04 // Interactive Formula Compendium // CALCULUS-ALGEBRA∫_a^b f(x) dx = F(b) − F(a), where F'(x) = f(x)
Fundamental Theorem of Calculus (FTC Part 1 & Part 2) — Reference Sheet
The Fundamental Theorem of Calculus states that if f is continuous on [a, b] and F is any antiderivative satisfying F'(x) = f(x), then the definite integral ∫_a^b f(x)dx equals the net boundary change F(b) − F(a).
Formula Reference
Main Formula
∫_a^b f(x) dx = F(b) − F(a), where F'(x) = f(x)
Rearranged Forms
d/dx [ ∫_a^x f(t) dt ] = f(x) (FTC Part 1)
d/dx [ ∫_{g(x)}^{h(x)} f(t) dt ] = f(h(x))·h'(x) − f(g(x))·g'(x) (Leibniz Rule)
f_avg = [ 1 / (b − a) ] · ∫_a^b f(x) dx (Mean Value for Integrals)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| I | Definite Power Integral ∫_a^b c·xⁿ dx | dimensionless |
| a | Lower Limit of Integration (a) | 1 |
| b | Upper Limit of Integration (b) | 1 |
| c | Coefficient (c) | 1 |
| n | Power Exponent (n ≠ −1) | 1 |
Inputs & Parameters
[1]
[1]
[1]
[1]
Calculated Result (Definite Power Integral ∫_a^b c·xⁿ dx)
I =8.666667dimensionless
Formula Notes & Background
- Partition [a, b] into subintervals [x_{i−1}, x_i] and express F(b) − F(a) as a telescoping sum Σ [F(x_i) − F(x_{i−1})].
- Apply the Mean Value Theorem on each subinterval: F(x_i) − F(x_{i−1}) = F'(c_i) Δx_i = f(c_i) Δx_i.
- Take the limit as max Δx_i → 0 so the Riemann sum Σ f(c_i) Δx_i converges to ∫_a^b f(x) dx = F(b) − F(a).
Step-by-Step Examples
#1: Compute the exact area under f(x) = x² from a = 1 to b = 3.
f(x) = x²Antiderivative F(x) = x³ / 3a = 1, b = 3
∫₁³ x² dx = F(3) − F(1) = 3³/3 − 1³/3 = 27/3 − 1/3
→ 26/3 ≈ 8.66667
#2: Find the derivative d/dx [ ∫₁^(x²) cos(t) dt ] using FTC Part 1 and the Chain Rule.
f(t) = cos(t)Upper limit h(x) = x² ⇒ h'(x) = 2x
d/dx = f(h(x)) · h'(x) = cos(x²) · (2x)
→ 2x cos(x²)
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Frequently Asked Questions
Why does the constant of integration +C cancel out in a definite integral?+
When evaluating [F(x) + C]_a^b = (F(b) + C) − (F(a) + C), the +C and −C cancel identically for any antiderivative.
What happens if the integrand has a vertical asymptote inside [a, b]?+
FTC Part 2 requires f(x) to be continuous on the closed interval [a, b]; if an interior singularity exists, you must split the integral at the discontinuity and evaluate improper limits.
What is the physical interpretation of ∫_a^b v(t) dt?+
Since velocity v(t) = s'(t) is the derivative of position, FTC Part 2 states that ∫_a^b v(t) dt = s(b) − s(a), the net displacement over [a, b].