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Lesson 3: The Definite Integral and FTC Part 1

Estimated time: 35-45 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Definition of the Definite Integral

The Definite Integral

If f is a function defined on [a, b], the definite integral of f from a to b is:

ab f(x) dx = limn→∞i=1n f(xi*) Δx

provided this limit exists (in which case f is called integrable on [a, b]).

Notation: In ∫ab f(x) dx, the number a is the lower limit, b is the upper limit, f(x) is the integrand, and dx indicates the variable of integration.

Integrability Theorem

If f is continuous on [a, b], or if f is bounded and has only finitely many discontinuities on [a, b], then f is integrable on [a, b].

Signed Area Interpretation

The definite integral computes signed area:

Therefore ∫ab f(x) dx = (area above x-axis) − (area below x-axis).

Example 1: Signed Area Using Geometry

Evaluate ∫−22 x dx using geometry.

Solution: The graph of y = x is a line through the origin. On [−2, 0] it forms a triangle below the x-axis with area (1/2)(2)(2) = 2. On [0, 2] it forms a triangle above with area 2.

Signed area: 2 − 2 = 0.

This makes sense: by symmetry, the positive and negative areas cancel.

Example 2: Geometric Evaluation

Evaluate ∫03 (2x + 1) dx using geometry.

Solution: The graph is a line from (0, 1) to (3, 7). The region is a trapezoid with parallel sides 1 and 7, height 3.

Area = (1/2)(1 + 7)(3) = 12.

Properties of Definite Integrals

Key Properties

  1. aa f(x) dx = 0
  2. ba f(x) dx = −∫ab f(x) dx (reversing limits changes the sign)
  3. ab c · f(x) dx = c · ∫ab f(x) dx (constant multiple)
  4. ab [f(x) ± g(x)] dx = ∫ab f(x) dx ± ∫ab g(x) dx (sum/difference)
  5. ab f(x) dx + ∫bc f(x) dx = ∫ac f(x) dx (additivity over intervals)

Example 3: Using Properties

Given ∫14 f(x) dx = 6 and ∫14 g(x) dx = −2, find ∫14 [3f(x) − 2g(x)] dx.

Solution: = 3 ∫14 f(x) dx − 2 ∫14 g(x) dx = 3(6) − 2(−2) = 18 + 4 = 22.

Example 4: Additivity

Given ∫05 f(x) dx = 10 and ∫03 f(x) dx = 7, find ∫35 f(x) dx.

Solution: By the additivity property: ∫03 f(x) dx + ∫35 f(x) dx = ∫05 f(x) dx.

So ∫35 f(x) dx = 10 − 7 = 3.

The Fundamental Theorem of Calculus, Part 1

This is one of the most important results in all of mathematics. It provides a practical way to evaluate definite integrals without computing limits of Riemann sums.

FTC Part 1 (Evaluation Theorem)

If f is continuous on [a, b] and F is any antiderivative of f (that is, F' = f), then:

ab f(x) dx = F(b) − F(a)

We write F(x) |ab = F(b) − F(a).

This is remarkable: instead of computing tedious limits of sums, we just need to find an antiderivative and plug in the endpoints.

Example 5: Applying FTC Part 1

Evaluate ∫13 x² dx.

Step 1: An antiderivative of x² is F(x) = x³/3.

Step 2: F(3) − F(1) = 27/3 − 1/3 = 9 − 1/3 = 26/3.

Example 6: FTC with Trig

Evaluate ∫0π/2 cos x dx.

Step 1: Antiderivative: F(x) = sin x.

Step 2: sin(π/2) − sin(0) = 1 − 0 = 1.

More Evaluation Theorem Examples

Example 7: Exponential Function

Evaluate ∫01 ex dx.

Solution: F(x) = ex. So ∫01 ex dx = e1 − e0 = e − 1 ≈ 1.718.

Example 8: Polynomial

Evaluate ∫−12 (3x² − 4x + 1) dx.

Step 1: F(x) = x³ − 2x² + x.

Step 2: F(2) − F(−1) = (8 − 8 + 2) − (−1 − 2 − 1) = 2 − (−4) = 6.

Comparison Properties

These inequalities are useful for estimating integrals:

Key Takeaways

Check Your Understanding

1. Evaluate ∫04 (x + 2) dx using the FTC.

Answer: F(x) = x²/2 + 2x. F(4) − F(0) = (8 + 8) − 0 = 16.

2. Evaluate ∫0π sin x dx.

Answer: F(x) = −cos x. F(π) − F(0) = −cos(π) − (−cos 0) = −(−1) + 1 = 2.

3. If ∫06 f(x) dx = 10 and ∫46 f(x) dx = 3, find ∫04 f(x) dx.

Answer: By additivity: ∫04 f dx + ∫46 f dx = ∫06 f dx. So ∫04 f dx = 10 − 3 = 7.

4. Evaluate ∫1e (1/x) dx.

Answer: F(x) = ln x. F(e) − F(1) = ln e − ln 1 = 1 − 0 = 1.

Ready for More?

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In Lesson 4, you will learn FTC Part 2 (differentiating an integral) and the Net Change Theorem.

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