Source: Chapter 5, Section 5.2 | Calculus (Texas A&M University)
Tags: Fundamental Theorem of Calculus, FTC, FTC II, anti-derivative, indefinite integral, definite integral, basic integrals table, antidifferentiation, evaluation theorem
The Fundamental Theorem of Calculus (FTC II) connects differentiation and integration: to evaluate a definite integral ∫ₐᵇ f(x) dx, find any anti-derivative F of f and compute F(b) − F(a). This eliminates the need for Riemann sum limits entirely. Knowing your table of basic anti-derivatives is essential.
Fundamental Theorem of Calculus, Part II (FTC II)
If f is continuous on [a, b] and F is any anti-derivative of f (meaning F′ = f), then ∫ₐᵇ f(x) dx = F(b) − F(a). Sometimes written with the evaluation bar: [F(x)]ₐᵇ = F(b) − F(a).
Anti-derivative
A function F such that F′(x) = f(x). Anti-derivatives are not unique: if F is an anti-derivative of f, so is F + C for any constant C. This is why the indefinite integral includes "+ C."
Definite integral
∫ₐᵇ f(t) dt. Produces a number (the signed area under f from a to b). Has specific limits of integration.
Indefinite integral
∫ f(t) dt. Produces a function (a family of anti-derivatives). No limits of integration. Always includes the constant of integration "+ C."
If f is continuous on [a, b] and F′ = f, then:
∫ₐᵇ f(x) dx = F(b) − F(a)
You can use any anti-derivative F of f. The constant of integration cancels out: [F(b) + C] − [F(a) + C] = F(b) − F(a).
The whole point: computing a definite integral reduces to finding an anti-derivative and plugging in two numbers. No more tedious Riemann sum limits.
∫ₐᵇ f(t) dt = a number (signed area)
∫ f(t) dt = a function (anti-derivative family, with + C)
Confusing these is one of the most common errors on exams. The definite integral has no "+ C." The indefinite integral always has one.
This table is the backbone of integration at this level. Memorise it.
Function g(x) | Anti-derivative ∫ g(x) dx |
|---|---|
xⁿ (n ≠ −1) | xⁿ⁺¹/(n+1) + C |
1 | x + C |
1/x = x⁻¹ | ln|x| + C |
1/(1+x²) | arctan(x) + C |
1/√(1−x²) | arcsin(x) + C |
sin x | −cos x + C |
cos x | sin x + C |
sec² x | tan x + C |
sec x · tan x | sec x + C |
eˣ | eˣ + C |
aˣ | aˣ / ln a + C |
(a) ∫₁² (1/x²) dx
Rewrite as ∫₁² x⁻² dx. Anti-derivative: −x⁻¹ = −1/x.
[−1/x]₁² = (−1/2) − (−1/1) = −1/2 + 1 = 1/2.
(b) ∫₋₁⁰ (5x² − 4x + 3) dx
Anti-derivative: (5/3)x³ − 2x² + 3x.
Evaluate at 0: 0. Evaluate at −1: (5/3)(−1) − 2(1) + 3(−1) = −5/3 − 2 − 3 = −29/3.
Result: 0 − (−29/3) = 29/3.
(c) ∫_{ln 3}^{ln 6} 8eᵗ dt
Anti-derivative: 8eᵗ.
[8eᵗ]_{ln 3}^{ln 6} = 8e^{ln 6} − 8e^{ln 3} = 8(6) − 8(3) = 48 − 24 = 24.
(d) ∫₁² (x⁶ − x²)/x⁷ dx
Simplify the integrand first: x⁶/x⁷ − x²/x⁷ = x⁻¹ − x⁻⁵.
Anti-derivative: ln|x| − x⁻⁴/(−4) = ln|x| + (1/4)x⁻⁴.
Evaluate at 2: ln 2 + 1/64. Evaluate at 1: 0 + 1/4.
Result: ln 2 + 1/64 − 1/4 = ln 2 − 15/64.
(e) ∫₀² (w³ − 1)² dw
Expand first: w⁶ − 2w³ + 1.
Anti-derivative: w⁷/7 − w⁴/2 + w.
Evaluate at 2: 128/7 − 8 + 2 = 128/7 − 6. Evaluate at 0: 0.
Result: 128/7 − 6 = 86/7.
(f) ∫₃³ x sin x dx
The limits are equal. The integral is 0 immediately, regardless of the integrand.
(g) ∫₋₂⁰ |x² − 1| dx
You cannot directly anti-differentiate an absolute value. The approach: determine where x² − 1 changes sign, split the integral at those points, and remove the absolute value on each piece.
x² − 1 = 0 at x = ±1. On [−2, −1]: x² − 1 ≥ 0, so |x² − 1| = x² − 1. On [−1, 0]: x² − 1 ≤ 0, so |x² − 1| = −(x² − 1) = 1 − x².
Split: ∫₋₂⁻¹ (x² − 1) dx + ∫₋₁⁰ (1 − x²) dx. Evaluate each using FTC.
∫₁² (1/x³) dx is fine, since 1/x³ is continuous on [1, 2].
∫₋₁² (1/x³) dx is problematic: 1/x³ is not continuous at x = 0, which lies in [−1, 2]. You cannot blindly apply FTC across a discontinuity. This type of integral requires improper integral techniques (covered later).
FTC II (the evaluation formula):
∫ₐᵇ f(x) dx = F(b) − F(a), where F′ = f
Power rule for integration (n ≠ −1):
∫ xⁿ dx = xⁿ⁺¹/(n+1) + C
⚠️ Definite integral = number. Indefinite integral = function + C. Forgetting "+ C" on indefinite integrals is a common mark loss.
⚠️ Always simplify the integrand before integrating. Divide out common factors, expand products, rewrite as power functions where possible.
⚠️ Absolute value integrals require splitting at the zeros of the expression inside | |. You must determine the sign on each sub-interval.
⚠️ FTC requires continuity on the entire interval [a, b]. If the integrand blows up inside the interval (like 1/x³ at x = 0), FTC does not apply directly.
⚠️ ∫₃³ (anything) dx = 0. Same-limit integrals are a classic exam freebie.
Q: Evaluate ∫₀¹ (3x² + 2x) dx.
A: Anti-derivative: x³ + x². At x = 1: 1 + 1 = 2. At x = 0: 0. Answer: 2.
Q: What is the difference between ∫₀² eˣ dx and ∫ eˣ dx?
A: The first is a definite integral and equals a number: e² − e⁰ = e² − 1. The second is an indefinite integral and equals a function: eˣ + C.
Q: Why can you not evaluate ∫₋₁¹ (1/x²) dx by writing [−1/x]₋₁¹ = (−1) − (1) = −2?
A: Because 1/x² is not continuous at x = 0, which lies inside [−1, 1]. FTC requires continuity on the whole interval. This is an improper integral.
Q: Find ∫₁ᵉ (1/x) dx.
A: Anti-derivative: ln|x|. At x = e: ln e = 1. At x = 1: ln 1 = 0. Answer: 1.
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