Calculus I - Week 10
Introduction to Antiderivatives and Integration
The second half of calculus. If the derivative asks 'how fast is this changing?', the antiderivative asks 'what function was being differentiated?'. This week introduces antiderivatives, indefinite integrals, and the power rule for integration — setting the stage for the Fundamental Theorem of Calculus.
Learning Objectives
- Define an antiderivative and the indefinite integral. - Explain why antiderivatives include a constant +C. - Apply the power rule for antiderivatives (reverse power rule). - Integrate sums and differences term by term. - Integrate constant multiples. - Use antiderivatives to find position from velocity and velocity from acceleration. - Apply initial conditions to determine the constant of integration.
1. The Big Question: Going Backward
Differentiation asks: given f(x), what is f'(x)? Antidifferentiation asks: given f'(x), what is f(x)? If f'(x) = 2x, then f(x) = x^2 works. But so does x^2 + 5, or x^2 - pi. All of these have derivative 2x. This is why antiderivatives always include an arbitrary constant +C.
2. The Indefinite Integral
The notation for the antiderivative is the indefinite integral: ∫ f(x) dx = F(x) + C where F'(x) = f(x). The integral sign ∫ is a stretched S (standing for sum). The dx indicates we are integrating with respect to x. f(x) is called the integrand. C is the constant of integration.
3. The Power Rule for Integration
If n ≠ -1: ∫ x^n dx = x^(n+1) / (n+1) + C In words: increase the exponent by 1, then divide by the new exponent. Examples: ∫ x^3 dx = x^4/4 + C ∫ x^5 dx = x^6/6 + C ∫ 1 dx = ∫ x^0 dx = x + C
4. Constant Multiple and Sum Rules
∫ k*f(x) dx = k * ∫ f(x) dx (constant pulls out) ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx (sum splits) Examples: ∫ 5x^3 dx = 5 * x^4/4 + C = 5x^4/4 + C ∫ (3x^2 + 2x + 1) dx = x^3 + x^2 + x + C
5. Checking Antiderivatives by Differentiating
Always check: differentiate the answer and verify you get the integrand. Example: ∫ (6x^2 - 4x + 7) dx = 2x^3 - 2x^2 + 7x + C Check: d/dx [2x^3 - 2x^2 + 7x + C] = 6x^2 - 4x + 7. ✓
6. Integrating Trig Functions
Since d/dx[sin(x)] = cos(x): ∫ cos(x) dx = sin(x) + C Since d/dx[cos(x)] = -sin(x): ∫ sin(x) dx = -cos(x) + C (note the negative) Since d/dx[tan(x)] = sec^2(x): ∫ sec^2(x) dx = tan(x) + C These come directly from reading the derivative table backward.
7. Integrating Power Functions with Negative and Fractional Exponents
The power rule works for all n ≠ -1. ∫ x^(-2) dx = x^(-1) / (-1) + C = -1/x + C ∫ x^(1/2) dx = x^(3/2) / (3/2) + C = (2/3)x^(3/2) + C ∫ x^(-1/2) dx = x^(1/2) / (1/2) + C = 2*sqrt(x) + C Verify each by differentiating.
8. Initial Value Problems
An initial condition determines which antiderivative is correct. Example: Find f(x) given f'(x) = 2x + 1 and f(0) = 3. Integrate: f(x) = x^2 + x + C. Apply f(0) = 3: 0 + 0 + C = 3, so C = 3. Answer: f(x) = x^2 + x + 3.
9. Physics: Velocity and Position
If a(t) is acceleration, then: v(t) = ∫ a(t) dt (velocity is antiderivative of acceleration) s(t) = ∫ v(t) dt (position is antiderivative of velocity) Example: A ball is thrown upward with v(0) = 20 m/s. Gravity gives a(t) = -10 m/s^2. v(t) = ∫ -10 dt = -10t + C. v(0)=20 → C=20. v(t) = -10t + 20. s(t) = ∫ (-10t + 20) dt = -5t^2 + 20t + C. s(0)=0 → C=0. s(t) = -5t^2 + 20t.
10. Common Mistakes
- Forgetting the +C in indefinite integrals. - Dividing by (n+1) but leaving the exponent as n (not n+1). - Applying the power rule to n = -1. (∫ 1/x dx = ln|x| + C, not x^0.) - Integrating sin(x) as cos(x) instead of -cos(x). - Not applying an initial condition to find C when given one.