Calculus I - Week 11

The Definite Integral and the Fundamental Theorem of Calculus

The definite integral computes the signed area between a curve and the x-axis over an interval. The Fundamental Theorem of Calculus is one of the greatest results in mathematics: it reveals that differentiation and integration are inverse operations, turning area calculations into algebraic ones.

Learning Objectives

- Interpret the definite integral as signed area. - Use Riemann sums to approximate definite integrals. - State both parts of the Fundamental Theorem of Calculus. - Evaluate definite integrals using antiderivatives. - Use the FTC Part 1 to differentiate an integral with a variable upper limit. - Compute net displacement and total distance from velocity. - Interpret negative area and signed area in context.

1. The Area Problem

Finding the area under a curve is hard because the curve is not straight. The approach: divide the interval [a, b] into n thin rectangles, approximate the area of each rectangle, then sum them all. A Riemann sum is: S_n = sum[i=1 to n] f(x_i) * Delta_x where Delta_x = (b - a)/n and x_i is a sample point in each subinterval. As n → infinity, the sum becomes exact.

2. The Definite Integral

The definite integral is the limit of Riemann sums: ∫[a to b] f(x) dx = lim[n→∞] sum[i=1 to n] f(x_i) Delta_x Geometrically: it is the net signed area between f(x) and the x-axis from a to b. - Area above the x-axis counts positive. - Area below the x-axis counts negative.

3. Properties of Definite Integrals

∫[a to b] c dx = c(b - a) (integral of a constant) ∫[a to b] [f + g] dx = ∫f dx + ∫g dx (linearity) ∫[a to b] c*f dx = c * ∫[a to b] f dx (constant factor) ∫[a to a] f dx = 0 (same endpoints) ∫[b to a] f dx = -∫[a to b] f dx (reverse limits) ∫[a to b] f dx + ∫[b to c] f dx = ∫[a to c] f dx (additivity)

4. The Fundamental Theorem of Calculus — Part 2

If F is any antiderivative of f (meaning F' = f), then: ∫[a to b] f(x) dx = F(b) - F(a) We write F(b) - F(a) as [F(x)] from a to b. Example: ∫[1 to 3] x^2 dx = [x^3/3] from 1 to 3 = 27/3 - 1/3 = 26/3.

5. More Definite Integral Examples

Example 1: ∫[0 to 2] (3x^2 + 1) dx = [x^3 + x] from 0 to 2 = (8 + 2) - (0) = 10. Example 2: ∫[0 to pi] sin(x) dx = [-cos(x)] from 0 to pi = (-cos(pi)) - (-cos(0)) = 1 + 1 = 2. Example 3: ∫[1 to 4] sqrt(x) dx = [(2/3)x^(3/2)] from 1 to 4 = (2/3)(8) - (2/3)(1) = 16/3 - 2/3 = 14/3.

6. The Fundamental Theorem of Calculus — Part 1

If g(x) = ∫[a to x] f(t) dt, then g'(x) = f(x). In words: the derivative of an integral (with variable upper limit) is just the integrand evaluated at x. Examples: d/dx ∫[0 to x] t^3 dt = x^3 d/dx ∫[1 to x] cos(t) dt = cos(x) With chain rule: d/dx ∫[0 to x^2] sin(t) dt = sin(x^2) * 2x

7. Net Displacement vs. Total Distance

If v(t) is velocity: Net displacement = ∫[a to b] v(t) dt (signed, can be zero if you return) Total distance = ∫[a to b] |v(t)| dt (always positive) To compute total distance: find where v changes sign, split the integral at those points, and add absolute values. Example: v(t) = t - 2 on [0, 4]. Net displacement = ∫[0 to 4] (t-2) dt = [t^2/2 - 2t] from 0 to 4 = (8-8) - 0 = 0. Total distance = ∫[0 to 2] (2-t) dt + ∫[2 to 4] (t-2) dt = 2 + 2 = 4.

8. Area Between Curves Preview

If f(x) >= g(x) on [a, b], the area between the curves is: ∫[a to b] [f(x) - g(x)] dx Example: Area between y = x^2 and y = x on [0, 1]. On [0,1]: x >= x^2. ∫[0 to 1] (x - x^2) dx = [x^2/2 - x^3/3] from 0 to 1 = 1/2 - 1/3 = 1/6.

9. Physics: Work Done by a Variable Force

If force F(x) varies with position x, the work done from x=a to x=b is: W = ∫[a to b] F(x) dx Example: A spring force is F(x) = kx (Hooke's Law). Work to compress the spring from 0 to d: W = ∫[0 to d] kx dx = k * d^2/2. For k=200 N/m and d=0.1 m: W = 200 * 0.01/2 = 1 joule.

10. Common Mistakes

- Forgetting to subtract F(a) — writing just F(b) without the lower limit. - Mixing up FTC Part 1 (differentiate an integral) and Part 2 (evaluate a definite integral). - Confusing net displacement and total distance when velocity goes negative. - Dropping the absolute value when computing total distance. - Forgetting the chain rule in FTC Part 1 when the upper limit is not just x.