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Fundamental Theorem of Calculus (Definite Integral)

Connects differentiation and integration, allowing a definite integral to be evaluated directly from an antiderivative at the interval endpoints.

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Fundamental Theorem CalculatorDefinite Integral Evaluation

abf'(x) dx = F(b)F(a)
Integral = F(b)F(a)  ·  Antiderivative difference
⟹ SolveF(a), F(b), Integral
a = b = |F is antiderivative of f
Please fix the errors above.
Solve for:
Examples:
Definite Integral
F(a): F(b): ∫:
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Integral Magnitude
Small (< 5) Moderate (5–20) Large (> 20)
= F(b)F(a)  ·  The definite integral equals the change in the antiderivative.

Interpretation

∫ₐᵇ f'(x) dx = F(b) − F(a). Evaluation of definite integral using antiderivative. Bridge between area and anti‑differentiation. Core calculus theorem.

Integral from a to b of f'(x) dx = F(b) - F(a)
Fundamental Theorem of Calculus (Definite Integral)

Variables

SymbolQuantityUnit
FAntiderivative of f
a, bLower and upper limits of integration

What it means

Same as id=61. This theorem is the central result that makes calculus practical. It states that if F is an antiderivative of f, then the definite integral of f from a to b equals F(b) − F(a). This allows us to compute areas under curves, total accumulated change, and many other quantities without summing infinitely many rectangles. It is the key to solving ordinary differential equations and is the theoretical basis for numerical integration. The theorem is also used in probability (computing cumulative distribution functions) and in signal processing (Fourier transforms). Understanding this theorem is fundamental for any quantitative discipline.

Worked example

FTC (Definite Integral) – Two Examples

Real‑World
Scenario: A physicist computes the work done (integral of force) from x=1 to x=3 when F(x)=2x.
ParameterValue
F(b)F(3)=9
F(a)F(1)=1
1∫₁³ 2x dx = [x²]₁³ = 9 − 1 = 8
Result 8 ✓ Work done
Scenario: A student finds the area under y=cos(x) from 0 to π/2.
ParameterValue
F(b)sin(π/2)=1
F(a)sin(0)=0
1∫₀^{π/2} cos(x) dx = [sin(x)]₀^{π/2} = 1 − 0 = 1
Result 1 ✓ Area
Insight: The FTC gives the net area/accumulation of a function over an interval.

Common mistakes

  • Same as ID 61 – this is a duplicate entry.
  • Note: Ensure you use the correct antiderivative and evaluate at bounds.

Applications

The Fundamental Theorem of Calculus (definite integral form) states that the definite integral of a derivative equals the difference of the original function at the endpoints. This is the workhorse of calculus, used to compute net change, area, and average values. Engineers apply it to calculate displacement from velocity, to design control systems (e.g., PID controllers), and to analyse signals in frequency domain. It is also used in calculating work, energy, and fluid flow. In economics, it helps compute cumulative surplus and total cost. This theorem is essential for all quantitative fields, providing a powerful link between differential and integral calculus, and enabling practical solutions to real‑world accumulation problems.

  • Displacement and velocity analysis in kinematics
  • Work, energy, and power calculations in mechanics
  • Signal integration in control systems and signal processing
  • Fluid flow rate and total flow volume computation
  • Economic surplus and consumer/producer welfare calculations

Frequently Asked Questions

Q01What is the Fundamental Theorem of Calculus for definite integrals?
A01

It states that if f is continuous on [a,b] and F is any antiderivative of f, then ∫ₐᵇ f(x) dx = F(b) – F(a). This allows evaluation of definite integrals without limits of Riemann sums.

Q02What is the common mistake when using this theorem?
A02

Reversing the order of subtraction: computing F(a) – F(b) instead of F(b) – F(a), which flips the sign of the result.

Q03How do you apply the theorem to functions with discontinuities?
A03

If the function has jump discontinuities, the integral may be split into intervals where the function is continuous. The theorem applies to each piece.

Q04What is the relationship between the theorem and the area under a curve?
A04

The definite integral gives the net signed area. The theorem provides a way to compute that area using antiderivatives.

Q05What is the constant of integration in definite integrals?
A05

It cancels out when subtracting F(b) – F(a), so we can ignore it.

Q06How do you evaluate ∫₀^1 x² dx using the theorem?
A06

F(x) = x³/3. Then F(1) – F(0) = 1/3 – 0 = 1/3.

Q07What is the theorem for a function that is not continuous?
A07

If f is piecewise continuous, the theorem applies to each piece, and the total integral is the sum of the pieces.

Q08How does the theorem extend to improper integrals?
A08

Evaluate the proper integral and take the limit. For example, ∫₀^∞ e^(–x) dx = lim_{b→∞} (1 – e^(–b)) = 1.

Q09What are the applications of the theorem?
A09

  • Computing areas, volumes, and work.
  • Solving differential equations.
  • Probability and statistics.

Q10What are the prerequisites for the theorem?
A10

The function f must be integrable on [a,b] (which is guaranteed if f is continuous or piecewise continuous).