = integral of speed. Total work done = integral of force over distance.
• Economics: total revenue over a time period = integral of the revenue rate.
• Biology: total growth of a population over time = integral of the growth rate.
• Engineering: total energy consumed = integral of power over time.

The FTC is what makes all of this practical. Without it, you’d have to compute limits of Riemann sums every time. With it, you just find an antiderivative and subtract.

You’ve Got This

The Fundamental Theorem of Calculus is the bridge between the two halves of calculus. Derivatives tell you how fast something is changing. Integrals tell you how much it changed in total. The FTC says: to find the total change, find an antiderivative and evaluate at the endpoints. F(b) - F(a). That’s the whole computation.

Quiz

The definite integral represents:

• A.The derivative at a point
• B.Net signed area under the curve
• C.The slope of the tangent line
• D.Only positive area

The FTC Part 1 says the definite integral equals:

• A.F(a) - F(b)
• B.F'(b) - F'(a)
• C.F(b) - F(a)
• D.F(x) + C

The integral of sin(x) from 0 to pi/2 equals:

• A.0
• B.2
• C.-1
• D.1

FTC Part 2 says the derivative of an integral (with variable upper bound) is:

• A.Zero
• B.The original integrand
• C.The antiderivative
• D.The average value

Why is the FTC so important?

• A.It only works for polynomials
• B.It replaces the need for derivatives
• C.It connects differentiation and integration as inverse operations
• D.It only applies to trig functions